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Exercise 5.B.1

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Proof. (a) We will prove the contrapositive.

Suppose $I - T$ is not invertible. There exists non-zero $v \in V$ such that $Iv - Tv = 0$ , which implies $Tv = v$ . Applying $T$ to both sides of this last equation, yields $TTv = Tv = v$ . Continuing this, we see that for any positive integer $n$ , $T^{n} v = v$ . Therefore $T^{n}$ can never be $0$ .

Now suppose $T^{n} = 0$ . We have

\begin{aligned} (I - T) (I + T + \dots + T^{n - 1}) & = (I - T) \sum_{k = 0}^{n - 1} T^{k} \\ = \sum_{k = 0}^{n - 1} T^{k} - \sum_{k = 0}^{n - 1} T^{k + 1} \\ = I + \sum_{k = 1}^{n - 1} T^{k} - \sum_{k = 1}^{n} T^{k} \\ = I + \sum_{k = 1}^{n - 1} T^{k} - T^{n} - \sum_{k = 1}^{n - 1} T^{k} \\ = I - T^{n} \\ = I \end{aligned}

Therefore $I - T$ is an inverse of $I + T + \dots + T^{n - 1}$ , which implies the desired result.

(b) I wouldn’t. □

celiopassos

2017-10-06 00:00

Exercise 5.B.1

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