TITLE: Show that the determinant of $A$ is equal to the product of its eigenvalues
QUESTION [135 upvotes]: Show that the determinant of a matrix $A$ is equal to the product of its eigenvalues $\lambda_i$.

So I'm having a tough time figuring this one out. I know that I have to work with the characteristic polynomial of the matrix $\det(A-\lambda I)$. But, when considering an $n \times n$ matrix, I do not know how to work out the proof. Should I just use the determinant formula for any $n \times n$ matrix? I'm guessing not, because that is quite complicated. Any insights would be great.

REPLY [2 votes]: Instead of assuming that the matrix is diagonalisable, as done in some of the previous answers, we can use the Jordan form.
That is, every matrix has an associated Jordan form through a similarity transformation:
$$J=M^{-1}AM$$
for a certain invertible $M$. Since $J$ is triangular, its determinant is simply the product of its diagonal entries, which also happen to be the eigenvalues of $A$. That is,
$$\det(J)=\prod_i{\lambda_i(A)}$$
Where $\lambda_i(A)$ denotes the $i$th eigenvalue of $A$. 
Computing the determinants, we have
$$\det(J)=\det(M^{-1})\det(A)\det(M)$$
The determinants of $M$ and $M^{-1}$ cancel to give $1$, and so
$$\det(J)=\det(A)$$
Combining our second and fourth equations we have the result:
$$\det(A)=\prod_i{\lambda_i(A)}=\lambda_1\lambda_2...\lambda_n$$
.
TL;DR:
$\det(A)=\det(M^{-1})\det(J)\det(M)$
$\implies \det(A) = \det(J)$.
And since $J$ is triangular and has eigenvalues along its diagonal, $\det(A)=\det(J)=\lambda_1\lambda_2...\lambda_n$.