TITLE: Formal proof of $\det(I + tA) = \prod\limits_{i=1}^n (1 + t\lambda_i)$
QUESTION [6 upvotes]: I'm looking for a formal proof for:
$$\det(I + tA) = \prod\limits_{i=1}^n (1 + t\lambda_i).$$
I'm very new to matrix theory therefore please forgive me if you find this elementary. Your help in this matter is appreciated.
Thanks,
Edit:
$A$ is a symmetric $n\times n$ matrix and $\lambda_i$ are the eigenvalues of $A$.

REPLY [3 votes]: Since you are using the term symmetric, I am going to assume that the matrix in question is real. There is no real need to invoke triangularizability or any machinery. We'll simply work with basic determinant properties. 
This holds generally for any real matrix with $n$ real eigenvalues, a symmetric matrix is simply a matrix which happens to satisfy this criteria.
Suppose that $A$ is $n\times n$ and let $p(t) = \det(tI - A)$ denote the characteristic polynomial of $A$. By assumption, $p(t)$ splits as
$$p(t) = \prod_{i=1}^n(t - \lambda_i)$$
We can manipulate the determinant as
$$\det(tI-A) = t^n\det\left(I-\frac{1}{t}A\right)$$
Now letting $s = \frac{1}{t}$, notice that
$$\det(I - sA) = s^{n}p(s^{-1})=s^n\prod_{i=1}^n\left(\frac{1}{s} - \lambda_i\right)$$
which upon splitting the $s^n$ term over the terms of the product, becomes
$$\det(I - sA)=\prod_{i=1}^n\left(1 - s\lambda_i\right)$$