Is every diagonalizable matrix is an exponentialSymmetric matrix is always diagonalizable?Is the following matrix diagonalizable?Showing that if $AB=BA$ then $A$ and $B$ are simultaneously diagonalizableTrue of False: If $A$ is an $ntimes n$ diagonalizable matrix, then $0$ can not be in eigenvalue of $A$.Diagonalizable Matrices and Triangular MatricesDetermine if a matrix is diagonalizableDoes every diagonalizable matrix have eigenvectors which form a basis?Matrix exponential of any matrixIs there a matrix $B$ such that $B^2=A$, with $A$ being diagonalizable?Let $A$ be a diagonalizable matrix, show that $A^-1 = A$
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Is every diagonalizable matrix is an exponential
Symmetric matrix is always diagonalizable?Is the following matrix diagonalizable?Showing that if $AB=BA$ then $A$ and $B$ are simultaneously diagonalizableTrue of False: If $A$ is an $ntimes n$ diagonalizable matrix, then $0$ can not be in eigenvalue of $A$.Diagonalizable Matrices and Triangular MatricesDetermine if a matrix is diagonalizableDoes every diagonalizable matrix have eigenvectors which form a basis?Matrix exponential of any matrixIs there a matrix $B$ such that $B^2=A$, with $A$ being diagonalizable?Let $A$ be a diagonalizable matrix, show that $A^-1 = A$
$begingroup$
Is every diagonalizable matrix is an exponential?
I know it is true in $SL_2(Bbb C)$ and I think it is true in $M_n(Bbb C)$ because if $M=PDP^-1$, we might be able to write D as $exp(E)$ for some $Ein M_n(Bbb C)$ as the exponential is surjective from $Bbb C$ onto $Bbb C^*$ and all eigenvalues of M are non zero because they are distinct.
Thank you for your help.
linear-algebra lie-groups diagonalization matrix-exponential
edited 2 days ago
José Carlos Santos
173k23133241
asked 2 days ago
PerelManPerelMan
732414
$endgroup$
add a comment |
$begingroup$
Is every diagonalizable matrix is an exponential?
I know it is true in $SL_2(Bbb C)$ and I think it is true in $M_n(Bbb C)$ because if $M=PDP^-1$, we might be able to write D as $exp(E)$ for some $Ein M_n(Bbb C)$ as the exponential is surjective from $Bbb C$ onto $Bbb C^*$ and all eigenvalues of M are non zero because they are distinct.
Thank you for your help.
linear-algebra lie-groups diagonalization matrix-exponential
edited 2 days ago
José Carlos Santos
173k23133241
asked 2 days ago
PerelManPerelMan
732414
$endgroup$
2
$begingroup$
What if $M=O$ (the zero matrix)?
$endgroup$
– Minus One-Twelfth
2 days ago
$begingroup$
Right! I missed this case, so I should add the condition X diagonalizable + inversible $Leftrightarrow$ X is an exponential. Thank you!
$endgroup$
– PerelMan
2 days ago
add a comment |
$begingroup$
Is every diagonalizable matrix is an exponential?
I know it is true in $SL_2(Bbb C)$ and I think it is true in $M_n(Bbb C)$ because if $M=PDP^-1$, we might be able to write D as $exp(E)$ for some $Ein M_n(Bbb C)$ as the exponential is surjective from $Bbb C$ onto $Bbb C^*$ and all eigenvalues of M are non zero because they are distinct.
Thank you for your help.
linear-algebra lie-groups diagonalization matrix-exponential
edited 2 days ago
José Carlos Santos
173k23133241
asked 2 days ago
PerelManPerelMan
732414
$endgroup$
Is every diagonalizable matrix is an exponential?
I know it is true in $SL_2(Bbb C)$ and I think it is true in $M_n(Bbb C)$ because if $M=PDP^-1$, we might be able to write D as $exp(E)$ for some $Ein M_n(Bbb C)$ as the exponential is surjective from $Bbb C$ onto $Bbb C^*$ and all eigenvalues of M are non zero because they are distinct.
Thank you for your help.
linear-algebra lie-groups diagonalization matrix-exponential
linear-algebra lie-groups diagonalization matrix-exponential
edited 2 days ago
José Carlos Santos
173k23133241
asked 2 days ago
PerelManPerelMan
732414
edited 2 days ago
José Carlos Santos
173k23133241
asked 2 days ago
PerelManPerelMan
732414
edited 2 days ago
José Carlos Santos
173k23133241
edited 2 days ago
José Carlos Santos
173k23133241
edited 2 days ago
José Carlos Santos
173k23133241
173k23133241
asked 2 days ago
PerelManPerelMan
732414
asked 2 days ago
PerelManPerelMan
732414
asked 2 days ago
PerelManPerelMan
732414
732414
2
$begingroup$
What if $M=O$ (the zero matrix)?
$endgroup$
– Minus One-Twelfth
2 days ago
$begingroup$
Right! I missed this case, so I should add the condition X diagonalizable + inversible $Leftrightarrow$ X is an exponential. Thank you!
$endgroup$
– PerelMan
2 days ago
add a comment |
2
$begingroup$
What if $M=O$ (the zero matrix)?
$endgroup$
– Minus One-Twelfth
2 days ago
$begingroup$
Right! I missed this case, so I should add the condition X diagonalizable + inversible $Leftrightarrow$ X is an exponential. Thank you!
$endgroup$
– PerelMan
2 days ago
2
2
$begingroup$
What if $M=O$ (the zero matrix)?
$endgroup$
– Minus One-Twelfth
2 days ago
$begingroup$
What if $M=O$ (the zero matrix)?
$endgroup$
– Minus One-Twelfth
2 days ago
$begingroup$
Right! I missed this case, so I should add the condition X diagonalizable + inversible $Leftrightarrow$ X is an exponential. Thank you!
$endgroup$
– PerelMan
2 days ago
$begingroup$
Right! I missed this case, so I should add the condition X diagonalizable + inversible $Leftrightarrow$ X is an exponential. Thank you!
$endgroup$
– PerelMan
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
A diagonalizable matrix is an exponential (over $mathbb C$) if and only if it is not a singular matrix. Of course, no singular matrix can be an exponential, since $det e^A=e^operatornametrAneq0$. On the other hand, if $A$ is diagonalizable, then it is similar to a diagonal matrix$$beginbmatrixd_1&0&0&ldots&0\0&d_2&0&ldots&0\0&0&d_3&ldots&0\vdots&vdots&vdots&ddots&vdots\0&0&0&ldots&d_nendbmatrix.$$Then $A$ is non-singular if and only if every $d_k$ is non--zero. So, let $lambda_k$ be a logarithm of $d_k$ and$$expleft(beginbmatrixlambda_1&0&0&ldots&0\0&lambda_2&0&ldots&0\0&0&lambda_3&ldots&0\vdots&vdots&vdots&ddots&vdots\0&0&0&ldots&lambda_nendbmatrixright)=beginbmatrixd_1&0&0&ldots&0\0&d_2&0&ldots&0\0&0&d_3&ldots&0\vdots&vdots&vdots&ddots&vdots\0&0&0&ldots&d_nendbmatrix.$$So, $A$ is exponential.
answered 2 days ago
José Carlos SantosJosé Carlos Santos
173k23133241
$endgroup$
add a comment |
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$begingroup$
A diagonalizable matrix is an exponential (over $mathbb C$) if and only if it is not a singular matrix. Of course, no singular matrix can be an exponential, since $det e^A=e^operatornametrAneq0$. On the other hand, if $A$ is diagonalizable, then it is similar to a diagonal matrix$$beginbmatrixd_1&0&0&ldots&0\0&d_2&0&ldots&0\0&0&d_3&ldots&0\vdots&vdots&vdots&ddots&vdots\0&0&0&ldots&d_nendbmatrix.$$Then $A$ is non-singular if and only if every $d_k$ is non--zero. So, let $lambda_k$ be a logarithm of $d_k$ and$$expleft(beginbmatrixlambda_1&0&0&ldots&0\0&lambda_2&0&ldots&0\0&0&lambda_3&ldots&0\vdots&vdots&vdots&ddots&vdots\0&0&0&ldots&lambda_nendbmatrixright)=beginbmatrixd_1&0&0&ldots&0\0&d_2&0&ldots&0\0&0&d_3&ldots&0\vdots&vdots&vdots&ddots&vdots\0&0&0&ldots&d_nendbmatrix.$$So, $A$ is exponential.
answered 2 days ago
José Carlos SantosJosé Carlos Santos
173k23133241
$endgroup$
add a comment |
$begingroup$
A diagonalizable matrix is an exponential (over $mathbb C$) if and only if it is not a singular matrix. Of course, no singular matrix can be an exponential, since $det e^A=e^operatornametrAneq0$. On the other hand, if $A$ is diagonalizable, then it is similar to a diagonal matrix$$beginbmatrixd_1&0&0&ldots&0\0&d_2&0&ldots&0\0&0&d_3&ldots&0\vdots&vdots&vdots&ddots&vdots\0&0&0&ldots&d_nendbmatrix.$$Then $A$ is non-singular if and only if every $d_k$ is non--zero. So, let $lambda_k$ be a logarithm of $d_k$ and$$expleft(beginbmatrixlambda_1&0&0&ldots&0\0&lambda_2&0&ldots&0\0&0&lambda_3&ldots&0\vdots&vdots&vdots&ddots&vdots\0&0&0&ldots&lambda_nendbmatrixright)=beginbmatrixd_1&0&0&ldots&0\0&d_2&0&ldots&0\0&0&d_3&ldots&0\vdots&vdots&vdots&ddots&vdots\0&0&0&ldots&d_nendbmatrix.$$So, $A$ is exponential.
answered 2 days ago
José Carlos SantosJosé Carlos Santos
173k23133241
$endgroup$
add a comment |
$begingroup$
A diagonalizable matrix is an exponential (over $mathbb C$) if and only if it is not a singular matrix. Of course, no singular matrix can be an exponential, since $det e^A=e^operatornametrAneq0$. On the other hand, if $A$ is diagonalizable, then it is similar to a diagonal matrix$$beginbmatrixd_1&0&0&ldots&0\0&d_2&0&ldots&0\0&0&d_3&ldots&0\vdots&vdots&vdots&ddots&vdots\0&0&0&ldots&d_nendbmatrix.$$Then $A$ is non-singular if and only if every $d_k$ is non--zero. So, let $lambda_k$ be a logarithm of $d_k$ and$$expleft(beginbmatrixlambda_1&0&0&ldots&0\0&lambda_2&0&ldots&0\0&0&lambda_3&ldots&0\vdots&vdots&vdots&ddots&vdots\0&0&0&ldots&lambda_nendbmatrixright)=beginbmatrixd_1&0&0&ldots&0\0&d_2&0&ldots&0\0&0&d_3&ldots&0\vdots&vdots&vdots&ddots&vdots\0&0&0&ldots&d_nendbmatrix.$$So, $A$ is exponential.
answered 2 days ago
José Carlos SantosJosé Carlos Santos
173k23133241
$endgroup$
A diagonalizable matrix is an exponential (over $mathbb C$) if and only if it is not a singular matrix. Of course, no singular matrix can be an exponential, since $det e^A=e^operatornametrAneq0$. On the other hand, if $A$ is diagonalizable, then it is similar to a diagonal matrix$$beginbmatrixd_1&0&0&ldots&0\0&d_2&0&ldots&0\0&0&d_3&ldots&0\vdots&vdots&vdots&ddots&vdots\0&0&0&ldots&d_nendbmatrix.$$Then $A$ is non-singular if and only if every $d_k$ is non--zero. So, let $lambda_k$ be a logarithm of $d_k$ and$$expleft(beginbmatrixlambda_1&0&0&ldots&0\0&lambda_2&0&ldots&0\0&0&lambda_3&ldots&0\vdots&vdots&vdots&ddots&vdots\0&0&0&ldots&lambda_nendbmatrixright)=beginbmatrixd_1&0&0&ldots&0\0&d_2&0&ldots&0\0&0&d_3&ldots&0\vdots&vdots&vdots&ddots&vdots\0&0&0&ldots&d_nendbmatrix.$$So, $A$ is exponential.
answered 2 days ago
José Carlos SantosJosé Carlos Santos
173k23133241
edited 2 days ago
answered 2 days ago
José Carlos SantosJosé Carlos Santos
173k23133241
answered 2 days ago
José Carlos SantosJosé Carlos Santos
173k23133241
answered 2 days ago
José Carlos SantosJosé Carlos Santos
173k23133241
173k23133241
add a comment |
add a comment |
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$begingroup$
What if $M=O$ (the zero matrix)?
$endgroup$
– Minus One-Twelfth
2 days ago
$begingroup$
Right! I missed this case, so I should add the condition X diagonalizable + inversible $Leftrightarrow$ X is an exponential. Thank you!
$endgroup$
– PerelMan
2 days ago