Exercise 5.C.2

Proof. We give a counterexample. Let $V={ℝ}^2$ and $T(x,y)=(-y,x)$. $T$ is invertible, thus $V=\mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->T\oplus \mathrm{range}<!--\; FUNCTION\; APPLICATION\; -->T$. But, as shown in Example 5.8, $T$ has no eigenvalues, therefore $T$ is not diagonalizable.

Note: If we let $V$ be infinite-dimensional, then any invertible operator $T\in L(V)$ will prove it to be wrong, because we will have $V=\mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->T\oplus \mathrm{range}<!--\; FUNCTION\; APPLICATION\; -->T$, but $T$ being diagonalizable requires $V$ to be finite-dimensional. □