Exercise 5.C.7

Proof. Suppose $\lambda$ appears $n$ times on the diagonal of $A$. Let $v_1,\dots ,v_n,u_1,\dots ,u_m$ be a basis composed of the same basis vectors chosen to represent $A$ (possibly in different order) such that the $v$’s are eigenvectors of $\lambda$ and the remaining are corresponding eigenvectors to ${\lambda }_1,\dots ,{\lambda }_m$. Note that ${\lambda }_k\ne \lambda$ for each $k$, because we have $n$ $v$’s.

Let $v\in E(\lambda ,T)$. There are $a_1,\dots ,a_n,c_1,\dots ,c_m\in 𝔽$ such that

Then

Since ${\lambda }_k-\lambda \ne 0$, all the $c$’s are 0, implying that $v\in \mathrm{span}<!--\; FUNCTION\; APPLICATION\; -->(v_1,\dots <!--\; FUNCTION\; APPLICATION\; -->,v_n)$. Hence $E(\lambda ,T)\subset \mathrm{span}<!--\; FUNCTION\; APPLICATION\; -->(v_1,\dots <!--\; FUNCTION\; APPLICATION\; -->,v_n)$. The inclusion in the other directions is obvious, thus it becomes an equality. Moreover, $v_1,\dots ,v_n$ is linearly independent and, hence, a basis of $E(\lambda ,T)$. Thus $\dim <!--\; FUNCTION\; APPLICATION\; -->E(\lambda ,T)=n$. □