User:IssaRice/Linear algebra/Trace of matrix equals sum of eigenvalues

This is chapter 4 exercise 1.11 in Linear Algebra Done Wrong (p. 105).

Theorem. Let $A$ be a matrix. Then $t r a c e A = λ_{1} + \dots + λ_{n}$ , where $λ_{1}, \dots, λ_{n}$ are the eigenvalues of $A$ (counting multiplicities).

Proof. We already know that $det (A - λ I) = (λ_{1} - λ) \dots (λ_{n} - λ)$ . Multiplying out the right hand side, the $λ^{n - 1}$ term is $λ_{1} (- λ)^{n - 1} + λ_{2} (- λ)^{n - 1} + \dots + λ_{n} (- λ)^{n - 1} = (λ_{1} + \dots + λ_{n}) (- 1)^{n - 1} λ^{n - 1}$ .

Now consider equation 4.2 from the book (p. 89):

$det A = \sum_{σ \in P e r m (n)} a_{σ (1), 1} a_{σ (2), 2} \dots a_{σ (n), n} s i g n (σ)$

The matrix $A - λ I$ has the form:

$(\begin{matrix} a_{1, 1} - λ & * \\ ⋱ \\ a_{n, n} - λ \end{matrix})$

Taking the identity permutation we get the term $(a_{1, 1} - λ) \dots (a_{n, n} - λ)$ . All the rest of the permutations correspond to terms that can take at most $n - 2$ of the elements along the diagonal, so will result in a term (when multiplied out) that will have degree at most $n - 2$ (why $n - 2$ rather than $n - 1$ ? because if we take row k off in column j!=k, then when we get to column k, we can't take row j either, since a permutation can take at most one element from each row).

So we can write $det (A - λ I) = (a_{1, 1} - λ) \dots (a_{n, n} - λ) + q (λ)$ , where $q (λ)$ has degree at most $n - 2$ .

Now if we look at the $λ^{n - 1}$ term in $(a_{1, 1} - λ) \dots (a_{n, n} - λ) + q (λ)$ we get $(a_{1, 1} + \dots + a_{n, n}) (- 1)^{n - 1} λ^{n - 1}$ . This means that $λ_{1} + \dots + λ_{n} = a_{1, 1} + \dots + a_{n, n} = t r a c e A$ as required.

See also