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## Solution to Linear Algebra Hoffman & Kunze Chapter 7.5

#### Exercise 7.5.1

If $N$ is a nilpotent linear operator on $V$, show that for any polynomial $f$ the semi-simple part of $f(N)$ is a scalar multiple of the identity operator ($F$ is a subfield of $\C$).

Solution: Let $$f(x)=\sum_{k=0}^n a_kx^k,$$where $a_k\in F$. Let $$g(x)=\sum_{k=1}^n a_k x^{k-1}.$$Then $f(x)=a_0+xg(x)$. Hence $f(N)=a_0I+Ng(N)$. Clearly, $a_0I$ is semi-simple.

Note that $N$ is nilpotent, hence $N^m=0$ for some $m>0$. Since  $N$ and $g(N)$ commute, we have$(Ng(N))^m=N^m(g(N))^m=0.$Hence $Ng(N)$ is nilpotent. It is  also clear that $a_0I$ commutes with $Ng(N)$. Therefore, we get the desired decompostion $f(N)=a_0I+Ng(N)$ as in Theorem 13 of page 267, where $a_0I$ is the semi-simple part and $Ng(N)$ is the nilpotent part. We are done.

#### Exercise 7.5.2

Let $F$ be a subfield of the complex numbers, $V$ a finite-dimensional vector space over $F$, and $T$ a semi-simple linear operator on $V$. If $f$ is any polynomial over $F$, prove that $f(T)$ is semi-simple.

Solution: Since $F$ is a subfield of the complex numbers, by Corollary of page 265, $T$ is diagonalizable over $\C$. Hence $f(T)$ is also diagonalizable over $\C$. Again by Corollary of page 265, $f(T)$ is semi-simple.

#### Exercise 7.5.3

Let $T$ be a linear operator on a finite-dimensional space over a subfield of $\C$. Prove that $T$ is semi-simple if and only if the following is true: If $f$ is a polynomial and $f(T)$ is nilpotent, then $f(T)=0$.

Solution: "$\Leftarrow$" Consider the decomposition $T=S+N$ in Theorem 13, where $S$ is semi-simple. Moreover, $N$ is equal to $f(T)$ for some polynomial $f$ and is nilpotent. By assumption, we have $N=0$. Hence $T=S$ is semi-simple.

"$\Rightarrow$" Suppose $f$ is a polynomial and $f(T)$ is nilpotent. Since $T$ is semi-simple, by Exercise 7.5.2, $f(T)$ is semi-simple. Hence $f(T)$ is both semi-simple and nilpotent. Hence, both$$f(T)=0+f(T), \quad f(T)=f(T)+0$$are  the decompositions of $f(T)$ in the sense of Theorem 13, where the first operator is semi-simple and the second one is nilpotent. By the uniqueness of Theorem 13, we must have $f(T)=0$.