Sum of eigenvalues is the trace of the matrix

Question

I'm reading a proof of the sum of eigenvalues is the trace of the matrix. I understand that the $\lambda^{n-1}$ term of $p(\lambda) = (\lambda - \lambda_{1}) \cdots (\lambda - \lambda_{n})$ is $-(\lambda_{1} + \cdots + \lambda_{n})\lambda^{n-1}$ by expanding the polynomial. But I don't understand the argument of evaluating $\det(\lambda I - A)$ using permutations. Like literally this argument does not make much sense to me. I don't understand why and how we need to evaluate this determinant here.

Answer 1

Let us consider the case $n=3$ with an explanation without permutations.

You know that the expansion of a $3 \times 3$ determinant has six products of 3 entries coming from the 6 ways one has to :

(R) : select a single entry in each row and each column.

Let us list them in this way :

$$\begin{cases}A)&3 \text{ diagonal elements} & \left(\begin{smallmatrix}d&&\\&d&\\&&d \end{smallmatrix}\right)& \text{monomials } \lambda^3, -T\lambda^2, U\lambda, \text{V = constant}\\ B)&2 \text{ diagonal elements} &\text{ impossible !} \\ C)&1 \text{ diagonal element} &\left(\begin{smallmatrix}d&&\\& &*\\&*&\end{smallmatrix}\right), \left(\begin{smallmatrix}&&*\\&d&\\*&& \end{smallmatrix}\right),\left(\begin{smallmatrix}&*&\\*&&\\&&d \end{smallmatrix}\right) &\text{terms in } \lambda, \text{ and constants uniquely}\\ D)&\text{no diagonal element} &\left(\begin{smallmatrix}&*&\\& &*\\*&&\end{smallmatrix}\right), \ \left(\begin{smallmatrix}&&*\\*&&\\&*& \end{smallmatrix}\right)& \text{constant terms, i.e., without } \lambda\end{cases}$$

(where the last column gives the type of generated monomials when $d$ has the form $\lambda-a_{kk}$).

Comments :

Case A) generates a term in $\lambda^2$ (by expansion of $ddd=(\lambda-a_{11})(\lambda-a_{22})(\lambda-a_{33})$)

Case B) is impossible because when $2$ diagonal elements are already selected, there is no way to select a non-diagonal element complying with rule (R)).

Case C) generates products of entries having the form $c(\lambda-a_{kk})$ (where $c$ is a constant) ; no one provides a $k\lambda^2$.

Case D) generates constants.

Conclusion : the coefficient of $\lambda^2$ (the opposite of the trace) in the characteristic polynomial can only be issued from coefficient $-T$ in the first line.

Remark : We haven't considered the $\pm$ signs to be given to the different products ; in case A) in particular (but also in cases D)), the product is unchanged, whereas in case C), the different products must be multiplied by $-1$ (this is where the signature of permutations have their role to play...).

This proof for $n=3$ can be transformed in a straightforward way to a general $n$ (no separate way to get a $\lambda^{n-1}$ term).