Question:
A and B are matrices size $n\times n$
given $AB=BA$ and A has n eigenvalues, prove $B$ is diagonalizable.
I would have written what I tried to do, but It's really nothing worth reading..
Any help would appreciated, Thanks.
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8This is not true in general. $B=A$ satisfies $AB=BA$ but doesn't necessarily imply $A$ is diagonal.. Unless you mean $B$ is diagonalizable.. – Cameron Williams Mar 24 '14 at 16:36
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1Do you mean that $B$ is diagonalizable? – Git Gud Mar 24 '14 at 16:37
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1If you really mean diagonalizable, have a look here: http://math.stackexchange.com/questions/285546/if-ab-ba-show-that-b-is-diagonalizable – Martin Sleziak Mar 24 '14 at 16:45
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Yes, yes! sorry, diagonalizable... :) – Splash Mar 24 '14 at 16:48
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Hint: if $AB=BA$ the set of eigenvectors of $A$ are $B$-stable.
details:
indeed, if $Ax=ux$, then $ A(Bx) = BAx = u(Bx) $ and $Bx\propto x$, because $\{ y| Ay=uy \}$ is a line which contains $x$ (because $A$ has $n$ eigenvalues).
Let $(e_i)$ be a basis of eigenvectors of $A$. Then in this basis, $B$ is diagonal as well.
mookid
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What do "eigenvectors' set is stable" mean? And what does $;Bx\propto x;$ mean, too? – DonAntonio Mar 24 '14 at 16:43
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it means that is $v $ is an eigenvector of $A$ then $Bv$ too. The second one means that there is a scalar $l$ such as $Bx=lx$ – mookid Mar 24 '14 at 16:44
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1Uh? Did you mean that $;v;$ eigenvector of $;A\implies Bv;$ also an eigenvector of $;A;$ ? – DonAntonio Mar 24 '14 at 16:46
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1The french are going to drive me crazy with their notations and names...! – DonAntonio Mar 24 '14 at 16:46
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Look at after "details": I show that if $Ax=ux$ then $A(Bx)=u(Bx)$. In other words: the set ${ x: Ax=ux}$ is $B$-stable. – mookid Mar 24 '14 at 16:48