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Exercise 2.4.10

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1.

Since

AB = I_{n}

is invertible, we have

A

and

B

is invertible by the previous exercise.

2.

We have that

AB = I_{n}

and

A

is invertible. So we can conclude that

A^{- 1} = A^{- 1} I_{n} = A^{- 1} AB = B .

3.

Let

T

is a mapping from

V

W

and

U

is a mapping from

W

V

with dim

W =

dim

V

. If

TU

be the identity mapping, then both

T

and

U

are invertible. Furthermore

T^{- 1} = U

To prove this we may pick bases $α$ of $V$ and $β$ of $W$ and set $A = {[T]}_{α}^{β}$ and $B = {[U]}_{β}^{α}$ . Now apply the above arguments we have that $A$ and $B$ is invertible, so are $T$ and $U$ by Theorem 2.18.

jephianlin

2011-06-27 00:00

Exercise 2.4.10

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