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Solution to Linear Algebra Hoffman & Kunze Chapter 6.8

Exercise 6.8.3

If $V$ is the space of all polynomials of degree less than or equal to $n$ over a field $F$, prove that the differentiation operator on $V$ is nilpotent.

Solution: This is clear. Let $D$ be the differentiation operator, then $\deg (Df)=\deg f-1$ for polynomial $f$ such that $\deg f\geqslant 1$. Hence we have$D^{\deg f+1}f=D(D^{\deg f}f)=Dc=0,$where $c$ is a constant number as $\deg(D^{\deg f}f)=0$.

If $\deg f=0$, then we also have $D^{\deg f+1}f=Dc=0$ since $f$ is a constant.

Therefore, $D^{n+1}$ is a zero operator on the space of all polynomials of degree less than or equal to $n$ over a field $F$.

Exercise 6.8.4

Solution:

(a) Let $S_i$ be the set of all vectors $\alpha$ in such that $(T-c_iI)^m\alpha=0$ for some $m>0$. It is clear that $W_i\subset S_i$. We would like to show $S_i\subset W_i$. Take any $\alpha\in S_i$. Then we have $(T-c_iI)^m\alpha=0$ for some $m>0$. We write$\alpha=\alpha_1+\cdots+\alpha_k$where $\alpha_j\in W_j$. Then$0=(T-c_iI)^m\alpha_1+\cdots+(T-c_iI)^m\alpha_k.$Since $W_j$ is invariant under $T$, we have $(T-c_iI)^m\alpha_j\in W_j$ for all $j$. But $V=W_1\oplus \cdots\oplus W_k$, we have $(T-c_iI)^m\alpha_j=0$ for all $j$.

If $i\ne j$, then $(T-c_iI)^m$ and $(T-c_jI)^{r_j}$ are relatively prime. Hence there exist polynomials $f_{ij}$ and $g_{ij}$ such that $f_{ij}(T)(T-c_iI)^m+g_{ij}(T-c_jI)^{r_j}=I$. Hence we have$\alpha_j=I\alpha_j=f_{ij}(T)(T-c_iI)^m\alpha_j+g_{ij}(T-c_jI)^{r_j}\alpha_j=0.$Here $(T-c_iI)^m\alpha_j=0$ as we proved above and $(T-c_jI)^{r_j}\alpha_j=0$ by the definition of $W_j$. Hence all $\alpha_j=0$ if $j\ne i$. It follows that $\alpha=\alpha_i\in W_i$. Part (a) is done.

(b) See the hint and Exercise 6.8.15. We remark that$\det(xI-(A-cI))=\det((x+c)I-A).$Hence the characteristic polynomial of $A-cI$ is equal to $f_A(x+c)$, where $f_A$ is the characteristic polynomial of $A$.

Exercise 6.8.5

Solution: We write $T=D+N$ as in Theorem 13. Since $D$ is diagonalizable, so is $g(D)$. Note that $D$ is a polynomial in $T$, so is $g(D)$. Thus $g(T)$ commutes with $g(D)$. It follows that $g(D)$ commutes with $g(T)-g(D)$. Therefore, by Theorem 13, it suffices to show that $g(T)-g(D)$ is nilpotent.

Since $T$ and $D$ commute, we can write $$g(T)-g(D)=(T-D)h(T,D)=Nh(T,D)$$for some polynomial $h$ in $T$ and $D$. Clearly, $N$ commutes with $h(T,D)$. Because $N$ is nilpotent, there exists $m>0$ such that $$(g(T)-g(D))^m=(Nh(T,D))^m=N^mh^m(T,D)=0.$$We are done.

Exercise 6.8.6

Solution: We prove it by contradiction. Suppose $T$ is diagonalizable and nilpotent. By Theorem 13, $T=0+T=T+0.$Clearly, $0$ is also diagonalizable and nilpotent. Moreover, $0$ and $T$ commute. Therefore, both $0+T$ and $T+0$ are the decomposition in the sense of Theorem 13. Since the decomposition is unique, we must have $T=0$ which contradicts with the condtion $\mathrm{rank}(T)=1$.

Exercise 6.8.7

Solution: Let $v\in V$ be nonzero. Consider a projection $E$ onto the subspace spanned by $v$. Then $E$ is diagonalizable and hence commutes with $T$. We have$TEv=ETv\Longrightarrow Tv=\lambda v$for some $\lambda\in F$. Hence $v$ is an characteristic vector of $T$. Therefore, all nonzero vectors are characteristic vectors of $T$.

Take a basis $v_1,\dots,v_n$, then we have $Tv_i=\lambda_i v_i$ for some $\lambda_i\in F$. We show that $\lambda_i=\lambda_j$ for $i\ne j$. Since $v_i+v_j$ is nonzero, there exists $\lambda\in F$ such that $T(v_i+v_j)=\lambda(v_i+v_j)$. Therefore, we have$\lambda(v_i+v_j)=Tv_i+Tv_j=\lambda_iv_i+\lambda_j v_j\iff (\lambda-\lambda_i)v_i+(\lambda-\lambda_j)v_j=0.$But $v_i,v_j$ is linearly independent, we have $\lambda=\lambda_i$ and $\lambda=\lambda_j$. It follows that $\lambda_i$ are all the same. Hence $T$ is a scalar multiple of identity matrix.

Exercise 6.8.8

Solution: The following formula is well-known:$T^n(B)=\sum_{k=0}^n (-1)^k \binom{n}{k} A^{n-k}BA^k.$It is can be proved by induction on $n$.

Suppose $A^m=0$, then if $n=2m$, we must have$\max\{2m-k,k\}\geqslant m.$Hence $A^{2m-k}=0$ or $A^k=0$. Thus $A^{2m-k}BA^{k}=0$ for all $k$. Hence $T^{2m}=0$ and $T$ is also nilpotent.

Exercise 6.8.9

Solution: Consider $$A=\begin{pmatrix}0&1 & 0 & 0\\ 0&0& 0 & 0\\0&0 & 0 & 1\\0&0 & 0 & 0\end{pmatrix}$$and $$B=\begin{pmatrix}0&0 & 0 & 0\\ 0&0& 0 & 0\\0&0 & 0 & 1\\0&0 & 0 & 0\end{pmatrix}.$$Then is clear that the characteristic polynomials of $A$ and $B$ are $x^4$ while the minimal polynomials are the same which is $x^2$. But $A$ and $B$ are not similar. Since $2=\mathrm{rank}(A)\ne \mathrm{rank}(B)=1.$

Exercise 6.8.10

Solution: See Lemma of page 263.

Exercise 6.8.11

Solution: The matrices $D$ and $N$ may not commute. Let $A=\begin{pmatrix}1&1\\ 0&2\end{pmatrix}$, then $D=\begin{pmatrix}1&0\\ 0&2\end{pmatrix}$ and $N\begin{pmatrix}0&1\\ 0&0\end{pmatrix}$. It is easy to check that $DN\ne ND$.

Exercise 6.8.12

Not clear what to say.

Exercise 6.8.13

Let $T$ be a linear operator on $V$ with minimal polynomial of the form $p^n$, where $p$ is irreducible over the scalar field. Show that there is a vector $\alpha$ in $V$ such that the $T$-annihilator of $\alpha$ is $p^n$.

Solution: (Let us assume $V$ is finite-dimensional!) Let $v_1,\dots,v_m$ be a basis of $V$. Then for each $v_i$, the $T$-annihilator of $v_i$ is a divisor of $p^n$ and hence has the form $p^{k_i}$ for some positive integer $k_i$. Clearly, $k_i\leqslant n$.

We consider the set $\{k_1,\dots,k_m\}$. If $$\max\{k_1,\dots,k_m\}<n,$$then $k_i\leqslant n-1$ for all $i$. Hence $p(T)^{n-1}v_i=0$ for all $i$. Thus $P(T)^{n-1}v=0$ for all $v\in V$ as $v_1,\dots,v_m$ is a basis of $V$. Then the minimal polynomial of $T$ cannot be $p^n$, we get a contradiction. Therefore, we must have $\max\{k_1,\dots,k_m\}=n$. That it, $k_i=n$ for some $i$. We are done.

Exercise 6.8.14

Solution: Here, we use the notation from Theorem 12. By Theorem 12 (Primary Decomposition Theorem) and Exercise 6.8.13, we can take $\alpha_i\in W_i$ such that the $T$-annihilator of $\alpha_i$ is $p_i^{r_i}$.

Let $\alpha=\alpha_1+\cdots+\alpha_k$. Suppose $f(T)\alpha=0$ for some polynomial $f$. Then we have$0=f(T)\alpha=f(T)\alpha_1+\cdots+f(T)\alpha_k.$Since $W_i$ is invariant under $T$, we have $f(T)\alpha_i\in W_i$. Note that$V=W_1\oplus \cdots\oplus W_k.$We have $f(T)\alpha_i=0$ for all $i$. Since the $T$-annihilator of $\alpha_i$ is $p_i^{r_i}$, we conclude that $p_i^{r_i}|f$ for all $i$. Hence $p|f$. Therefore, the $T$-annihilator of $\alpha$ is $p$.

Exercise 6.8.15

If $N$ is a nilpotent linear operator on an $n$-dimensional vector space $V$, then the characteristic polynomial for $N$ is $x^n$.

Solution: Let us work on algebraically closed field. In general, we can enlarge the filed $F$ to its algebraic closure. So this argument works even $F$ is not algebraically closed. But I don't like this argument. This statement is clear if we know the Jordan form.

Since $N$ is nilpotent, we have $N^m=0$ for some positive integer $m$. Hence the minimal polynomial of $N$ is a monic divisor of $x^m$ and so has the form $x^k$. Recall Theorem 3 of page 193, the characteristic polynomial and the minimal polynomial has the same roots. Hence characteristic polynomial must be of the form $x^\ell$. But we also know that the degree of characteristic polynomial is exactly $\dim V=n$. Hence the characteristic polynomial for $N$ is $x^n$.