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.

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