Formal proof of $\det(I + tA) = \prod\limits_{i=1}^n (1 + t\lambda_i)$ 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 $A$. 
 A: This is true for every complex square matrix.
It does not need to be symmetric.
Since $\mathbb{C}$ is algebraically closed, there exists $P$ invertible such that $P^{-1}AP$ is upper triangular with diagonal $(\lambda_1,\ldots,\lambda_n)$ (the Schur decomposition).
Now $P^{-1}(I+tA)P$ is upper triangular with diagonal $(1+t\lambda_1,\ldots,1+t\lambda_n)$. 
So the formula follows (recall that $\det(BC)=\det(CB)$).
Note: the more I think about your question, the more I think EuYu'S answer is the right one. After all, the eigenvalues are the roots of the characteristic polynomial $p_A(\lambda)=\det(\lambda A-I)$ by definition. So your identity is actually essentially trivial. It is clearly true for $t=0$. Then factor out $t^n$ when $t\neq 0$, and it boils down to $p_A(-1/t)=\prod(-1/t-\lambda_i)$, which is true by definition of the eigenvalues (when $p_A$ splits).
A: Over the complex numbers, each matrix $A$ is equivalent to a matrix $M$ in a Jordan normal form, which is an almost diagonal matrix and the diagonal elements are just the eigenvalues $\lambda_i$ of $A$. Being equivalent means that there is an invertible matrix $B$ (its columns contain the coordinates of the corresponding eigenvectors) such that
$$A=BMB^{-1}$$
Then $\det A=1/\det B\cdot\det M\cdot \det B=\det M=\prod_i(\lambda_i)$.
Now use that $I+tA=B(I+tM)B^{-1}$.
A: 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
$$\det(I - sA)=\prod_{i=1}^n\left(1 - s\lambda_i\right)$$
