Prove that $p_{AB}(\lambda) = p_{BA}(\lambda)$

Question

Characteristic polynomial of square matrix $A \in \mathbb{R}^{n \times n}$ is defined like this $p_{A}(\lambda) = \det(\lambda I - A)$. I have to prove the following:

$$p_{AB}(\lambda) = p_{BA}(\lambda)$$

I know that it's easy to prove if just one of $A$ or $B$ is non-singular by using $AB = A^{-1}(AB)A$ and $p_{PAP^{-1}}(\lambda) = p_{A}$.

But I struggle to prove it for the case of both matrices being singular. I am trying to compare coefficients of $\lambda^i$ from both sides. And I see that it's true for $\lambda^n$, $\lambda^0$ and $\lambda^{n-1}$ since the coefficients for them are $1$, $(-1)^n \det(AB)$ or $(-1)^n \det(BA)$ which is the same and $-trAB$ or $-trBA$ which is also the same.

But for any $\lambda^i$ the formula of coefficient is $(-1)^{n-k} (\sum _{i_1<i_2...<i_k} \det R_{i_1,i_2,...,i_k})$ where $R_{i_1,i_2,...,i_k}$ is $AB$ or $BA$ with crossed $i_1,...,i_k$ rows and columns.

I'd like to add that I am just at the beginning of studying linear algebra so I'am not aware of fields, ranks or any concept past determinants.

I see it from this perspective: If $\lambda\neq 0$ is an eigenvalue of $AB$, then $B$ maps the generalized eigenspace of $AB$ corresponding to $\lambda$ bijectively to that of $BA$. Hence, the sizes of the algebraic eigenspaces of $AB$ and $BA$ are the same for any eigenvalue $\lambda\neq 0$. As they all (including that for zero) should add up to $n$, that statement also holds for $\lambda = 0$. — amsmath, Sep 01 '19 at 20:40
Look here. There are much more in Related field in the link above. — A.Γ., Sep 01 '19 at 20:48
@amsmath, I don't get the answer here either, not aware of the "density" concept — Назар Петровский, Sep 01 '19 at 20:51
@A.Γ. Thank you. The answers there contain really nice arguments. — amsmath, Sep 01 '19 at 20:51

Prove that $p_{AB}(\lambda) = p_{BA}(\lambda)$

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