I saw these notes (page 5, question 6) on proving that
$$\det(A) = \lambda_1 \cdots \lambda_n$$
where $\lambda_1, \dots, \lambda_n$ are all the eigenvalues of $A \in \mathbb{R}^{n \times n}$. I am pasting the proof below for easier reference.
I followed the proof up until the point it set $\lambda = 0$. I don't understand the logic (paraphrased based on my interpretation) "because $\lambda$ is an abstract variable, we can set it to $0$." Doesn't setting $\lambda = 0$ implicitly mean that $\lambda = 0$ is an eigenvalue of $A$?
There is abuse by notation above. $\lambda_i$ in the proof are the eigenvalues; $\lambda$ is a parameter. Notice that when $\lambda = \lambda_i$ you get $\det (A-\lambda_iI) = 0$, meaning $\lambda_i$ is an eigenvalue indeed. A general case may involve $\lambda_i = 0$, and it is still true.