Bsrgvty

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!

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 $U$.

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 $a=-fracsqrt-52,;c=-fracsqrt-54$.

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
$V X V^-1 = beginbmatrix 2 & 4 \ 1 & 2endbmatrix$.

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
  $$
  X=frac 15A=beginbmatrix2&4\1&2endbmatrix .
  $$

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 $X=A$.