Given an algebraically closed field $\mathbb K$ and matrices $A, B \in \mathbb K^{n \times n}$ such that $A B = B A$, show that $A$ and $B$ are simultaneously triangularizable, i.e., show that there exists a matrix $T$ such that $T^{-1} A T$ and $T^{-1} B T$ are both upper triangular.
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3What have you tried ? – user577215664 Jul 05 '20 at 16:56
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The hint is to show that A and B have a the same eigenvector but i dont know how to approach this problem – spoonispooni Jul 05 '20 at 17:06
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Is it given that $\mathbb K$ is algebraically closed? Otherwise, I am not sure that the hint is true. – Dylan C. Beck Jul 05 '20 at 17:15
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Not mentioned but i would take that as a given yes – spoonispooni Jul 05 '20 at 17:16
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I think we mean the same thing sorry for my bad english i am from germany. It means that it is similar to a upper triangular matrix – spoonispooni Jul 05 '20 at 17:22
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Observe that any eigenspace of $A$ is $B$-invariant. Explicitly, given any vector $v$ such that $Av = \lambda v,$ we have that $A(Bv) = (AB)v = (BA)v = B(Av) = B(\lambda v) = \lambda Bv$ so that $Bv$ is either $0$ or an eigenvector of $A$ with respect to $\lambda.$ Given any nonzero eigenspace $W_\lambda$ of $A$ corresponding to the eigenvalue $\lambda$ of $A,$ we conclude that $B$ restricts to a linear operator $B|_{W_\lambda} : W_\lambda \to W_\lambda.$ Considering that $\mathbb K$ is algebraically closed, the characteristic polynomial of $B|_{W_\lambda}$ splits into (not necessarily distinct) linear factors, hence there exists a linear polynomial $x - \mu$ such that $B|_{W_\lambda} - \mu I$ is the zero operator on $W_\lambda.$ Consequently, there exists a nonzero vector $w$ in $W_\lambda$ such that $Bw = \mu w$ and $Aw = \lambda w.$ We conclude therefore that $A$ and $B$ have the same eigenvectors.
Can you finish the proof now that you have proven the hint?
Dylan C. Beck
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Now i need to show that the T matrix exists but again i´m missing an approach for that – spoonispooni Jul 05 '20 at 17:28
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I need to show that T exists at all. i sat too long on this problem i probably dont see the obvious right now – spoonispooni Jul 05 '20 at 17:33
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Of course, there exists a $T$ for $A$ and a $T'$ for $B$ because $\mathbb K$ is algebraically closed. I am wondering if you have proven this fact. If so, then the same $T$ will work for both $A$ and $B$ because $A$ and $B$ have the same eigenvectors, as I have proven in my solution. – Dylan C. Beck Jul 05 '20 at 17:38
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I recommend taking a look at Theorem 2.9 of these notes by Pete L. Clark. – Dylan C. Beck Jul 05 '20 at 17:59
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I still struggle to see why a matrix T exists if A and B are simultaneously triangularizable – spoonispooni Jul 05 '20 at 18:25
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We form the matrix $T$ by taking a basis of eigenvectors for $A.$ But the eigenvectors of $A$ are precisely the eigenvectors of $B,$ hence the matrix $T$ is a basis of eigenvectors for $B.$ – Dylan C. Beck Jul 05 '20 at 18:30
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yes i got that but why do they need to be simultaneously triangularizable – spoonispooni Jul 05 '20 at 18:39