Consider the case when the matrix $A$ is invertible.

Even if $A$ is not invertible, note that $A-\epsilon I$ is invertible matrix for sufficiently small $\epsilon$.
(See Problem Perturbation of a singular matrix is nonsingular for a proof of this fact.)

Take the limit $\epsilon \to 0$.

Proof.

We want to show that $|AB-\lambda I|=|BA-\lambda I|$, where $\lambda$ is an unknown and $I$ is the $n \times n$ identity matrix.

First let us consider the case when the matrix $A$ is invertible.
So suppose that $A$ is invertible.
Then we have
\begin{align*}
|AB-\lambda I| &=|A^{-1}(AB-\lambda I)A| =|BA-\lambda I|.
\end{align*}
Here we use the fact that $|A||A^{-1}|=|AA^{-1}|=|I|=1$.
Thus when $A$ is invertible, the claim is proved.

Now we consider the general case.
Whether or not $A$ is invertible, the matrix $A-\epsilon I$ is invertible for sufficiently small $\epsilon$.

(More precisely, if $\epsilon$ is smaller than absolute values of all nonzero eigenvalues of $A$. then $A-\epsilon I$ is invertible. See Problem Perturbation of a singular matrix is nonsingular for a proof.)

Then by the previous case, we have
\[ |(A-\epsilon I) B-\lambda I|=|B(A-\epsilon I)-\lambda I|.\]
Taking the limit $\epsilon \to 0$, we obtain
\[|AB-\lambda I|=|BA-\lambda I|.\]

Comment.

When you study hard linear algebra, you might have forgotten about calculus.
This problem suggests that for some problems in linear algebra, techniques from calculus (like limits) might help to solve them.

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