## The determinant and trace of a matrix are the product and sum of its eigenvalues, respectively

Let $A \in \mathsf{Mat}_n(F)$ be a matrix over a field $F$. Prove that the constant term of the characteristic polynomial $c_A(x)$ is $(-1)^n \mathsf{det}(A)$ and that the $x^{n-1}$ term is the negative of the sum of the diagonal entries of $A$ (called the trace of $A$). Deduce that $\mathsf{det}(A)$ is the product of the eigenvalues of $A$, and that $\mathsf{tr}(A)$ (the trace of $A$) is the sum of the eigenvalues of $A$.

Recall that $c_A(x) = \mathsf{det}(xI - A)$ $= \mathsf{det}([\delta_{i,j}x - a_{i,j}])$, where $\delta_{i,j}$ is 1 if $i = j$ and 0 otherwise. So $c_A(x) = \sum_{\sigma \in S_n} \varepsilon{\sigma} \prod_{i=1}^n (\delta_{\sigma(i),i} x - a_{\sigma(i), i})$.

Now the constant term of $c_A(x)$ is merely $c_A(0) = \sum_{\sigma \in S_n} \varepsilon{\sigma} \prod_{i=1}^n (-a_{\sigma(i), i})$ $= \sum_{\sigma \in S_n} \varepsilon{\sigma} (-1)^n \prod_{i=1}^n a_{\sigma(i), i}$ $= (-1)^n \mathsf{det}(A)$.

Consider now the polynomial $p_\sigma(x) = \prod_{i=1}^n (\delta_{\sigma(i), i}x - a_{\sigma(i), i})$. Note in particular that the degree of $p_\sigma$ is just the number of elements in $\{1,\ldots,n\}$ which are fixed by $\sigma$. No $\sigma$ can fix $n-1$ elements, so that except for $\sigma = 1$, the polynomial $p_\sigma$ does not contribute to the $x^{n-1}$ term of $c_A(x)$. That is, the $x^{n-1}$ term of $c_A(x)$ is just the $x^{n-1}$ term of $p_1(x) = \prod_{i=1}^n (x - a_{i,i})$, which is just $- \sum_{i=1}^n a_{i,i}$ as desired.

Now recall that the eigenvalues of $A$ are (by definition) the roots of $c_A(x)$. It then follows that the constant term of $c_A(x)$ is $(-1)^n$ times the product of the eigenvalues and that the $x^{n-1}$ term is the negative of the sum of the eigenvalues.