linear algebra - Help with understanding the proof for: $AB$ and $BA$ have the same characteristic polynomial (for square complex matrices)

I saw many proofs but they all use advanced techniques and are impossible to understand.
I'm looking for a proof that $AB$ and $BA$ have the same characteristic polynomial for any square matrices $A$ and $B$ over $\mathbb C$.

It's really easy when dealing with invertible matrices, but hard to prove for singular matrices.

I found several solutions that I could not understand:

This solution says

it is not too difficult to show that $AB$, and $BA$ have the same characteristic polynomial ... If the matrices are in $M_n(\mathbb C)$, you use the fact that $GL_n(\mathbb C)$ is dense in $M_n(\mathbb C)$ and the continuity of the function which maps a matrix to its characteristic polynomial. There are at least 5 other ways to proceed

I've bolded every term that I am not familiar with.

This solution I could not understand as well (it uses the limit definition when $\lambda$ approaches zero but I hardly understand how that solves the issue).

I'm looking for a simpler solution using more basic linear algebra.

Answer

My preferred proof is as follows: it suffices to note that for any $\lambda \neq 0$, we have by Sylvester's determinant identity that
$$
\det(\lambda I - AB) = \lambda^n\det\left(I - \frac 1{\lambda}AB\right) =

\lambda^n\det\left(I - \frac 1{\lambda}BA\right) =
\det(\lambda I - BA)
$$
Thus, the two polynomials on $\lambda$ are identical for all $\lambda \neq 0$. We may conclude that the polynomials are exactly the same.

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Here's the gist of your proof: for any $A,B$, there are sequences $(A_n)_{n \in \Bbb N}, (B_n)_{n \in \Bbb N}$ of invertible matrices such that $A_n \to A$ and $B_n \to B$ (the existence of such sequences is equivalent to density). We note that by the continuity of the function that maps a matrix to its characteristic polynomial (and by the continuity of matrix multiplication), we have
$$
\det(\lambda I - AB) = \lim_{n \to \infty} \det(\lambda I - A_nB_n)\\

\det(\lambda I - BA) = \lim_{n \to \infty} \det(\lambda I - B_nA_n)
$$
However, because the statement holds for invertible matrices, these two sequences are exactly the same. So, they have the same limit. So, $\det(\lambda I - AB) = \det(\lambda I - BA)$, which is what we wanted.