Find all real matrices such that $X^3-4X^2+5X=beginpmatrix 10 & 20 \ 5 & 10 endpmatrix$Find all real matrices such that $X^6 + 2X^4 + 10X = beginpmatrix 0 & -1 \ 1 & 0 endpmatrix$Proof of this result related to Fibonacci numbers: $beginpmatrix1&1\1&0endpmatrix^n=beginpmatrixF_n+1&F_n\F_n&F_n-1endpmatrix$?Prove that matrices $tinybeginpmatrix 2&-1 \ 0&2 \ endpmatrix,beginpmatrix 2& 0 \ 1&2 \ endpmatrix $ are similar. Error in my method?Matrices $beginpmatrix a &b \ c &d endpmatrixbeginpmatrix x\y endpmatrix=kbeginpmatrixx\y endpmatrix$How do I find the vector $Tbeginpmatrix 5 & 0 \ -10 & -13 endpmatrix$?Geometrical meaning of $ beginpmatrix a\ b endpmatrix mapsto beginpmatrix a&-b\ b&a endpmatrix$Find 2 by 3 matrix M such that M $beginpmatrixa \ b \ cendpmatrix= beginpmatrixd \ eendpmatrix$ whenever $(ax^2+bx + c)'=dx+e$Diagonalize $f(A)= beginpmatrix 1 & 0 \ -1 & 3 endpmatrix A $Knowing $M^2 + M = beginpmatrix 1 & 1 \ 1 & 1 endpmatrix$, find the eigenvalues of $M$Find all real matrices such that $X^6 + 2X^4 + 10X = beginpmatrix 0 & -1 \ 1 & 0 endpmatrix$

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Find all real matrices such that $X^3-4X^2+5X=beginpmatrix 10 & 20 \ 5 & 10 endpmatrix$


Find all real matrices such that $X^6 + 2X^4 + 10X = beginpmatrix 0 & -1 \ 1 & 0 endpmatrix$Proof of this result related to Fibonacci numbers: $beginpmatrix1&1\1&0endpmatrix^n=beginpmatrixF_n+1&F_n\F_n&F_n-1endpmatrix$?Prove that matrices $tinybeginpmatrix 2&-1 \ 0&2 \ endpmatrix,beginpmatrix 2& 0 \ 1&2 \ endpmatrix $ are similar. Error in my method?Matrices $beginpmatrix a &b \ c &d endpmatrixbeginpmatrix x\y endpmatrix=kbeginpmatrixx\y endpmatrix$How do I find the vector $Tbeginpmatrix 5 & 0 \ -10 & -13 endpmatrix$?Geometrical meaning of $ beginpmatrix a\ b endpmatrix mapsto beginpmatrix a&-b\ b&a endpmatrix$Find 2 by 3 matrix M such that M $beginpmatrixa \ b \ cendpmatrix= beginpmatrixd \ eendpmatrix$ whenever $(ax^2+bx + c)'=dx+e$Diagonalize $f(A)= beginpmatrix 1 & 0 \ -1 & 3 endpmatrix A $Knowing $M^2 + M = beginpmatrix 1 & 1 \ 1 & 1 endpmatrix$, find the eigenvalues of $M$Find all real matrices such that $X^6 + 2X^4 + 10X = beginpmatrix 0 & -1 \ 1 & 0 endpmatrix$






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5















$begingroup$


The following question come from the 1998 Romanian Mathematical Competition:




Find all matrices in $M_2(mathbb R)$ such that $$X^3-4X^2+5X=beginpmatrix 10 & 20 \ 5 & 10 endpmatrix$$




Can you guy please help me? Thanks a lot!










share|cite|improve this question











$endgroup$










  • 2




    $begingroup$
    First you can conclude that $n=2$ and $Xneq 0$. Also the determinant of the RHS is zero.
    $endgroup$
    – Dietrich Burde
    Sep 27 at 18:11







  • 1




    $begingroup$
    You could assume the right hand side is $textdiag(20,0)$ with eigenvectors $e_1,e_2$.
    $endgroup$
    – copper.hat
    Sep 27 at 18:25







  • 1




    $begingroup$
    I don't know if this helps, but $X=beginpmatrix2&4\1&2endpmatrix$ is a solution.
    $endgroup$
    – kneidell
    Sep 27 at 18:30










  • $begingroup$
    @kneidell It does help, because you've found the only solution.
    $endgroup$
    – Dietrich Burde
    Sep 27 at 18:55

















5















$begingroup$


The following question come from the 1998 Romanian Mathematical Competition:




Find all matrices in $M_2(mathbb R)$ such that $$X^3-4X^2+5X=beginpmatrix 10 & 20 \ 5 & 10 endpmatrix$$




Can you guy please help me? Thanks a lot!










share|cite|improve this question











$endgroup$










  • 2




    $begingroup$
    First you can conclude that $n=2$ and $Xneq 0$. Also the determinant of the RHS is zero.
    $endgroup$
    – Dietrich Burde
    Sep 27 at 18:11







  • 1




    $begingroup$
    You could assume the right hand side is $textdiag(20,0)$ with eigenvectors $e_1,e_2$.
    $endgroup$
    – copper.hat
    Sep 27 at 18:25







  • 1




    $begingroup$
    I don't know if this helps, but $X=beginpmatrix2&4\1&2endpmatrix$ is a solution.
    $endgroup$
    – kneidell
    Sep 27 at 18:30










  • $begingroup$
    @kneidell It does help, because you've found the only solution.
    $endgroup$
    – Dietrich Burde
    Sep 27 at 18:55













5













5









5


1



$begingroup$


The following question come from the 1998 Romanian Mathematical Competition:




Find all matrices in $M_2(mathbb R)$ such that $$X^3-4X^2+5X=beginpmatrix 10 & 20 \ 5 & 10 endpmatrix$$




Can you guy please help me? Thanks a lot!










share|cite|improve this question











$endgroup$




The following question come from the 1998 Romanian Mathematical Competition:




Find all matrices in $M_2(mathbb R)$ such that $$X^3-4X^2+5X=beginpmatrix 10 & 20 \ 5 & 10 endpmatrix$$




Can you guy please help me? Thanks a lot!







linear-algebra matrices contest-math matrix-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Sep 28 at 11:01









Rodrigo de Azevedo

14.3k4 gold badges23 silver badges69 bronze badges




14.3k4 gold badges23 silver badges69 bronze badges










asked Sep 27 at 18:08









aDmaLaDmaL

1732 bronze badges




1732 bronze badges










  • 2




    $begingroup$
    First you can conclude that $n=2$ and $Xneq 0$. Also the determinant of the RHS is zero.
    $endgroup$
    – Dietrich Burde
    Sep 27 at 18:11







  • 1




    $begingroup$
    You could assume the right hand side is $textdiag(20,0)$ with eigenvectors $e_1,e_2$.
    $endgroup$
    – copper.hat
    Sep 27 at 18:25







  • 1




    $begingroup$
    I don't know if this helps, but $X=beginpmatrix2&4\1&2endpmatrix$ is a solution.
    $endgroup$
    – kneidell
    Sep 27 at 18:30










  • $begingroup$
    @kneidell It does help, because you've found the only solution.
    $endgroup$
    – Dietrich Burde
    Sep 27 at 18:55












  • 2




    $begingroup$
    First you can conclude that $n=2$ and $Xneq 0$. Also the determinant of the RHS is zero.
    $endgroup$
    – Dietrich Burde
    Sep 27 at 18:11







  • 1




    $begingroup$
    You could assume the right hand side is $textdiag(20,0)$ with eigenvectors $e_1,e_2$.
    $endgroup$
    – copper.hat
    Sep 27 at 18:25







  • 1




    $begingroup$
    I don't know if this helps, but $X=beginpmatrix2&4\1&2endpmatrix$ is a solution.
    $endgroup$
    – kneidell
    Sep 27 at 18:30










  • $begingroup$
    @kneidell It does help, because you've found the only solution.
    $endgroup$
    – Dietrich Burde
    Sep 27 at 18:55







2




2




$begingroup$
First you can conclude that $n=2$ and $Xneq 0$. Also the determinant of the RHS is zero.
$endgroup$
– Dietrich Burde
Sep 27 at 18:11





$begingroup$
First you can conclude that $n=2$ and $Xneq 0$. Also the determinant of the RHS is zero.
$endgroup$
– Dietrich Burde
Sep 27 at 18:11





1




1




$begingroup$
You could assume the right hand side is $textdiag(20,0)$ with eigenvectors $e_1,e_2$.
$endgroup$
– copper.hat
Sep 27 at 18:25





$begingroup$
You could assume the right hand side is $textdiag(20,0)$ with eigenvectors $e_1,e_2$.
$endgroup$
– copper.hat
Sep 27 at 18:25





1




1




$begingroup$
I don't know if this helps, but $X=beginpmatrix2&4\1&2endpmatrix$ is a solution.
$endgroup$
– kneidell
Sep 27 at 18:30




$begingroup$
I don't know if this helps, but $X=beginpmatrix2&4\1&2endpmatrix$ is a solution.
$endgroup$
– kneidell
Sep 27 at 18:30












$begingroup$
@kneidell It does help, because you've found the only solution.
$endgroup$
– Dietrich Burde
Sep 27 at 18:55




$begingroup$
@kneidell It does help, because you've found the only solution.
$endgroup$
– Dietrich Burde
Sep 27 at 18:55










5 Answers
5






active

oldest

votes


















4

















$begingroup$

There is a matrix $U$ such that
$$U^-1beginpmatrix 10 & 20 \ 5 & 10 endpmatrix U=beginpmatrix 20 & 0 \ 0&0 endpmatrix.$$



Let $Y=U^-1XU$, then $Y^3-4Y^2+5Y=beginpmatrix 20 & 0 \0 & 0 endpmatrix$. The matrix $Y$ then commutes with $beginpmatrix 20 & 0 \ 0&0 endpmatrix$ and so is diagonal. Let $Y=beginpmatrix a & 0 \ 0&b endpmatrix.$



Then $a^3-4a^2+5a=20,b^3-4b^2+5b=0.$ The only real solutions are $a=4,b=0.$



Then $X$ is uniquely determined as $UYU^-1$. If we did not know that $X=beginpmatrix 2 & 4 cr 1 & 2 endpmatrix$ we could find it using the matrix $U$.






share|cite|improve this answer












$endgroup$





















    3

















    $begingroup$

    Let $X=beginpmatrix a & b cr c & d endpmatrix$. Then the matrix equation immediately gives that $d=a$ and $b=4c$. Now it is easy to solve the last equations in $a$ and $c$. The only real solution is $a=2$ and $c=1$, and hence
    $$
    X=beginpmatrix 2 & 4 cr 1 & 2 endpmatrix.
    $$

    For complex numbers there are several other solutions, e.g.,
    $a=fraci + 62,; c= frac2- i 4$, or $a=-fracsqrt-52,;c=-fracsqrt-54$.






    share|cite|improve this answer












    $endgroup$













    • $begingroup$
      I had decided the equation was very messy when multiplied out - did you spot a clever way of simplifying it?
      $endgroup$
      – S. Dolan
      Sep 27 at 18:47










    • $begingroup$
      Well done then hence upvote! With your method would diagonalising have been irrelevant?
      $endgroup$
      – S. Dolan
      Sep 27 at 18:51










    • $begingroup$
      Yes, diagonalising would make no difference. The four equations do not become easier then.
      $endgroup$
      – Dietrich Burde
      Sep 27 at 18:52







    • 1




      $begingroup$
      I can see $a=d$ or $a^2+ad+d^2+bc+5=0$. Is that how you proceeded?
      $endgroup$
      – S. Dolan
      Sep 27 at 19:06










    • $begingroup$
      I think the key to a quick solution is noting that $X$ cannot have conjugate pair eigenvalues. For me, diagonalisation was what made this clear.
      $endgroup$
      – copper.hat
      Sep 27 at 19:55


















    3

















    $begingroup$

    Let $p$ be the polynomial in question and $R$ be the right hand side. Note that $R$ is equivalent to $D=operatornamediag (20,0)$.



    Let $V^-1RV = D$, then since $V^-1p(X)V = p(V^-1XV) = D$, we can look for solutions
    to $p(X)=D$ and then conjugate back to get the original solutions.



    Note that $De_1 = 20 e_1, D e_2 = 0$. Hence $p(X)e_1 = 20e_1$, $p(X)e_2 = 0$.



    If $lambda$ is an eigenvalue of $X$ then $p(lambda)$ is an eigenvalue of $p(X)$,
    hence $X$ has distinct eigenvalues and $p(lambda_1) = 20, p(lambda_1) = 0$.
    Hence $e_1,e_2$ are eigenvectors of $X$ (this is the key here).



    In particular, $X$ is diagonal, so the problem reduces to solving $p(x) = 0$ (roots $0, 2 pm i$) to get $X_22$ and $p(x)=20$ (roots $4,pm sqrt5i$) to get $X_11$ and seeing what combinations work.



    Since the matrix is real, we see that $X$ must have roots $4,0$ and so $X = operatornamediag (4,0)$.



    To finish, we need to conjugate, if we let
    $V= beginbmatrix 2 & -2 \ 1 & 1endbmatrix$, then
    $V X V^-1 = beginbmatrix 2 & 4 \ 1 & 2endbmatrix$.






    share|cite|improve this answer












    $endgroup$





















      1

















      $begingroup$

      Here is an other possibility to proceed. The Romanian NMO "should not know" the linear algebra related to diagonalization and/or Jordan forms for matrices, but for $2times 2$ matrices it is a standard idea to use Cayley-Hamilton, since having the trace $t$ and the determinant $d$ of a matrix $A$ it is an exercise for matrix operations (theoretically also done in the classes) to check $A^2-tA+d=0$. In this sense, we may work as follows, using as much as possibly the arithmetics of the polynomial ring $Bbb R[x]$.




      The given matrix $A$ with entries $10, 20, 5, 10$ has trace $20$, and determinant zero. Let $g$ be the characteristic polynomial of $A$, so $g(A) = A^2-20A=0$. The unknown matrix $X$ satisfies for the polynomial $f(x)=x^3-4x^2+5x$ the given relation $f(x)=A$. So
      $$
      beginaligned
      h(x):=g(f(x))
      &=x^2(x^2 - 4x + 5)^2 - 20x(x^2 - 4x + 5)
      \
      &=(x^2 - 4x + 5)(x^2 + 5)(x - 4)x
      endaligned
      $$

      annihilates $X$.



      Let $pinBbb R[x]$ be the (monic) minimal polynomial of $X$.



      It has degree two, (else we get a contradiction with $f(X)=A$,) so it divides $h$.



      • The first factor is excluded immediately as a value for $p$, because this would imply $A=f(X)=Xp(X)=0$.


      • The second factor is also excluded as a value for $p$, because else $f(x)-20=(x^2+5)(x-4)$ has the factor $(x^2+5)$, so $A-20I=(f-20)(A) =0$, again a contradiction.


      • It follows $p(x)=x(x-4)=x^2-4x$. The rest obtained by division with rest of $f(x)=colorgrayx^3-4x^2+5x$ by $p(x)=x^2-4x$ is $5x$, so we obtain:
        $$
        A=f(X)=5X .
        $$

        This brings the only matrix operation in the game
        $$
        X=frac 15A=beginbmatrix2&4\1&2endbmatrix .
        $$






      share|cite|improve this answer










      $endgroup$





















        1

















        $begingroup$

        dan_fulea mentioned in another answer that the contestants are not expected to know diagonalisation or Jordan form. So, I will give a more elementary solution below. Let
        $$
        A=uv^T=pmatrix2\ 1pmatrix1&2.
        $$

        The equation in question is equivalent to
        $$
        X^3-4X^2+5X=5A.tag1
        $$

        One can easily verify that $A^2=4A$ and $X=A$ is a solution to $(1)$. In general, if $X$ satisfies $(1)$, we must have $XA=AX$, i.e. $Xuv^T=uv^TX$. Therefore $Xu=ku$ and $v^TX=kv^T$ for some common real factor $k$, and $XA=AX=kA$. It follows from $(1)$ that
        beginaligned
        X^3A-4X^2A+5XA&=5A^2,\
        k^3A-4k^2A+5kA&=20A,\
        k^3-4k^2+5k-20&=0,\
        (k-4)(k^2+5)&=0.
        endaligned

        Therefore $k=4$ and $XA=AX=4A$. Since $A^2=4A$, if we put $Y=X-A$, we get
        $YA=AY=0$ or $Yuv^T=uv^TY=0$. Hence $Y$ must be a real scalar multiple of
        $$
        B=pmatrix2\ -1pmatrix1&-2
        $$

        and $X=A+bB$ for some real scalar $b$. As $X=A$ is a solution to $(1)$, $AB=BA=0$ and $B^2=4B$, if we substitute $X=A+bB$ into $(1)$, we get
        beginaligned
        b^3B^3-4b^2B^2+5bB&=0,\
        16b^3-16b^2+5b&=0,\
        b(16^2-16b+5)&=0,\
        b&=0.
        endaligned

        Hence the only solution to $(1)$ is given by $X=A$.






        share|cite|improve this answer












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          5 Answers
          5






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          5 Answers
          5






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          4

















          $begingroup$

          There is a matrix $U$ such that
          $$U^-1beginpmatrix 10 & 20 \ 5 & 10 endpmatrix U=beginpmatrix 20 & 0 \ 0&0 endpmatrix.$$



          Let $Y=U^-1XU$, then $Y^3-4Y^2+5Y=beginpmatrix 20 & 0 \0 & 0 endpmatrix$. The matrix $Y$ then commutes with $beginpmatrix 20 & 0 \ 0&0 endpmatrix$ and so is diagonal. Let $Y=beginpmatrix a & 0 \ 0&b endpmatrix.$



          Then $a^3-4a^2+5a=20,b^3-4b^2+5b=0.$ The only real solutions are $a=4,b=0.$



          Then $X$ is uniquely determined as $UYU^-1$. If we did not know that $X=beginpmatrix 2 & 4 cr 1 & 2 endpmatrix$ we could find it using the matrix $U$.






          share|cite|improve this answer












          $endgroup$


















            4

















            $begingroup$

            There is a matrix $U$ such that
            $$U^-1beginpmatrix 10 & 20 \ 5 & 10 endpmatrix U=beginpmatrix 20 & 0 \ 0&0 endpmatrix.$$



            Let $Y=U^-1XU$, then $Y^3-4Y^2+5Y=beginpmatrix 20 & 0 \0 & 0 endpmatrix$. The matrix $Y$ then commutes with $beginpmatrix 20 & 0 \ 0&0 endpmatrix$ and so is diagonal. Let $Y=beginpmatrix a & 0 \ 0&b endpmatrix.$



            Then $a^3-4a^2+5a=20,b^3-4b^2+5b=0.$ The only real solutions are $a=4,b=0.$



            Then $X$ is uniquely determined as $UYU^-1$. If we did not know that $X=beginpmatrix 2 & 4 cr 1 & 2 endpmatrix$ we could find it using the matrix $U$.






            share|cite|improve this answer












            $endgroup$
















              4















              4











              4







              $begingroup$

              There is a matrix $U$ such that
              $$U^-1beginpmatrix 10 & 20 \ 5 & 10 endpmatrix U=beginpmatrix 20 & 0 \ 0&0 endpmatrix.$$



              Let $Y=U^-1XU$, then $Y^3-4Y^2+5Y=beginpmatrix 20 & 0 \0 & 0 endpmatrix$. The matrix $Y$ then commutes with $beginpmatrix 20 & 0 \ 0&0 endpmatrix$ and so is diagonal. Let $Y=beginpmatrix a & 0 \ 0&b endpmatrix.$



              Then $a^3-4a^2+5a=20,b^3-4b^2+5b=0.$ The only real solutions are $a=4,b=0.$



              Then $X$ is uniquely determined as $UYU^-1$. If we did not know that $X=beginpmatrix 2 & 4 cr 1 & 2 endpmatrix$ we could find it using the matrix $U$.






              share|cite|improve this answer












              $endgroup$



              There is a matrix $U$ such that
              $$U^-1beginpmatrix 10 & 20 \ 5 & 10 endpmatrix U=beginpmatrix 20 & 0 \ 0&0 endpmatrix.$$



              Let $Y=U^-1XU$, then $Y^3-4Y^2+5Y=beginpmatrix 20 & 0 \0 & 0 endpmatrix$. The matrix $Y$ then commutes with $beginpmatrix 20 & 0 \ 0&0 endpmatrix$ and so is diagonal. Let $Y=beginpmatrix a & 0 \ 0&b endpmatrix.$



              Then $a^3-4a^2+5a=20,b^3-4b^2+5b=0.$ The only real solutions are $a=4,b=0.$



              Then $X$ is uniquely determined as $UYU^-1$. If we did not know that $X=beginpmatrix 2 & 4 cr 1 & 2 endpmatrix$ we could find it using the matrix $U$.







              share|cite|improve this answer















              share|cite|improve this answer




              share|cite|improve this answer








              edited Sep 28 at 7:12

























              answered Sep 27 at 18:30









              S. DolanS. Dolan

              10.5k1 gold badge4 silver badges19 bronze badges




              10.5k1 gold badge4 silver badges19 bronze badges


























                  3

















                  $begingroup$

                  Let $X=beginpmatrix a & b cr c & d endpmatrix$. Then the matrix equation immediately gives that $d=a$ and $b=4c$. Now it is easy to solve the last equations in $a$ and $c$. The only real solution is $a=2$ and $c=1$, and hence
                  $$
                  X=beginpmatrix 2 & 4 cr 1 & 2 endpmatrix.
                  $$

                  For complex numbers there are several other solutions, e.g.,
                  $a=fraci + 62,; c= frac2- i 4$, or $a=-fracsqrt-52,;c=-fracsqrt-54$.






                  share|cite|improve this answer












                  $endgroup$













                  • $begingroup$
                    I had decided the equation was very messy when multiplied out - did you spot a clever way of simplifying it?
                    $endgroup$
                    – S. Dolan
                    Sep 27 at 18:47










                  • $begingroup$
                    Well done then hence upvote! With your method would diagonalising have been irrelevant?
                    $endgroup$
                    – S. Dolan
                    Sep 27 at 18:51










                  • $begingroup$
                    Yes, diagonalising would make no difference. The four equations do not become easier then.
                    $endgroup$
                    – Dietrich Burde
                    Sep 27 at 18:52







                  • 1




                    $begingroup$
                    I can see $a=d$ or $a^2+ad+d^2+bc+5=0$. Is that how you proceeded?
                    $endgroup$
                    – S. Dolan
                    Sep 27 at 19:06










                  • $begingroup$
                    I think the key to a quick solution is noting that $X$ cannot have conjugate pair eigenvalues. For me, diagonalisation was what made this clear.
                    $endgroup$
                    – copper.hat
                    Sep 27 at 19:55















                  3

















                  $begingroup$

                  Let $X=beginpmatrix a & b cr c & d endpmatrix$. Then the matrix equation immediately gives that $d=a$ and $b=4c$. Now it is easy to solve the last equations in $a$ and $c$. The only real solution is $a=2$ and $c=1$, and hence
                  $$
                  X=beginpmatrix 2 & 4 cr 1 & 2 endpmatrix.
                  $$

                  For complex numbers there are several other solutions, e.g.,
                  $a=fraci + 62,; c= frac2- i 4$, or $a=-fracsqrt-52,;c=-fracsqrt-54$.






                  share|cite|improve this answer












                  $endgroup$













                  • $begingroup$
                    I had decided the equation was very messy when multiplied out - did you spot a clever way of simplifying it?
                    $endgroup$
                    – S. Dolan
                    Sep 27 at 18:47










                  • $begingroup$
                    Well done then hence upvote! With your method would diagonalising have been irrelevant?
                    $endgroup$
                    – S. Dolan
                    Sep 27 at 18:51










                  • $begingroup$
                    Yes, diagonalising would make no difference. The four equations do not become easier then.
                    $endgroup$
                    – Dietrich Burde
                    Sep 27 at 18:52







                  • 1




                    $begingroup$
                    I can see $a=d$ or $a^2+ad+d^2+bc+5=0$. Is that how you proceeded?
                    $endgroup$
                    – S. Dolan
                    Sep 27 at 19:06










                  • $begingroup$
                    I think the key to a quick solution is noting that $X$ cannot have conjugate pair eigenvalues. For me, diagonalisation was what made this clear.
                    $endgroup$
                    – copper.hat
                    Sep 27 at 19:55













                  3















                  3











                  3







                  $begingroup$

                  Let $X=beginpmatrix a & b cr c & d endpmatrix$. Then the matrix equation immediately gives that $d=a$ and $b=4c$. Now it is easy to solve the last equations in $a$ and $c$. The only real solution is $a=2$ and $c=1$, and hence
                  $$
                  X=beginpmatrix 2 & 4 cr 1 & 2 endpmatrix.
                  $$

                  For complex numbers there are several other solutions, e.g.,
                  $a=fraci + 62,; c= frac2- i 4$, or $a=-fracsqrt-52,;c=-fracsqrt-54$.






                  share|cite|improve this answer












                  $endgroup$



                  Let $X=beginpmatrix a & b cr c & d endpmatrix$. Then the matrix equation immediately gives that $d=a$ and $b=4c$. Now it is easy to solve the last equations in $a$ and $c$. The only real solution is $a=2$ and $c=1$, and hence
                  $$
                  X=beginpmatrix 2 & 4 cr 1 & 2 endpmatrix.
                  $$

                  For complex numbers there are several other solutions, e.g.,
                  $a=fraci + 62,; c= frac2- i 4$, or $a=-fracsqrt-52,;c=-fracsqrt-54$.







                  share|cite|improve this answer















                  share|cite|improve this answer




                  share|cite|improve this answer








                  edited Sep 27 at 18:55

























                  answered Sep 27 at 18:43









                  Dietrich BurdeDietrich Burde

                  90.8k6 gold badges50 silver badges112 bronze badges




                  90.8k6 gold badges50 silver badges112 bronze badges














                  • $begingroup$
                    I had decided the equation was very messy when multiplied out - did you spot a clever way of simplifying it?
                    $endgroup$
                    – S. Dolan
                    Sep 27 at 18:47










                  • $begingroup$
                    Well done then hence upvote! With your method would diagonalising have been irrelevant?
                    $endgroup$
                    – S. Dolan
                    Sep 27 at 18:51










                  • $begingroup$
                    Yes, diagonalising would make no difference. The four equations do not become easier then.
                    $endgroup$
                    – Dietrich Burde
                    Sep 27 at 18:52







                  • 1




                    $begingroup$
                    I can see $a=d$ or $a^2+ad+d^2+bc+5=0$. Is that how you proceeded?
                    $endgroup$
                    – S. Dolan
                    Sep 27 at 19:06










                  • $begingroup$
                    I think the key to a quick solution is noting that $X$ cannot have conjugate pair eigenvalues. For me, diagonalisation was what made this clear.
                    $endgroup$
                    – copper.hat
                    Sep 27 at 19:55
















                  • $begingroup$
                    I had decided the equation was very messy when multiplied out - did you spot a clever way of simplifying it?
                    $endgroup$
                    – S. Dolan
                    Sep 27 at 18:47










                  • $begingroup$
                    Well done then hence upvote! With your method would diagonalising have been irrelevant?
                    $endgroup$
                    – S. Dolan
                    Sep 27 at 18:51










                  • $begingroup$
                    Yes, diagonalising would make no difference. The four equations do not become easier then.
                    $endgroup$
                    – Dietrich Burde
                    Sep 27 at 18:52







                  • 1




                    $begingroup$
                    I can see $a=d$ or $a^2+ad+d^2+bc+5=0$. Is that how you proceeded?
                    $endgroup$
                    – S. Dolan
                    Sep 27 at 19:06










                  • $begingroup$
                    I think the key to a quick solution is noting that $X$ cannot have conjugate pair eigenvalues. For me, diagonalisation was what made this clear.
                    $endgroup$
                    – copper.hat
                    Sep 27 at 19:55















                  $begingroup$
                  I had decided the equation was very messy when multiplied out - did you spot a clever way of simplifying it?
                  $endgroup$
                  – S. Dolan
                  Sep 27 at 18:47




                  $begingroup$
                  I had decided the equation was very messy when multiplied out - did you spot a clever way of simplifying it?
                  $endgroup$
                  – S. Dolan
                  Sep 27 at 18:47












                  $begingroup$
                  Well done then hence upvote! With your method would diagonalising have been irrelevant?
                  $endgroup$
                  – S. Dolan
                  Sep 27 at 18:51




                  $begingroup$
                  Well done then hence upvote! With your method would diagonalising have been irrelevant?
                  $endgroup$
                  – S. Dolan
                  Sep 27 at 18:51












                  $begingroup$
                  Yes, diagonalising would make no difference. The four equations do not become easier then.
                  $endgroup$
                  – Dietrich Burde
                  Sep 27 at 18:52





                  $begingroup$
                  Yes, diagonalising would make no difference. The four equations do not become easier then.
                  $endgroup$
                  – Dietrich Burde
                  Sep 27 at 18:52





                  1




                  1




                  $begingroup$
                  I can see $a=d$ or $a^2+ad+d^2+bc+5=0$. Is that how you proceeded?
                  $endgroup$
                  – S. Dolan
                  Sep 27 at 19:06




                  $begingroup$
                  I can see $a=d$ or $a^2+ad+d^2+bc+5=0$. Is that how you proceeded?
                  $endgroup$
                  – S. Dolan
                  Sep 27 at 19:06












                  $begingroup$
                  I think the key to a quick solution is noting that $X$ cannot have conjugate pair eigenvalues. For me, diagonalisation was what made this clear.
                  $endgroup$
                  – copper.hat
                  Sep 27 at 19:55




                  $begingroup$
                  I think the key to a quick solution is noting that $X$ cannot have conjugate pair eigenvalues. For me, diagonalisation was what made this clear.
                  $endgroup$
                  – copper.hat
                  Sep 27 at 19:55











                  3

















                  $begingroup$

                  Let $p$ be the polynomial in question and $R$ be the right hand side. Note that $R$ is equivalent to $D=operatornamediag (20,0)$.



                  Let $V^-1RV = D$, then since $V^-1p(X)V = p(V^-1XV) = D$, we can look for solutions
                  to $p(X)=D$ and then conjugate back to get the original solutions.



                  Note that $De_1 = 20 e_1, D e_2 = 0$. Hence $p(X)e_1 = 20e_1$, $p(X)e_2 = 0$.



                  If $lambda$ is an eigenvalue of $X$ then $p(lambda)$ is an eigenvalue of $p(X)$,
                  hence $X$ has distinct eigenvalues and $p(lambda_1) = 20, p(lambda_1) = 0$.
                  Hence $e_1,e_2$ are eigenvectors of $X$ (this is the key here).



                  In particular, $X$ is diagonal, so the problem reduces to solving $p(x) = 0$ (roots $0, 2 pm i$) to get $X_22$ and $p(x)=20$ (roots $4,pm sqrt5i$) to get $X_11$ and seeing what combinations work.



                  Since the matrix is real, we see that $X$ must have roots $4,0$ and so $X = operatornamediag (4,0)$.



                  To finish, we need to conjugate, if we let
                  $V= beginbmatrix 2 & -2 \ 1 & 1endbmatrix$, then
                  $V X V^-1 = beginbmatrix 2 & 4 \ 1 & 2endbmatrix$.






                  share|cite|improve this answer












                  $endgroup$


















                    3

















                    $begingroup$

                    Let $p$ be the polynomial in question and $R$ be the right hand side. Note that $R$ is equivalent to $D=operatornamediag (20,0)$.



                    Let $V^-1RV = D$, then since $V^-1p(X)V = p(V^-1XV) = D$, we can look for solutions
                    to $p(X)=D$ and then conjugate back to get the original solutions.



                    Note that $De_1 = 20 e_1, D e_2 = 0$. Hence $p(X)e_1 = 20e_1$, $p(X)e_2 = 0$.



                    If $lambda$ is an eigenvalue of $X$ then $p(lambda)$ is an eigenvalue of $p(X)$,
                    hence $X$ has distinct eigenvalues and $p(lambda_1) = 20, p(lambda_1) = 0$.
                    Hence $e_1,e_2$ are eigenvectors of $X$ (this is the key here).



                    In particular, $X$ is diagonal, so the problem reduces to solving $p(x) = 0$ (roots $0, 2 pm i$) to get $X_22$ and $p(x)=20$ (roots $4,pm sqrt5i$) to get $X_11$ and seeing what combinations work.



                    Since the matrix is real, we see that $X$ must have roots $4,0$ and so $X = operatornamediag (4,0)$.



                    To finish, we need to conjugate, if we let
                    $V= beginbmatrix 2 & -2 \ 1 & 1endbmatrix$, then
                    $V X V^-1 = beginbmatrix 2 & 4 \ 1 & 2endbmatrix$.






                    share|cite|improve this answer












                    $endgroup$
















                      3















                      3











                      3







                      $begingroup$

                      Let $p$ be the polynomial in question and $R$ be the right hand side. Note that $R$ is equivalent to $D=operatornamediag (20,0)$.



                      Let $V^-1RV = D$, then since $V^-1p(X)V = p(V^-1XV) = D$, we can look for solutions
                      to $p(X)=D$ and then conjugate back to get the original solutions.



                      Note that $De_1 = 20 e_1, D e_2 = 0$. Hence $p(X)e_1 = 20e_1$, $p(X)e_2 = 0$.



                      If $lambda$ is an eigenvalue of $X$ then $p(lambda)$ is an eigenvalue of $p(X)$,
                      hence $X$ has distinct eigenvalues and $p(lambda_1) = 20, p(lambda_1) = 0$.
                      Hence $e_1,e_2$ are eigenvectors of $X$ (this is the key here).



                      In particular, $X$ is diagonal, so the problem reduces to solving $p(x) = 0$ (roots $0, 2 pm i$) to get $X_22$ and $p(x)=20$ (roots $4,pm sqrt5i$) to get $X_11$ and seeing what combinations work.



                      Since the matrix is real, we see that $X$ must have roots $4,0$ and so $X = operatornamediag (4,0)$.



                      To finish, we need to conjugate, if we let
                      $V= beginbmatrix 2 & -2 \ 1 & 1endbmatrix$, then
                      $V X V^-1 = beginbmatrix 2 & 4 \ 1 & 2endbmatrix$.






                      share|cite|improve this answer












                      $endgroup$



                      Let $p$ be the polynomial in question and $R$ be the right hand side. Note that $R$ is equivalent to $D=operatornamediag (20,0)$.



                      Let $V^-1RV = D$, then since $V^-1p(X)V = p(V^-1XV) = D$, we can look for solutions
                      to $p(X)=D$ and then conjugate back to get the original solutions.



                      Note that $De_1 = 20 e_1, D e_2 = 0$. Hence $p(X)e_1 = 20e_1$, $p(X)e_2 = 0$.



                      If $lambda$ is an eigenvalue of $X$ then $p(lambda)$ is an eigenvalue of $p(X)$,
                      hence $X$ has distinct eigenvalues and $p(lambda_1) = 20, p(lambda_1) = 0$.
                      Hence $e_1,e_2$ are eigenvectors of $X$ (this is the key here).



                      In particular, $X$ is diagonal, so the problem reduces to solving $p(x) = 0$ (roots $0, 2 pm i$) to get $X_22$ and $p(x)=20$ (roots $4,pm sqrt5i$) to get $X_11$ and seeing what combinations work.



                      Since the matrix is real, we see that $X$ must have roots $4,0$ and so $X = operatornamediag (4,0)$.



                      To finish, we need to conjugate, if we let
                      $V= beginbmatrix 2 & -2 \ 1 & 1endbmatrix$, then
                      $V X V^-1 = beginbmatrix 2 & 4 \ 1 & 2endbmatrix$.







                      share|cite|improve this answer















                      share|cite|improve this answer




                      share|cite|improve this answer








                      edited Sep 27 at 20:00

























                      answered Sep 27 at 19:49









                      copper.hatcopper.hat

                      135k6 gold badges63 silver badges175 bronze badges




                      135k6 gold badges63 silver badges175 bronze badges
























                          1

















                          $begingroup$

                          Here is an other possibility to proceed. The Romanian NMO "should not know" the linear algebra related to diagonalization and/or Jordan forms for matrices, but for $2times 2$ matrices it is a standard idea to use Cayley-Hamilton, since having the trace $t$ and the determinant $d$ of a matrix $A$ it is an exercise for matrix operations (theoretically also done in the classes) to check $A^2-tA+d=0$. In this sense, we may work as follows, using as much as possibly the arithmetics of the polynomial ring $Bbb R[x]$.




                          The given matrix $A$ with entries $10, 20, 5, 10$ has trace $20$, and determinant zero. Let $g$ be the characteristic polynomial of $A$, so $g(A) = A^2-20A=0$. The unknown matrix $X$ satisfies for the polynomial $f(x)=x^3-4x^2+5x$ the given relation $f(x)=A$. So
                          $$
                          beginaligned
                          h(x):=g(f(x))
                          &=x^2(x^2 - 4x + 5)^2 - 20x(x^2 - 4x + 5)
                          \
                          &=(x^2 - 4x + 5)(x^2 + 5)(x - 4)x
                          endaligned
                          $$

                          annihilates $X$.



                          Let $pinBbb R[x]$ be the (monic) minimal polynomial of $X$.



                          It has degree two, (else we get a contradiction with $f(X)=A$,) so it divides $h$.



                          • The first factor is excluded immediately as a value for $p$, because this would imply $A=f(X)=Xp(X)=0$.


                          • The second factor is also excluded as a value for $p$, because else $f(x)-20=(x^2+5)(x-4)$ has the factor $(x^2+5)$, so $A-20I=(f-20)(A) =0$, again a contradiction.


                          • It follows $p(x)=x(x-4)=x^2-4x$. The rest obtained by division with rest of $f(x)=colorgrayx^3-4x^2+5x$ by $p(x)=x^2-4x$ is $5x$, so we obtain:
                            $$
                            A=f(X)=5X .
                            $$

                            This brings the only matrix operation in the game
                            $$
                            X=frac 15A=beginbmatrix2&4\1&2endbmatrix .
                            $$






                          share|cite|improve this answer










                          $endgroup$


















                            1

















                            $begingroup$

                            Here is an other possibility to proceed. The Romanian NMO "should not know" the linear algebra related to diagonalization and/or Jordan forms for matrices, but for $2times 2$ matrices it is a standard idea to use Cayley-Hamilton, since having the trace $t$ and the determinant $d$ of a matrix $A$ it is an exercise for matrix operations (theoretically also done in the classes) to check $A^2-tA+d=0$. In this sense, we may work as follows, using as much as possibly the arithmetics of the polynomial ring $Bbb R[x]$.




                            The given matrix $A$ with entries $10, 20, 5, 10$ has trace $20$, and determinant zero. Let $g$ be the characteristic polynomial of $A$, so $g(A) = A^2-20A=0$. The unknown matrix $X$ satisfies for the polynomial $f(x)=x^3-4x^2+5x$ the given relation $f(x)=A$. So
                            $$
                            beginaligned
                            h(x):=g(f(x))
                            &=x^2(x^2 - 4x + 5)^2 - 20x(x^2 - 4x + 5)
                            \
                            &=(x^2 - 4x + 5)(x^2 + 5)(x - 4)x
                            endaligned
                            $$

                            annihilates $X$.



                            Let $pinBbb R[x]$ be the (monic) minimal polynomial of $X$.



                            It has degree two, (else we get a contradiction with $f(X)=A$,) so it divides $h$.



                            • The first factor is excluded immediately as a value for $p$, because this would imply $A=f(X)=Xp(X)=0$.


                            • The second factor is also excluded as a value for $p$, because else $f(x)-20=(x^2+5)(x-4)$ has the factor $(x^2+5)$, so $A-20I=(f-20)(A) =0$, again a contradiction.


                            • It follows $p(x)=x(x-4)=x^2-4x$. The rest obtained by division with rest of $f(x)=colorgrayx^3-4x^2+5x$ by $p(x)=x^2-4x$ is $5x$, so we obtain:
                              $$
                              A=f(X)=5X .
                              $$

                              This brings the only matrix operation in the game
                              $$
                              X=frac 15A=beginbmatrix2&4\1&2endbmatrix .
                              $$






                            share|cite|improve this answer










                            $endgroup$
















                              1















                              1











                              1







                              $begingroup$

                              Here is an other possibility to proceed. The Romanian NMO "should not know" the linear algebra related to diagonalization and/or Jordan forms for matrices, but for $2times 2$ matrices it is a standard idea to use Cayley-Hamilton, since having the trace $t$ and the determinant $d$ of a matrix $A$ it is an exercise for matrix operations (theoretically also done in the classes) to check $A^2-tA+d=0$. In this sense, we may work as follows, using as much as possibly the arithmetics of the polynomial ring $Bbb R[x]$.




                              The given matrix $A$ with entries $10, 20, 5, 10$ has trace $20$, and determinant zero. Let $g$ be the characteristic polynomial of $A$, so $g(A) = A^2-20A=0$. The unknown matrix $X$ satisfies for the polynomial $f(x)=x^3-4x^2+5x$ the given relation $f(x)=A$. So
                              $$
                              beginaligned
                              h(x):=g(f(x))
                              &=x^2(x^2 - 4x + 5)^2 - 20x(x^2 - 4x + 5)
                              \
                              &=(x^2 - 4x + 5)(x^2 + 5)(x - 4)x
                              endaligned
                              $$

                              annihilates $X$.



                              Let $pinBbb R[x]$ be the (monic) minimal polynomial of $X$.



                              It has degree two, (else we get a contradiction with $f(X)=A$,) so it divides $h$.



                              • The first factor is excluded immediately as a value for $p$, because this would imply $A=f(X)=Xp(X)=0$.


                              • The second factor is also excluded as a value for $p$, because else $f(x)-20=(x^2+5)(x-4)$ has the factor $(x^2+5)$, so $A-20I=(f-20)(A) =0$, again a contradiction.


                              • It follows $p(x)=x(x-4)=x^2-4x$. The rest obtained by division with rest of $f(x)=colorgrayx^3-4x^2+5x$ by $p(x)=x^2-4x$ is $5x$, so we obtain:
                                $$
                                A=f(X)=5X .
                                $$

                                This brings the only matrix operation in the game
                                $$
                                X=frac 15A=beginbmatrix2&4\1&2endbmatrix .
                                $$






                              share|cite|improve this answer










                              $endgroup$



                              Here is an other possibility to proceed. The Romanian NMO "should not know" the linear algebra related to diagonalization and/or Jordan forms for matrices, but for $2times 2$ matrices it is a standard idea to use Cayley-Hamilton, since having the trace $t$ and the determinant $d$ of a matrix $A$ it is an exercise for matrix operations (theoretically also done in the classes) to check $A^2-tA+d=0$. In this sense, we may work as follows, using as much as possibly the arithmetics of the polynomial ring $Bbb R[x]$.




                              The given matrix $A$ with entries $10, 20, 5, 10$ has trace $20$, and determinant zero. Let $g$ be the characteristic polynomial of $A$, so $g(A) = A^2-20A=0$. The unknown matrix $X$ satisfies for the polynomial $f(x)=x^3-4x^2+5x$ the given relation $f(x)=A$. So
                              $$
                              beginaligned
                              h(x):=g(f(x))
                              &=x^2(x^2 - 4x + 5)^2 - 20x(x^2 - 4x + 5)
                              \
                              &=(x^2 - 4x + 5)(x^2 + 5)(x - 4)x
                              endaligned
                              $$

                              annihilates $X$.



                              Let $pinBbb R[x]$ be the (monic) minimal polynomial of $X$.



                              It has degree two, (else we get a contradiction with $f(X)=A$,) so it divides $h$.



                              • The first factor is excluded immediately as a value for $p$, because this would imply $A=f(X)=Xp(X)=0$.


                              • The second factor is also excluded as a value for $p$, because else $f(x)-20=(x^2+5)(x-4)$ has the factor $(x^2+5)$, so $A-20I=(f-20)(A) =0$, again a contradiction.


                              • It follows $p(x)=x(x-4)=x^2-4x$. The rest obtained by division with rest of $f(x)=colorgrayx^3-4x^2+5x$ by $p(x)=x^2-4x$ is $5x$, so we obtain:
                                $$
                                A=f(X)=5X .
                                $$

                                This brings the only matrix operation in the game
                                $$
                                X=frac 15A=beginbmatrix2&4\1&2endbmatrix .
                                $$







                              share|cite|improve this answer













                              share|cite|improve this answer




                              share|cite|improve this answer










                              answered Sep 27 at 22:20









                              dan_fuleadan_fulea

                              12k1 gold badge7 silver badges18 bronze badges




                              12k1 gold badge7 silver badges18 bronze badges
























                                  1

















                                  $begingroup$

                                  dan_fulea mentioned in another answer that the contestants are not expected to know diagonalisation or Jordan form. So, I will give a more elementary solution below. Let
                                  $$
                                  A=uv^T=pmatrix2\ 1pmatrix1&2.
                                  $$

                                  The equation in question is equivalent to
                                  $$
                                  X^3-4X^2+5X=5A.tag1
                                  $$

                                  One can easily verify that $A^2=4A$ and $X=A$ is a solution to $(1)$. In general, if $X$ satisfies $(1)$, we must have $XA=AX$, i.e. $Xuv^T=uv^TX$. Therefore $Xu=ku$ and $v^TX=kv^T$ for some common real factor $k$, and $XA=AX=kA$. It follows from $(1)$ that
                                  beginaligned
                                  X^3A-4X^2A+5XA&=5A^2,\
                                  k^3A-4k^2A+5kA&=20A,\
                                  k^3-4k^2+5k-20&=0,\
                                  (k-4)(k^2+5)&=0.
                                  endaligned

                                  Therefore $k=4$ and $XA=AX=4A$. Since $A^2=4A$, if we put $Y=X-A$, we get
                                  $YA=AY=0$ or $Yuv^T=uv^TY=0$. Hence $Y$ must be a real scalar multiple of
                                  $$
                                  B=pmatrix2\ -1pmatrix1&-2
                                  $$

                                  and $X=A+bB$ for some real scalar $b$. As $X=A$ is a solution to $(1)$, $AB=BA=0$ and $B^2=4B$, if we substitute $X=A+bB$ into $(1)$, we get
                                  beginaligned
                                  b^3B^3-4b^2B^2+5bB&=0,\
                                  16b^3-16b^2+5b&=0,\
                                  b(16^2-16b+5)&=0,\
                                  b&=0.
                                  endaligned

                                  Hence the only solution to $(1)$ is given by $X=A$.






                                  share|cite|improve this answer












                                  $endgroup$


















                                    1

















                                    $begingroup$

                                    dan_fulea mentioned in another answer that the contestants are not expected to know diagonalisation or Jordan form. So, I will give a more elementary solution below. Let
                                    $$
                                    A=uv^T=pmatrix2\ 1pmatrix1&2.
                                    $$

                                    The equation in question is equivalent to
                                    $$
                                    X^3-4X^2+5X=5A.tag1
                                    $$

                                    One can easily verify that $A^2=4A$ and $X=A$ is a solution to $(1)$. In general, if $X$ satisfies $(1)$, we must have $XA=AX$, i.e. $Xuv^T=uv^TX$. Therefore $Xu=ku$ and $v^TX=kv^T$ for some common real factor $k$, and $XA=AX=kA$. It follows from $(1)$ that
                                    beginaligned
                                    X^3A-4X^2A+5XA&=5A^2,\
                                    k^3A-4k^2A+5kA&=20A,\
                                    k^3-4k^2+5k-20&=0,\
                                    (k-4)(k^2+5)&=0.
                                    endaligned

                                    Therefore $k=4$ and $XA=AX=4A$. Since $A^2=4A$, if we put $Y=X-A$, we get
                                    $YA=AY=0$ or $Yuv^T=uv^TY=0$. Hence $Y$ must be a real scalar multiple of
                                    $$
                                    B=pmatrix2\ -1pmatrix1&-2
                                    $$

                                    and $X=A+bB$ for some real scalar $b$. As $X=A$ is a solution to $(1)$, $AB=BA=0$ and $B^2=4B$, if we substitute $X=A+bB$ into $(1)$, we get
                                    beginaligned
                                    b^3B^3-4b^2B^2+5bB&=0,\
                                    16b^3-16b^2+5b&=0,\
                                    b(16^2-16b+5)&=0,\
                                    b&=0.
                                    endaligned

                                    Hence the only solution to $(1)$ is given by $X=A$.






                                    share|cite|improve this answer












                                    $endgroup$
















                                      1















                                      1











                                      1







                                      $begingroup$

                                      dan_fulea mentioned in another answer that the contestants are not expected to know diagonalisation or Jordan form. So, I will give a more elementary solution below. Let
                                      $$
                                      A=uv^T=pmatrix2\ 1pmatrix1&2.
                                      $$

                                      The equation in question is equivalent to
                                      $$
                                      X^3-4X^2+5X=5A.tag1
                                      $$

                                      One can easily verify that $A^2=4A$ and $X=A$ is a solution to $(1)$. In general, if $X$ satisfies $(1)$, we must have $XA=AX$, i.e. $Xuv^T=uv^TX$. Therefore $Xu=ku$ and $v^TX=kv^T$ for some common real factor $k$, and $XA=AX=kA$. It follows from $(1)$ that
                                      beginaligned
                                      X^3A-4X^2A+5XA&=5A^2,\
                                      k^3A-4k^2A+5kA&=20A,\
                                      k^3-4k^2+5k-20&=0,\
                                      (k-4)(k^2+5)&=0.
                                      endaligned

                                      Therefore $k=4$ and $XA=AX=4A$. Since $A^2=4A$, if we put $Y=X-A$, we get
                                      $YA=AY=0$ or $Yuv^T=uv^TY=0$. Hence $Y$ must be a real scalar multiple of
                                      $$
                                      B=pmatrix2\ -1pmatrix1&-2
                                      $$

                                      and $X=A+bB$ for some real scalar $b$. As $X=A$ is a solution to $(1)$, $AB=BA=0$ and $B^2=4B$, if we substitute $X=A+bB$ into $(1)$, we get
                                      beginaligned
                                      b^3B^3-4b^2B^2+5bB&=0,\
                                      16b^3-16b^2+5b&=0,\
                                      b(16^2-16b+5)&=0,\
                                      b&=0.
                                      endaligned

                                      Hence the only solution to $(1)$ is given by $X=A$.






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                                      $endgroup$



                                      dan_fulea mentioned in another answer that the contestants are not expected to know diagonalisation or Jordan form. So, I will give a more elementary solution below. Let
                                      $$
                                      A=uv^T=pmatrix2\ 1pmatrix1&2.
                                      $$

                                      The equation in question is equivalent to
                                      $$
                                      X^3-4X^2+5X=5A.tag1
                                      $$

                                      One can easily verify that $A^2=4A$ and $X=A$ is a solution to $(1)$. In general, if $X$ satisfies $(1)$, we must have $XA=AX$, i.e. $Xuv^T=uv^TX$. Therefore $Xu=ku$ and $v^TX=kv^T$ for some common real factor $k$, and $XA=AX=kA$. It follows from $(1)$ that
                                      beginaligned
                                      X^3A-4X^2A+5XA&=5A^2,\
                                      k^3A-4k^2A+5kA&=20A,\
                                      k^3-4k^2+5k-20&=0,\
                                      (k-4)(k^2+5)&=0.
                                      endaligned

                                      Therefore $k=4$ and $XA=AX=4A$. Since $A^2=4A$, if we put $Y=X-A$, we get
                                      $YA=AY=0$ or $Yuv^T=uv^TY=0$. Hence $Y$ must be a real scalar multiple of
                                      $$
                                      B=pmatrix2\ -1pmatrix1&-2
                                      $$

                                      and $X=A+bB$ for some real scalar $b$. As $X=A$ is a solution to $(1)$, $AB=BA=0$ and $B^2=4B$, if we substitute $X=A+bB$ into $(1)$, we get
                                      beginaligned
                                      b^3B^3-4b^2B^2+5bB&=0,\
                                      16b^3-16b^2+5b&=0,\
                                      b(16^2-16b+5)&=0,\
                                      b&=0.
                                      endaligned

                                      Hence the only solution to $(1)$ is given by $X=A$.







                                      share|cite|improve this answer















                                      share|cite|improve this answer




                                      share|cite|improve this answer








                                      edited Sep 28 at 2:14

























                                      answered Sep 28 at 1:52









                                      user1551user1551

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