Exercise 9.A.16

Answers

Proof. Suppose $T$ is such that $T^{2} = - I$ . Then clearly $T$ does not have an eigenvalue. Then 9.19 implies that $\dim V$ is even.

Conversely, suppose $V$ has even dimension. Let $v_{1}, \dots, v_{n}, u_{1}, \dots, u_{n}$ be a basis of $V$ . Define $T \in L (V)$ by

T v_{j} = - u_{j}, T u_{j} = v_{j}

for each $j = 1, \dots, n$ . We have

T^{2} v_{j} = - T u_{j} = - v_{j} and T^{2} u_{j} = T v_{j} = - u_{j} .

Thus $T^{2} = - I$ . □

celiopassos

2017-10-06 00:00