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AB diagonalizable then BA also diagonalizable

Simultaneously Diagonalizable ProofCommuting matrices are simultaneously triangularizableFind a non diagonalizable matrix that commutes with a given matrixAll linear combinations diagonalizable over $mathbbC$ implies commuting.Simultaneously diagonalizable without distinct eigenvalues$A$ is diagonalizable, if $A,B$ have then same eigenvalues, then $B$ is also diagonalizableGiven two diagonalizable matrices that commute (AB = BA),is AB necessarily diagonalizable?Why do the nilpotent and diagonalizable parts commute?When is the product of two non simultaneously diagonalizable matrices (one of them nonsingular but none of them positive definite) diagonalizable?$A^2$ is diagonalizable and so is $A$

3

$begingroup$

If A and B don't commute are there counterexamples that AB is diagonalizable but BA not?

I read that if AB=BA then both AB and BA are diagonalizable.

linear-algebra

share|cite|improve this question

asked 2 days ago

craftcraft

232

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  If $AB=BA$ then $AB$, $BA$ are either both diagonalisable or neither is diagonalisable. I hope that's obvious.
  $endgroup$
  – Lord Shark the Unknown
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @LordSharktheUnknown They are asking what can happen if $AB neq BA$.
  $endgroup$
  – kccu
  2 days ago
  

  
  
  
  

add a comment | 

3

$begingroup$

If A and B don't commute are there counterexamples that AB is diagonalizable but BA not?

I read that if AB=BA then both AB and BA are diagonalizable.

linear-algebra

share|cite|improve this question

asked 2 days ago

craftcraft

232

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  If $AB=BA$ then $AB$, $BA$ are either both diagonalisable or neither is diagonalisable. I hope that's obvious.
  $endgroup$
  – Lord Shark the Unknown
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @LordSharktheUnknown They are asking what can happen if $AB neq BA$.
  $endgroup$
  – kccu
  2 days ago
  

  
  
  
  

add a comment | 

$begingroup$

If A and B don't commute are there counterexamples that AB is diagonalizable but BA not?

I read that if AB=BA then both AB and BA are diagonalizable.

linear-algebra

share|cite|improve this question

asked 2 days ago

craftcraft

232

$endgroup$

share|cite|improve this question

asked 2 days ago

craftcraft

232

share|cite|improve this question

asked 2 days ago

craftcraft

232

asked 2 days ago

craftcraft

232

asked 2 days ago

craftcraft

232

2 Answers
2

active

oldest

votes

8

$begingroup$

Try $A = beginbmatrix0&0\1&0endbmatrix$ and $B = beginbmatrix0&0\0&1endbmatrix$

share|cite|improve this answer

answered 2 days ago

Michael BiroMichael Biro

11.6k21831

$endgroup$

add a comment | 

4

$begingroup$

This is true in the special case when both $A$ and $B$ are invertible.

If we denote by $E_C(lambda)$ the eigenspace of the matrix $C$ relative to the eigenvalue $lambda$, we can see that
$$
vmapsto Bv
$$
induces an injective linear map $E_AB(lambda)to E_BA(lambda)$, so $dim E_AB(lambda)ledim E_BA(lambda)$, where $lambda$ is any eigenvalue of $AB$. By symmetry, the two eigenspaces have the same dimension.

Also this proves that $AB$ and $BA$ have the same eigenvalues. If $AB$ is diagonalizable, then the geometric multiplicities of its eigenvalues sum up to $n$ (the size of the matrices $A$ and $B$). Thus the same happens for $BA$ and $BA$ is diagonalizable.

If $A$ or $B$ is not invertible, we can still see that $AB$ and $BA$ have the same nonzero eigenvalues. Indeed, if $lambda$ is a nonzero eigenvalue of $AB$, with eigenvector $v$, then $ABv=lambda v$ (which implies $Bvne0$, so $(BA)(Bv)=lambda(Bv)$ and therefore $lambda$ is an eigenvalue of $BA$. By symmetry, $AB$ and $BA$ share the nonzero eigenvalues. Also, $0$ is an eigenvalue of both, because neither is invertible. However, in this case we can't control the geometric multiplicities, as Michael Biro's example shows.

share|cite|improve this answer

edited 2 days ago

Shalop

9,43811130

answered 2 days ago

egregegreg

185k1486206

$endgroup$

add a comment | 

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8

$begingroup$

Try $A = beginbmatrix0&0\1&0endbmatrix$ and $B = beginbmatrix0&0\0&1endbmatrix$

share|cite|improve this answer

answered 2 days ago

Michael BiroMichael Biro

11.6k21831

$endgroup$

add a comment | 

8

$begingroup$

Try $A = beginbmatrix0&0\1&0endbmatrix$ and $B = beginbmatrix0&0\0&1endbmatrix$

share|cite|improve this answer

answered 2 days ago

Michael BiroMichael Biro

11.6k21831

$endgroup$

add a comment | 

$begingroup$

Try $A = beginbmatrix0&0\1&0endbmatrix$ and $B = beginbmatrix0&0\0&1endbmatrix$

share|cite|improve this answer

answered 2 days ago

Michael BiroMichael Biro

11.6k21831

$endgroup$

share|cite|improve this answer

answered 2 days ago

Michael BiroMichael Biro

11.6k21831

answered 2 days ago

Michael BiroMichael Biro

11.6k21831

answered 2 days ago

Michael BiroMichael Biro

11.6k21831

4

$begingroup$

This is true in the special case when both $A$ and $B$ are invertible.

If we denote by $E_C(lambda)$ the eigenspace of the matrix $C$ relative to the eigenvalue $lambda$, we can see that
$$
vmapsto Bv
$$
induces an injective linear map $E_AB(lambda)to E_BA(lambda)$, so $dim E_AB(lambda)ledim E_BA(lambda)$, where $lambda$ is any eigenvalue of $AB$. By symmetry, the two eigenspaces have the same dimension.

Also this proves that $AB$ and $BA$ have the same eigenvalues. If $AB$ is diagonalizable, then the geometric multiplicities of its eigenvalues sum up to $n$ (the size of the matrices $A$ and $B$). Thus the same happens for $BA$ and $BA$ is diagonalizable.

If $A$ or $B$ is not invertible, we can still see that $AB$ and $BA$ have the same nonzero eigenvalues. Indeed, if $lambda$ is a nonzero eigenvalue of $AB$, with eigenvector $v$, then $ABv=lambda v$ (which implies $Bvne0$, so $(BA)(Bv)=lambda(Bv)$ and therefore $lambda$ is an eigenvalue of $BA$. By symmetry, $AB$ and $BA$ share the nonzero eigenvalues. Also, $0$ is an eigenvalue of both, because neither is invertible. However, in this case we can't control the geometric multiplicities, as Michael Biro's example shows.

share|cite|improve this answer

edited 2 days ago

Shalop

9,43811130

answered 2 days ago

egregegreg

185k1486206

$endgroup$

add a comment | 

4

$begingroup$

This is true in the special case when both $A$ and $B$ are invertible.

If we denote by $E_C(lambda)$ the eigenspace of the matrix $C$ relative to the eigenvalue $lambda$, we can see that
$$
vmapsto Bv
$$
induces an injective linear map $E_AB(lambda)to E_BA(lambda)$, so $dim E_AB(lambda)ledim E_BA(lambda)$, where $lambda$ is any eigenvalue of $AB$. By symmetry, the two eigenspaces have the same dimension.

Also this proves that $AB$ and $BA$ have the same eigenvalues. If $AB$ is diagonalizable, then the geometric multiplicities of its eigenvalues sum up to $n$ (the size of the matrices $A$ and $B$). Thus the same happens for $BA$ and $BA$ is diagonalizable.

If $A$ or $B$ is not invertible, we can still see that $AB$ and $BA$ have the same nonzero eigenvalues. Indeed, if $lambda$ is a nonzero eigenvalue of $AB$, with eigenvector $v$, then $ABv=lambda v$ (which implies $Bvne0$, so $(BA)(Bv)=lambda(Bv)$ and therefore $lambda$ is an eigenvalue of $BA$. By symmetry, $AB$ and $BA$ share the nonzero eigenvalues. Also, $0$ is an eigenvalue of both, because neither is invertible. However, in this case we can't control the geometric multiplicities, as Michael Biro's example shows.

share|cite|improve this answer

edited 2 days ago

Shalop

9,43811130

answered 2 days ago

egregegreg

185k1486206

$endgroup$

add a comment | 

$begingroup$

This is true in the special case when both $A$ and $B$ are invertible.

If we denote by $E_C(lambda)$ the eigenspace of the matrix $C$ relative to the eigenvalue $lambda$, we can see that
$$
vmapsto Bv
$$
induces an injective linear map $E_AB(lambda)to E_BA(lambda)$, so $dim E_AB(lambda)ledim E_BA(lambda)$, where $lambda$ is any eigenvalue of $AB$. By symmetry, the two eigenspaces have the same dimension.

Also this proves that $AB$ and $BA$ have the same eigenvalues. If $AB$ is diagonalizable, then the geometric multiplicities of its eigenvalues sum up to $n$ (the size of the matrices $A$ and $B$). Thus the same happens for $BA$ and $BA$ is diagonalizable.

If $A$ or $B$ is not invertible, we can still see that $AB$ and $BA$ have the same nonzero eigenvalues. Indeed, if $lambda$ is a nonzero eigenvalue of $AB$, with eigenvector $v$, then $ABv=lambda v$ (which implies $Bvne0$, so $(BA)(Bv)=lambda(Bv)$ and therefore $lambda$ is an eigenvalue of $BA$. By symmetry, $AB$ and $BA$ share the nonzero eigenvalues. Also, $0$ is an eigenvalue of both, because neither is invertible. However, in this case we can't control the geometric multiplicities, as Michael Biro's example shows.

share|cite|improve this answer

edited 2 days ago

Shalop

9,43811130

answered 2 days ago

egregegreg

185k1486206

$endgroup$

share|cite|improve this answer

edited 2 days ago

Shalop

9,43811130

answered 2 days ago

egregegreg

185k1486206

edited 2 days ago

Shalop

9,43811130

edited 2 days ago

Shalop

9,43811130

answered 2 days ago

egregegreg

185k1486206

answered 2 days ago

egregegreg

185k1486206