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.
Prove that commuting matrices over an algebraically closed field are simultaneously triangularizable.
linear algebramatricestriangularization
Best Answer
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?