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Let $S$ and $T$ be two linear maps from $V$ to $V$ ($V$ complex vector space) such that $ST=TS$. I need to prove that there exists a basis with respect to which both the matrices are in upper-triangular form. How to proceed ?

Because $V$ is a complex vector space there exists a basis wrt to which any linear map can be upper triangularized. But how to prove that both $S$ and $T$ can be upper triangularized wrt to the same basis.

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You just need to know the right lingo to google. This is a standard result, often stated as "commuting matrices over $\mathbb{C}$ are simultaneously triangularizable", and, in fact, unitarily simultaneously triangularizable. The proof can be found here: planetmath.org/… –  Alex Youcis Sep 29 '13 at 11:05

1 Answer 1

Step 1: If $S$ and $T$ are commuting linear maps, then for any eigenvalue $\lambda$, let $V_{\lambda}$ be the $\lambda$-eigenspace for $S$. Then $V_{\lambda}$ is $T$-invariant: $T V_{\lambda} \subset V_{\lambda}$.

Step 2: Since the scalar field is $\mathbb{C}$, the restriction of $T$ to $V_{\lambda}$ has an eigenvector $v$, which is thus a simultaneous eigenvector for $S$ and $T$. Hence any two commuting endomorphisms of a finite-dimensional $\mathbb{C}$-vector space have a common eigenvector.

Step 3: Revisit your favorite proof of the triangularizability of $S$. In that proof, whenever you see an eigenvector for $S$, use Step 2 to get a simultaneous eigenvector for $S$ and $T$. Verify that the proof for $S$ works to give a simultaneous triangularization for $S$ and $T$.

Remark: For any scalar field $F$, commuting triangularizable endomorphisms of a finite-dimensional $F$-vector space are simultaneously triangularizable. To modify the above proof to yield this, we need to recall that an endomorphism is triangularizable iff its minimal polynomial is split, i.e., factors into a product of linear polynomials. If so, then the minimal polynomial of the endomorphism restricted to any invariant subspace divides the original minimal polynomial so it is also split, hence in Step 2 above the required eigenvector will exist over $F$.

(This remark was partially motivated by Alex Youcis's comment. In my view, talking about unitary triangularizability is bringing in extra structure that is not necessary to solve the problem. Unless you care specifically about unitary triangularizability, of course, but the OP did not mention that.)

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