# Proofs in Linear Algebra via Topology

I'm watching the lecture series by Tadashi Tokieda on Topology and Geometry on YouTube. In the second lecture he shows how one can prove, using a topological argument, that given two $n\times n$ matrices $P$ and $Q$, the matrix products $PQ$ and $QP$ share the same eigenvalues. Then he claims that one can also prove that $\det(e^L)=e^{\operatorname{tr} L}$ and the Cayley-Hamilton theorem topologically as well.

This has me interested. Is there some book/ reference work that discusses linear algebra in the context of topology? Ideally I'd like a book (at maybe the 1st year grad level) on linear algebra that uses topological arguments to prove linear algebraic statements. Does anyone know of such of work?

For example, if $L$ is diagonalizable with eigenvalues $\lambda_1, \dots \lambda_n$, then it's clear that $(L - \lambda_1) \dots (L - \lambda_n) = 0$, which is the Cayley-Hamilton theorem for $L$. But the Cayley-Hamilton theorem is a "continuous" fact: for an $n \times n$ matrix it asserts that $n^2$ polynomial functions of the $n^2$ entries of $L$ vanish. And the diagonalizable matrices are dense (over $\mathbb{C}$). Hence we get Cayley-Hamilton in general.
Similarly, the claim that $PQ$ and $QP$ have the same characteristic polynomial (equivalently, the same eigenvalues, with the same arithmetic multiplicities; this is a bit stronger than what you wrote) is clear if, say, $P$ is invertible. But this is a "continuous" fact: for $n \times n$ matrices it asserts that $n$ polynomial functions of the $2n^2$ entries of $P$ and $Q$ vanish. And the invertible matrices are dense. Hence we get the claim in general. See also this blog post for other proofs and generalizations.
But I think $\det (e^L) = e^{\text{tr}(L)}$ is a bad example; the density reduction doesn't really buy you anything here. It's clear that $\text{tr}(L)$ is the sum of the eigenvalues of $L$ and that $\det (e^L)$ is the product of the exponentials of the eigenvalues of $L$ whether or not $L$ is diagonalizable, because we can upper-triangularize $L$ (e.g. bring it into Jordan normal form) instead of diagonalizing it. Note that this is not good enough to prove Cayley-Hamilton.
• It's worth mentioning that there's a very short and clean proof that matrices can be upper-triangularized over an algebraically closed field; this is much easier than Jordan normal form. If $T : V \to V$ is a linear operator on a f.d. vector space $V$ over such a field, then $T$ has an eigenvector $v_1$. Also $T$ acting on the quotient $V/\text{span}(v_1)$ has an eigenvector $v_2$. Also $T$ acting on the quotient $V/\text{span}(v_1, v_2)$ has an eigenvector $v_3$. And so on. After lifting all of these to elements of $V$ you get a basis in which $T$ is upper triangular. – Qiaochu Yuan Nov 27 '15 at 7:58