Is every diagonalizable matrix is an exponential












1












$begingroup$


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







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  • 2




    $begingroup$
    What if $M=O$ (the zero matrix)?
    $endgroup$
    – Minus One-Twelfth
    14 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
    13 hours ago


















1












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    What if $M=O$ (the zero matrix)?
    $endgroup$
    – Minus One-Twelfth
    14 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
    13 hours ago
















1












1








1





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 14 hours ago









José Carlos Santos

173k23133241




173k23133241










asked 14 hours ago









PerelManPerelMan

723413




723413








  • 2




    $begingroup$
    What if $M=O$ (the zero matrix)?
    $endgroup$
    – Minus One-Twelfth
    14 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
    13 hours ago
















  • 2




    $begingroup$
    What if $M=O$ (the zero matrix)?
    $endgroup$
    – Minus One-Twelfth
    14 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
    13 hours ago










2




2




$begingroup$
What if $M=O$ (the zero matrix)?
$endgroup$
– Minus One-Twelfth
14 hours ago




$begingroup$
What if $M=O$ (the zero matrix)?
$endgroup$
– Minus One-Twelfth
14 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
13 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
13 hours ago












1 Answer
1






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4












$begingroup$

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.






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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.






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      4












      $begingroup$

      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.






      share|cite|improve this answer











      $endgroup$
















        4












        4








        4





        $begingroup$

        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.






        share|cite|improve this answer











        $endgroup$



        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.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 14 hours ago

























        answered 14 hours ago









        José Carlos SantosJosé Carlos Santos

        173k23133241




        173k23133241






























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