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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 $A^{-1} = A$

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

edited 12 hours ago

José Carlos Santos

173k23133241

asked 12 hours ago

PerelManPerelMan

723413

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  What if $M=O$ (the zero matrix)?
  $endgroup$
  – Minus One-Twelfth
  12 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
  11 hours ago
  
  
  

  
  
  
  

add a comment | 

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

edited 12 hours ago

José Carlos Santos

173k23133241

asked 12 hours ago

PerelManPerelMan

723413

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  What if $M=O$ (the zero matrix)?
  $endgroup$
  – Minus One-Twelfth
  12 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
  11 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

share|cite|improve this question

edited 12 hours ago

José Carlos Santos

173k23133241

asked 12 hours ago

PerelManPerelMan

723413

$endgroup$

share|cite|improve this question

edited 12 hours ago

José Carlos Santos

173k23133241

asked 12 hours ago

PerelManPerelMan

723413

share|cite|improve this question

edited 12 hours ago

José Carlos Santos

173k23133241

asked 12 hours ago

PerelManPerelMan

723413

edited 12 hours ago

José Carlos Santos

173k23133241

edited 12 hours ago

José Carlos Santos

173k23133241

asked 12 hours ago

PerelManPerelMan

723413

asked 12 hours ago

PerelManPerelMan

723413

1 Answer
1

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oldest

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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

edited 12 hours ago

answered 12 hours ago

José Carlos SantosJosé Carlos Santos

173k23133241

$endgroup$

add a comment | 

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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

edited 12 hours ago

answered 12 hours ago

José Carlos SantosJosé Carlos Santos

173k23133241

$endgroup$

add a comment | 

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

edited 12 hours ago

answered 12 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 $det e^A=e^{operatorname{tr}A}neq0$. 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

edited 12 hours ago

answered 12 hours ago

José Carlos SantosJosé Carlos Santos

173k23133241

$endgroup$

share|cite|improve this answer

edited 12 hours ago

answered 12 hours ago

José Carlos SantosJosé Carlos Santos

173k23133241

answered 12 hours ago

José Carlos SantosJosé Carlos Santos

173k23133241

answered 12 hours ago

José Carlos SantosJosé Carlos Santos

173k23133241