Exercise 5.C.3

Proof. Obviously (a) implies (b) and, by 1.45, it also implies (c).

Suppose (b) holds. Combining 3.22 and 2.43, we have

Therefore $\dim <!--\; FUNCTION\; APPLICATION\; -->(\mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->T\cap \mathrm{range}<!--\; FUNCTION\; APPLICATION\; -->T)=0$, implying (c) is true.

Suppose (c) holds. By 1.45, $\mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->T+\mathrm{range}<!--\; FUNCTION\; APPLICATION\; -->T$ is a direct sum. Then, by 2.43 and 3.22, we have

Since $\mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->T\oplus \mathrm{range}<!--\; FUNCTION\; APPLICATION\; -->T$ is a subspace of $V$, it follows that $\mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->T\oplus \mathrm{range}<!--\; FUNCTION\; APPLICATION\; -->T=V$, implying (a) and completing the proof. □