Exercise 10.A.1

Answers

Proof. Suppose $M (T, (v_{1}, \dots, v_{n}))$ is invertible. Then there exists an $n$ -by- $n$ matrix $A$ such that

I = M (T, (v_{1}, \dots, v_{n})) A .

Define $S \in L (V)$ such that

M (S, (v_{1}, \dots, v_{n})) = A .

Then

M (ST, (v_{1}, \dots, v_{n})) = M (S, (v_{1}, \dots, v_{n})) M (T, (v_{1}, \dots, v_{n})) = M (I, (v_{1}, \dots, v_{n})) .

The equation above shows that $ST = I$ . Exercise 10 in section 3D now implies that $T$ is invertible and $S = T^{- 1}$ .

Conversely, suppose $T$ is invertible. Then

I = M (T - 1 T, (v_{1}, \dots, v_{n})) = M (T^{- 1}, (v_{1}, \dots, v_{n})) M (T, (v_{1}, \dots, v_{n}))

and

I = M (TT - 1, (v_{1}, \dots, v_{n})) = M (T^{- 1}, (v_{1}, \dots, v_{n})) M (T^{- 1}, (v_{1}, \dots, v_{n})),

which shows that $M (T, (v_{1}, \dots, v_{n}))$ is invertible. □

celiopassos

2017-10-06 00:00