Proofs in Linear Algebra via Topology

These topological arguments involve the same basic idea: it's often easy to prove things for a subset of matrices which are dense in the space of all matrices. Any "continuous" fact (e.g. the assertion that two continuous functions are equal) can be proven for all matrices by proving it for this dense subset.

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.