Is every diagonalizable matrix is an exponential
$begingroup$
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 Apr 6 at 14:00
José Carlos Santos
173k23133242
asked Apr 6 at 13:37
PerelManPerelMan
751414
$endgroup$
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 Apr 6 at 14:00
José Carlos Santos
173k23133242
asked Apr 6 at 13:37
PerelManPerelMan
751414
$endgroup$
2
$begingroup$
What if $M=O$ (the zero matrix)?
$endgroup$
– Minus One-Twelfth
Apr 6 at 13:41
$begingroup$
Right! I missed this case, so I should add the condition X diagonalizable + inversible $Leftrightarrow$ X is an exponential. Thank you!
$endgroup$
– PerelMan
Apr 6 at 14:14
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 Apr 6 at 14:00
José Carlos Santos
173k23133242
asked Apr 6 at 13:37
PerelManPerelMan
751414
$endgroup$
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 Apr 6 at 14:00
José Carlos Santos
173k23133242
asked Apr 6 at 13:37
PerelManPerelMan
751414
edited Apr 6 at 14:00
José Carlos Santos
173k23133242
asked Apr 6 at 13:37
PerelManPerelMan
751414
edited Apr 6 at 14:00
José Carlos Santos
173k23133242
edited Apr 6 at 14:00
José Carlos Santos
173k23133242
edited Apr 6 at 14:00
José Carlos Santos
173k23133242
173k23133242
asked Apr 6 at 13:37
PerelManPerelMan
751414
asked Apr 6 at 13:37
PerelManPerelMan
751414
asked Apr 6 at 13:37
PerelManPerelMan
751414
751414
2
$begingroup$
What if $M=O$ (the zero matrix)?
$endgroup$
– Minus One-Twelfth
Apr 6 at 13:41
$begingroup$
Right! I missed this case, so I should add the condition X diagonalizable + inversible $Leftrightarrow$ X is an exponential. Thank you!
$endgroup$
– PerelMan
Apr 6 at 14:14
add a comment |
2
$begingroup$
What if $M=O$ (the zero matrix)?
$endgroup$
– Minus One-Twelfth
Apr 6 at 13:41
$begingroup$
Right! I missed this case, so I should add the condition X diagonalizable + inversible $Leftrightarrow$ X is an exponential. Thank you!
$endgroup$
– PerelMan
Apr 6 at 14:14
2
2
$begingroup$
What if $M=O$ (the zero matrix)?
$endgroup$
– Minus One-Twelfth
Apr 6 at 13:41
$begingroup$
What if $M=O$ (the zero matrix)?
$endgroup$
– Minus One-Twelfth
Apr 6 at 13:41
$begingroup$
Right! I missed this case, so I should add the condition X diagonalizable + inversible $Leftrightarrow$ X is an exponential. Thank you!
$endgroup$
– PerelMan
Apr 6 at 14:14
$begingroup$
Right! I missed this case, so I should add the condition X diagonalizable + inversible $Leftrightarrow$ X is an exponential. Thank you!
$endgroup$
– PerelMan
Apr 6 at 14:14
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 $det e^A=e^{operatorname{tr}A}neq0$. On the other hand, if $A$ is diagonalizable, then it is similar to a diagonal matrix$$begin{bmatrix}d_1&0&0&ldots&0\0&d_2&0&ldots&0\0&0&d_3&ldots&0\vdots&vdots&vdots&ddots&vdots\0&0&0&ldots&d_nend{bmatrix}.$$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$$expleft(begin{bmatrix}lambda_1&0&0&ldots&0\0&lambda_2&0&ldots&0\0&0&lambda_3&ldots&0\vdots&vdots&vdots&ddots&vdots\0&0&0&ldots&lambda_nend{bmatrix}right)=begin{bmatrix}d_1&0&0&ldots&0\0&d_2&0&ldots&0\0&0&d_3&ldots&0\vdots&vdots&vdots&ddots&vdots\0&0&0&ldots&d_nend{bmatrix}.$$So, $A$ is exponential.
answered Apr 6 at 13:41
José Carlos SantosJosé Carlos Santos
173k23133242
$endgroup$
add a comment |
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1 Answer
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1 Answer
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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 $det e^A=e^{operatorname{tr}A}neq0$. On the other hand, if $A$ is diagonalizable, then it is similar to a diagonal matrix$$begin{bmatrix}d_1&0&0&ldots&0\0&d_2&0&ldots&0\0&0&d_3&ldots&0\vdots&vdots&vdots&ddots&vdots\0&0&0&ldots&d_nend{bmatrix}.$$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$$expleft(begin{bmatrix}lambda_1&0&0&ldots&0\0&lambda_2&0&ldots&0\0&0&lambda_3&ldots&0\vdots&vdots&vdots&ddots&vdots\0&0&0&ldots&lambda_nend{bmatrix}right)=begin{bmatrix}d_1&0&0&ldots&0\0&d_2&0&ldots&0\0&0&d_3&ldots&0\vdots&vdots&vdots&ddots&vdots\0&0&0&ldots&d_nend{bmatrix}.$$So, $A$ is exponential.
answered Apr 6 at 13:41
José Carlos SantosJosé Carlos Santos
173k23133242
$endgroup$
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 $det e^A=e^{operatorname{tr}A}neq0$. On the other hand, if $A$ is diagonalizable, then it is similar to a diagonal matrix$$begin{bmatrix}d_1&0&0&ldots&0\0&d_2&0&ldots&0\0&0&d_3&ldots&0\vdots&vdots&vdots&ddots&vdots\0&0&0&ldots&d_nend{bmatrix}.$$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$$expleft(begin{bmatrix}lambda_1&0&0&ldots&0\0&lambda_2&0&ldots&0\0&0&lambda_3&ldots&0\vdots&vdots&vdots&ddots&vdots\0&0&0&ldots&lambda_nend{bmatrix}right)=begin{bmatrix}d_1&0&0&ldots&0\0&d_2&0&ldots&0\0&0&d_3&ldots&0\vdots&vdots&vdots&ddots&vdots\0&0&0&ldots&d_nend{bmatrix}.$$So, $A$ is exponential.
answered Apr 6 at 13:41
José Carlos SantosJosé Carlos Santos
173k23133242
$endgroup$
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 $det e^A=e^{operatorname{tr}A}neq0$. On the other hand, if $A$ is diagonalizable, then it is similar to a diagonal matrix$$begin{bmatrix}d_1&0&0&ldots&0\0&d_2&0&ldots&0\0&0&d_3&ldots&0\vdots&vdots&vdots&ddots&vdots\0&0&0&ldots&d_nend{bmatrix}.$$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$$expleft(begin{bmatrix}lambda_1&0&0&ldots&0\0&lambda_2&0&ldots&0\0&0&lambda_3&ldots&0\vdots&vdots&vdots&ddots&vdots\0&0&0&ldots&lambda_nend{bmatrix}right)=begin{bmatrix}d_1&0&0&ldots&0\0&d_2&0&ldots&0\0&0&d_3&ldots&0\vdots&vdots&vdots&ddots&vdots\0&0&0&ldots&d_nend{bmatrix}.$$So, $A$ is exponential.
answered Apr 6 at 13:41
José Carlos SantosJosé Carlos Santos
173k23133242
$endgroup$
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 $det e^A=e^{operatorname{tr}A}neq0$. On the other hand, if $A$ is diagonalizable, then it is similar to a diagonal matrix$$begin{bmatrix}d_1&0&0&ldots&0\0&d_2&0&ldots&0\0&0&d_3&ldots&0\vdots&vdots&vdots&ddots&vdots\0&0&0&ldots&d_nend{bmatrix}.$$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$$expleft(begin{bmatrix}lambda_1&0&0&ldots&0\0&lambda_2&0&ldots&0\0&0&lambda_3&ldots&0\vdots&vdots&vdots&ddots&vdots\0&0&0&ldots&lambda_nend{bmatrix}right)=begin{bmatrix}d_1&0&0&ldots&0\0&d_2&0&ldots&0\0&0&d_3&ldots&0\vdots&vdots&vdots&ddots&vdots\0&0&0&ldots&d_nend{bmatrix}.$$So, $A$ is exponential.
answered Apr 6 at 13:41
José Carlos SantosJosé Carlos Santos
173k23133242
edited Apr 6 at 13:50
answered Apr 6 at 13:41
José Carlos SantosJosé Carlos Santos
173k23133242
answered Apr 6 at 13:41
José Carlos SantosJosé Carlos Santos
173k23133242
answered Apr 6 at 13:41
José Carlos SantosJosé Carlos Santos
173k23133242
173k23133242
add a comment |
add a comment |
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2
$begingroup$
What if $M=O$ (the zero matrix)?
$endgroup$
– Minus One-Twelfth
Apr 6 at 13:41
$begingroup$
Right! I missed this case, so I should add the condition X diagonalizable + inversible $Leftrightarrow$ X is an exponential. Thank you!
$endgroup$
– PerelMan
Apr 6 at 14:14