Exercise 5.C.1

Proof. If $\mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->T=\{0\}$ (because it implies surjectivity) or $\mathrm{range}<!--\; FUNCTION\; APPLICATION\; -->T=\{0\}$ the result is obvious. Assume both contain non-zero vectors.

Since $\mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->T\ne \{0\}$, we have that $0$ is an eigenvalue of $T$ and that $E(0,T)=\mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->T$. Let ${\lambda }_1,\dots ,{\lambda }_m$ denote the other distinct eigenvalues of $T$. Now 5.41 implies that $V=\mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->T\oplus E({\lambda }_1,T)\oplus \cdots \oplus E({\lambda }_m,T)$. We will prove that $\mathrm{range}<!--\; FUNCTION\; APPLICATION\; -->T=E({\lambda }_1,T)\oplus \cdots \oplus E({\lambda }_m,T)$.

Suppose $v\in E({\lambda }_1,T)\oplus \cdots \oplus E({\lambda }_m,T)$. Then $v=v_1+\cdots +v_m$ for some $v_1,\dots ,v_m$, where each $v_j\in E({\lambda }_j,T)$. Moreover, $v=T(\frac{1}{{\lambda }_1}{v}_1)+\cdots +T(\frac{1}{{\lambda }_m}{v}_m)$, which implies that $v\in \mathrm{range}<!--\; FUNCTION\; APPLICATION\; -->(T)$. Hence

For the inclusion in the other direction, suppose $v\in \mathrm{range}<!--\; FUNCTION\; APPLICATION\; -->T$. Note that $\mathrm{range}<!--\; FUNCTION\; APPLICATION\; -->T$ stays the same when we restrict $T$ to $E({\lambda }_1,T)\oplus \cdots \oplus E({\lambda }_m,T)$. Then $v=T(v_1+\cdots +v_m)$ for some $v_1,\dots ,v_m$, where each $v_j\in E({\lambda }_j,T)$. Therefore $v={\lambda }_1{v}_1+\cdots +{\lambda }_m{v}_m$ and, because ${\lambda }_j{v}_j\in E({\lambda }_j,T)$ for each $j$, this proves that $v\in E({\lambda }_1,T)\oplus \cdots \oplus E({\lambda }_m,T)$. Thus $\mathrm{range}<!--\; FUNCTION\; APPLICATION\; -->T\subset E({\lambda }_1,T)\oplus \cdots \oplus E({\lambda }_m,T)$, completing the proof. □