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 9 hours ago
José Carlos Santos
173k23133241
asked 10 hours ago
PerelManPerelMan
723413
$endgroup$
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 9 hours ago
José Carlos Santos
173k23133241
asked 10 hours ago
PerelManPerelMan
723413
$endgroup$
2
$begingroup$
What if $M=O$ (the zero matrix)?
$endgroup$
– Minus One-Twelfth
10 hours 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
9 hours 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 9 hours ago
José Carlos Santos
173k23133241
asked 10 hours ago
PerelManPerelMan
723413
$endgroup$
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 9 hours ago
José Carlos Santos
173k23133241
asked 10 hours ago
PerelManPerelMan
723413
edited 9 hours ago
José Carlos Santos
173k23133241
asked 10 hours ago
PerelManPerelMan
723413
edited 9 hours ago
José Carlos Santos
173k23133241
edited 9 hours ago
José Carlos Santos
173k23133241
edited 9 hours ago
José Carlos Santos
173k23133241
173k23133241
asked 10 hours ago
PerelManPerelMan
723413
asked 10 hours ago
PerelManPerelMan
723413
asked 10 hours ago
PerelManPerelMan
723413
723413
2
$begingroup$
What if $M=O$ (the zero matrix)?
$endgroup$
– Minus One-Twelfth
10 hours 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
9 hours ago
add a comment |
2
$begingroup$
What if $M=O$ (the zero matrix)?
$endgroup$
– Minus One-Twelfth
10 hours 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
9 hours ago
2
2
$begingroup$
What if $M=O$ (the zero matrix)?
$endgroup$
– Minus One-Twelfth
10 hours ago
$begingroup$
What if $M=O$ (the zero matrix)?
$endgroup$
– Minus One-Twelfth
10 hours 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
9 hours 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
9 hours 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 10 hours 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 10 hours 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 10 hours 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 10 hours 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 10 hours ago
José Carlos SantosJosé Carlos Santos
173k23133241
edited 10 hours ago
answered 10 hours ago
José Carlos SantosJosé Carlos Santos
173k23133241
answered 10 hours ago
José Carlos SantosJosé Carlos Santos
173k23133241
answered 10 hours 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
10 hours 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
9 hours ago