Exercise 4.1.11

Proof. First, recall the binomial theorem, the coefficient of ${\lambda }^{n-1}$ in $\det <!--\; FUNCTION\; APPLICATION\; -->(A-\mathit{\lambda I})=({\lambda }_1-\lambda )({\lambda }_2-\lambda )...({\lambda }_n-\lambda )$ is $C({\lambda }^{n-1})={(-1)}^{n-1}({\lambda }_1+{\lambda }_2+...+{\lambda }_n)$. Because to get the term ${\lambda }^{n-1}$, we need to pick $-\lambda n$ times from the total $n$ factors ${\lambda }_i-\lambda$. Then the last one pick is ${\lambda }_j$ from the factor whose $-\lambda$ is not picked. The resulting term is then ${(-1)}^{n-1}{\lambda }_j{\lambda }^{n-1}$. There are $n$ combinations and the sum of each term is $C({\lambda }^{n-1}){\lambda }^{n-1}$.

Then, we show $\det <!--\; FUNCTION\; APPLICATION\; -->(A-\mathit{\lambda I})$ can be represented as

That is to say in $\det <!--\; FUNCTION\; APPLICATION\; -->(A-\mathit{\lambda I})$, the term ${\lambda }^{n-1}$ are all from $(a_{1,1}-\lambda )(a_{2,2}-\lambda )...(a_{n,n}-\lambda )$. This holds because $\lambda$ only appears on the diagonal of $A-\mathit{\lambda I}$. Using the formal definition of determinant, if we pick $n-1$ diagonal term with $\lambda$, then the last pick must also be on the diagonal. There is no other way to generate ${\lambda }^{n-1}$. Thus $q(\lambda )$ is a polynomial of degree at most $n-2$. Then we know the coefficient of ${\lambda }^{n-1}$ also equals to $C({\lambda }^{n-1})={(-1)}^{n-1}(a_{1,1}+a_{2,2}+...+a_{n,n}).$

The coefficients derived from two different ways are identical, so we have ${\sum <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^n{a}_{i,i}={\sum }_{i=1}^n{\lambda }_i$, namely, the trace a matrix equals the sum of eigenvalues. □