When the problem says "repeated according to multiplicities" does that mean there is one value of lambda that is repeated $n$ times? I'm confused on why the determinant of $A$ must be the product of the $n$ eigenvalues of $A$. Couldn't lambda be any value? I only see how the determinant of $A$ would equal the $n$ eigenvalues if lambda is 0. Why is this result true when complex eigenvalues are considered?
Prove that the determinant of an $n × n$ matrix A is the product of the eigenvalues (counted according to their algebraic multiplicities). Hint: Write the characteristic polynomial as $p(\lambda) = (\lambda_1 − \lambda)(\lambda_2 − \lambda)· · ·(\lambda_n − \lambda)$.
Solution: If the eigenvalues of $A$ are $\lambda_1, . . . , \lambda_n$ (counted with algebraic multiplicity), then as the hint says, the characteristic polynomial of $A$ is $det(A − \lambda I) = (\lambda_1 − > \lambda)(\lambda_2 − \lambda)...(\lambda_n − \lambda)$. Plugging in $\lambda = 0$ yields $det A = \lambda_1\lambda_2... \lambda_n$.