Show that the determinant of $A$ is equal to the product of its eigenvalues 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.
Suppose that $\lambda_1, \ldots, \lambda_n$ are the eigenvalues of $A$. Then the $\lambda$s are also the roots of the characteristic polynomial, i.e.
$$\begin{array}{rcl} \det (A-\lambda I)=p(\lambda)&amp;=&amp;(-1)^n (\lambda - \lambda_1 )(\lambda - \lambda_2)\cdots (\lambda - \lambda_n) \\ &amp;=&amp;(-1) (\lambda - \lambda_1 )(-1)(\lambda - \lambda_2)\cdots (-1)(\lambda - \lambda_n) \\ &amp;=&amp;(\lambda_1 - \lambda )(\lambda_2 - \lambda)\cdots (\lambda_n - \lambda)
\end{array}$$
The first equality follows from the factorization of a polynomial given its roots; the leading (highest degree) coefficient $(-1)^n$ can be obtained by expanding the determinant along the diagonal.
Now, by setting $\lambda$ to zero (simply because it is a variable) we get on the left side $\det(A)$, and on the right side $\lambda_1 \lambda_2\cdots\lambda_n$, that is, we indeed obtain the desired result
$$ \det(A) = \lambda_1 \lambda_2\cdots\lambda_n$$
So the determinant of the matrix is equal to the product of its eigenvalues.
I think this is right...

Write A =
$$\begin{pmatrix}a_{11} &amp; a_{12}&amp;\cdots&amp;a_{1n}\\\ \vdots &amp; \vdots&amp;&amp;\vdots\\\ a_{n1}&amp; a_{n2}&amp;\cdots&amp; a_{nn} \end{pmatrix}$$

Let the n eigenvalues of A be $\lambda_1, \cdots , \lambda_n$. Finally, denote the characteristic polynomial of A by
$p(\lambda) = |\lambda I − A| = \lambda_n + c_{n−1}\lambda{n−1} + \cdots + c_1λ + c_0$. 

Note that since the eigenvalues of A are the zeros
of $p(\lambda)$, this implies that $p(\lambda)$ can be factorised as $p(\lambda) = (\lambda − \lambda_1)\cdots(\lambda − \lambda_n)$.
Consider the constant term of $p(λ), c_0$. The constant term of $p(\lambda)$ is given by $p(0)$, which can be
calculated in two ways:

Firstly, $p(0) = (0 − λ_1)\cdots(0 − λ_n) = (−1)^nλ_1 \cdots λ_n$. 

Secondly, $p(0) = |0I − A| = | − A| = (−1)^n
|A|$.

Therefore $c_0 = (−1)^nλ_1 \cdots λ_n = (−1)^n
|A|$, and so  $λ_1 \cdots λ_n = |A|$. 

That is, the product of the n
eigenvalues of A is the determinant of A.