AB diagonalizable then BA also diagonalizable Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)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 $A$
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AB diagonalizable then BA also diagonalizable
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)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 $A$
$begingroup$
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
asked Mar 31 at 17:38
craftcraft
233
$endgroup$
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
asked Mar 31 at 17:38
craftcraft
233
$endgroup$
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
Mar 31 at 17:41
1
$begingroup$
@LordSharktheUnknown They are asking what can happen if $AB neq BA$.
$endgroup$
– kccu
Mar 31 at 17:42
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
asked Mar 31 at 17:38
craftcraft
233
$endgroup$
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
linear-algebra
asked Mar 31 at 17:38
craftcraft
233
asked Mar 31 at 17:38
craftcraft
233
asked Mar 31 at 17:38
craftcraft
233
asked Mar 31 at 17:38
craftcraft
233
asked Mar 31 at 17:38
craftcraft
233
233
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
Mar 31 at 17:41
1
$begingroup$
@LordSharktheUnknown They are asking what can happen if $AB neq BA$.
$endgroup$
– kccu
Mar 31 at 17:42
add a comment |
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
Mar 31 at 17:41
1
$begingroup$
@LordSharktheUnknown They are asking what can happen if $AB neq BA$.
$endgroup$
– kccu
Mar 31 at 17:42
2
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
Mar 31 at 17:41
$begingroup$
If $AB=BA$ then $AB$, $BA$ are either both diagonalisable or neither is diagonalisable. I hope that's obvious.
$endgroup$
– Lord Shark the Unknown
Mar 31 at 17:41
1
1
$begingroup$
@LordSharktheUnknown They are asking what can happen if $AB neq BA$.
$endgroup$
– kccu
Mar 31 at 17:42
$begingroup$
@LordSharktheUnknown They are asking what can happen if $AB neq BA$.
$endgroup$
– kccu
Mar 31 at 17:42
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Try $A = beginbmatrix0&0\1&0endbmatrix$ and $B = beginbmatrix0&0\0&1endbmatrix$
answered Mar 31 at 17:43
Michael BiroMichael Biro
11.7k21831
$endgroup$
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.
edited Mar 31 at 22:35
Shalop
9,43811130
answered Mar 31 at 19:46
egregegreg
186k1486208
$endgroup$
add a comment |
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$begingroup$
Try $A = beginbmatrix0&0\1&0endbmatrix$ and $B = beginbmatrix0&0\0&1endbmatrix$
answered Mar 31 at 17:43
Michael BiroMichael Biro
11.7k21831
$endgroup$
add a comment |
$begingroup$
Try $A = beginbmatrix0&0\1&0endbmatrix$ and $B = beginbmatrix0&0\0&1endbmatrix$
answered Mar 31 at 17:43
Michael BiroMichael Biro
11.7k21831
$endgroup$
add a comment |
$begingroup$
Try $A = beginbmatrix0&0\1&0endbmatrix$ and $B = beginbmatrix0&0\0&1endbmatrix$
answered Mar 31 at 17:43
Michael BiroMichael Biro
11.7k21831
$endgroup$
Try $A = beginbmatrix0&0\1&0endbmatrix$ and $B = beginbmatrix0&0\0&1endbmatrix$
answered Mar 31 at 17:43
Michael BiroMichael Biro
11.7k21831
answered Mar 31 at 17:43
Michael BiroMichael Biro
11.7k21831
answered Mar 31 at 17:43
Michael BiroMichael Biro
11.7k21831
answered Mar 31 at 17:43
Michael BiroMichael Biro
11.7k21831
11.7k21831
add a comment |
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.
edited Mar 31 at 22:35
Shalop
9,43811130
answered Mar 31 at 19:46
egregegreg
186k1486208
$endgroup$
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.
edited Mar 31 at 22:35
Shalop
9,43811130
answered Mar 31 at 19:46
egregegreg
186k1486208
$endgroup$
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.
edited Mar 31 at 22:35
Shalop
9,43811130
answered Mar 31 at 19:46
egregegreg
186k1486208
$endgroup$
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.
edited Mar 31 at 22:35
Shalop
9,43811130
answered Mar 31 at 19:46
egregegreg
186k1486208
edited Mar 31 at 22:35
Shalop
9,43811130
edited Mar 31 at 22:35
Shalop
9,43811130
edited Mar 31 at 22:35
Shalop
9,43811130
9,43811130
answered Mar 31 at 19:46
egregegreg
186k1486208
answered Mar 31 at 19:46
egregegreg
186k1486208
answered Mar 31 at 19:46
egregegreg
186k1486208
186k1486208
add a comment |
add a comment |
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$begingroup$
If $AB=BA$ then $AB$, $BA$ are either both diagonalisable or neither is diagonalisable. I hope that's obvious.
$endgroup$
– Lord Shark the Unknown
Mar 31 at 17:41
1
$begingroup$
@LordSharktheUnknown They are asking what can happen if $AB neq BA$.
$endgroup$
– kccu
Mar 31 at 17:42