Exercise 8.A.3

Proof. We will prove $\mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->{(T-\mathit{\lambda I})}^n=\mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->{\left(T^{-1}-\frac{1}{\lambda }I\right)}^n$ for all nonnegative integers $n$ by induction on $n$.

It is easy to check that $\mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->(T-\mathit{\lambda I})=\mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->\left(T^{-1}-\frac{1}{\lambda }I\right)$ (see Exercise 9 in section 5C). Let $n>1$ and assume the result holds for all nonnegative integers less than $n$. Suppose $v\in \mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->{(T-\mathit{\lambda I})}^n$. Then

By the induction hypothesis

Thus

where the second equality follows from Theorem 1 in Chapter 5 notes.

Therefore

But

Hence

and so

which shows that $v\in \mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->(T^{-1}-\frac{1}{\lambda }I)$. Therefore $\mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->{(T-\mathit{\lambda I})}^n\subset \mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->{\left(T^{-1}-\frac{1}{\lambda }I\right)}^n$. To prove the inclusion in the other direction, it suffices to repeat the same thing replacing $\left(T-\mathit{\lambda I}\right)$ with $\left(T^{-1}-\frac{1}{\lambda }I\right)$ and vice versa.

Now, by 8.11, we have

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