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## Chapter 8 Exercise A

1. Solution: Since
$$T^2(w, z) = T(z, 0) = (0, 0),$$ it follows that $G(0, T) = V$. Therefore every vector in $\mathbb{C}^2$ is a generalized eigenvector of $T$.

2. Solution: The eigenvalues of $T$ are $i$ and $-i$. Since $\mathbb{C}^2$ has dimension $2$, the generalized eigenspaces are the eigenspaces themselves.

3. Solution: We will prove $\operatorname{null} (T – \lambda I)^n = \operatorname{null} \left(T^{-1} – \dfrac{1}{\lambda} I\right)^n$ for all nonnegative integers $n$ by induction on $n$.

It is easy to check that $\operatorname{null} (T – \lambda I) = \operatorname{null} \left(T^{-1} – \dfrac{1}{\lambda} I\right)$ (see Exercise 9 in section 5C). Let $n > 1$ and assume the result holds for all nonnegative integers less than $n$.

Suppose $v \in \operatorname{null}(T – \lambda I)^n$. Then
$$(T – \lambda I)v \in \operatorname{null}\left(T – \lambda I\right)^{n-1}.$$ By the induction hypothesis $$(T – \lambda I)v \in \operatorname{null}\left(T^{-1} – \frac{1}{\lambda} I\right)^{n-1}.$$ Thus $$0 = \left(T^{-1} – \frac{1}{\lambda} I\right)^{n-1}(T – \lambda I)v = (T – \lambda I)\left(T^{-1} – \frac{1}{\lambda} I\right)^{n-1}v,$$ where the second equality follows from Theorem 1 below.

Theorem 1. Suppose $T \in \mathcal{L}(V)$ is invertible and $p, q \in \mathcal{P}(\mathbb{F})$. Then $p(T^{-1}) q(T) = q(T) p(T^{-1})$.

Proof. The key idea used here is that $T$ commutes with $T^{-1}$, even when raised to different powers.
Suppose $p(z) = \sum_{j=0}^m a_j z^j$ and $q(z) = \sum_{k=0}^n b_k z^k$ for $z \in \mathbb{F}$. Then
\begin{aligned} p\left(T\right)q\left(T^{-1}\right) &= \left(\sum_{j=0}^m a_j T^j\right)\left(\sum_{k=0}^n b_k \left(T^{-1}\right)^k\right)\\ &= \sum_{j=0}^m \sum_{k=0}^n a_j b_k T^j \left(T^{-1}\right)^k\\ &= \sum_{j=0}^m \sum_{k=0}^n b_k a_j \left(T^{-1}\right)^k T^j\\ &= \sum_{k=0}^n \sum_{j=0}^m b_k a_j \left(T^{-1}\right)^k T^j\\ &= \left(\sum_{k=0}^n b_k \left(T^{-1}\right)^k\right)\left(\sum_{j=0}^m a_j T^j\right)\\ &= q\left(T^{-1}\right)p\left(T\right) \end{aligned}.

Therefore $$\left(T^{-1} – \frac{1}{\lambda}I\right)^{n-1}v \in \operatorname{null} (T – \lambda I).$$ But
$$\operatorname{null} (T – \lambda I) = \operatorname{null} \left(T^{-1} – \frac{1}{\lambda} I\right).$$ Hence
$$\left(T^{-1} – \frac{1}{\lambda}I\right)^{n-1}v \in \operatorname{null} \left(T^{-1} – \frac{1}{\lambda} I\right)$$ and so
$$0 = \left(T^{-1} – \frac{1}{\lambda} I\right) \left(T^{-1} – \frac{1}{\lambda}I\right)^{n-1}v = \left(T^{-1} – \frac{1}{\lambda}I\right)^n v,$$ which shows that $v \in \operatorname{null} (T^{-1} – \frac{1}{\lambda}I)$. Therefore $$\operatorname{null} (T – \lambda I)^n \subset \operatorname{null} \left(T^{-1} – \frac{1}{\lambda} I\right)^n.$$ To prove the inclusion in the other direction, it suffices to repeat the same thing replacing $\left(T – \lambda I\right)$ with $\left(T^{-1} – \frac{1}{\lambda}I\right)$ and vice versa.

Now, by 8.11, we have
$$G(\lambda, T) = \operatorname{null}(T – \lambda I)^{\operatorname{dim} V} = \operatorname{null}\left(T^{-1} – \frac{1}{\lambda} I\right)^{\operatorname{dim} V} = G\left(\frac{1}{\lambda}, T^{-1}\right).$$

4. Solution: Suppose $v \in G(\alpha, T) \cap G(\beta, T)$ and suppose by contradiction that $v \neq 0$. Then $v, v$ are generalized eigenvectors corresponding to distinct generalized eigenvalues of $T$. Now 8.13 implies that $v, v$ is linearly independent, which is clearly a contradiction. Therefore $v$ must be $0$.

5. Solution: Let $a_0, a_1, \dots, a_{m-1} \in \mathbb{F}$ such that
$$0 = a_0v + a_1Tv + \dots + a_{m-1}T^{m-1}v$$ Applying $T^{m-1}$ to both sides of the equation above yields
$$0 = a_0T^{m-1}v,$$ which shows that $a_0 = 0$. Therefore
$$0 = a_1Tv + \dots + a_{m-1}T^{m-1}v.$$ Applying $T^{m-2}$ yields
$$0 = a_1T^{m-1}v,$$ which shows that $a_1 = 0$. Continuing in this fashion, we see that $a_0 = a_1 = \dots = a_m = 0$. Thus $v, Tv, T^2v, \dots, T^{m-1}v$ is linearly independent.

6. Solution: Suppose by contradiction that $S \in \mathcal{L}(\mathbb{C}^3)$ is a square root of $T$. Note that $V = \operatorname{null} T^3$. We have
\begin{aligned} V &= \operatorname{null} T^3 = \operatorname{null} R^6\\ &= \operatorname{null} R^3 = \operatorname{null} RT\\ &\subset \operatorname{null} R^2T= \operatorname{null} T^2, \end{aligned} where the third line follows by 8.4. But this a contradiction, since $T^2(z_1, z_2, z_3) = (z_3, 0, 0)$, we see that $\operatorname{null} T^2 = \{(0, 0, z): z \in \mathbb{C}\}$, then we can’t have $V \subset \operatorname{null} T^2$.

7. Solution: This follows directly from 8.19 and 5.32.

8. Solution: False. Let $V = \mathbb{C}^2$. Define $S, T \in \mathcal{L}(\mathbb{C})$ by
\begin{aligned} S(z_1, z_2) &= (0, z_1)\\ T(z_1, z_2) &= (z_2, 0). \end{aligned} Both $S$ and $T$ are nilpotent, however $S + T$ is not (its square equals the identity).

9. Solution: We have
$$\operatorname{null} (TS)^{\operatorname{dim} V} = \operatorname{null} TS(TS)^{\operatorname{dim} V} = \operatorname{null} T(ST)^{\operatorname{dim} V}S = V,$$ where the first equality follows from 8.4 and the third because $(ST)^{\operatorname{dim} V} = 0$ (by 8.18). Thus $(TS)^{\operatorname{dim} V} = 0$ and so $TS$ is nilpotent.

10. Solution: If $T$ is not nilpotent, then $\operatorname{dim} \operatorname{null} T^n < n$ and , by the same reasoning used in 8.4, it follows that $\operatorname{null} T^{n-1} = \operatorname{null} T^n$. Thus, by 8.5, we have
$$V = \operatorname{null} T^{n-1} + \operatorname{range} T^n.$$ Since $\operatorname{range} T^n \subset \operatorname{range} T^{n-1}$, we must also have
$$V = \operatorname{null} T^{n-1} + \operatorname{range} T^{n-1}.$$ Then, by the Fundamental Theorem of Linear Maps (3.22),
$$\operatorname{dim} (\operatorname{null} T^{n-1} + \operatorname{range} T^{n-1}) = \operatorname{dim} V = \operatorname{dim} \operatorname{null} T^{n-1} + \operatorname{dim} \operatorname{range} T^{n-1}.$$ 3.78 now implies that $\operatorname{null} T^{n-1} + \operatorname{range} T^{n-1}$ is a direct sum.

12. Solution: Suppose $v_1, \dots, v_n$ is such basis. Then $Nv_1 = 0$, because the the first column of the matrix has $0$ in all its entries. The definition of matrix of linear map shows that $Nv_2 \in \operatorname{span}(v_1)$. But this implies that $N^2v_2 = 0$. Similarly, $Nv_3 \in \operatorname{span}(v_1, v_2)$, so $N^3v_3 = 0$.

Continuing like this, we see that $N^j v_j = 0$, for each $j = 1, \dots, n$. Therefore $N^n = 0$ and so $N$ is nilpotent.

13. Solution: It is easy when $\mathbb{F} = \mathbb{C}$, because then $V$ has a basis consisting of eigenvectors of $N$ and for each vector $v$ in this basis we have $0 = N^{\operatorname{dim} V} v = \lambda^{\operatorname{dim} V} v$ for the corresponding eigenvalue $\lambda$, which implies that $\lambda = 0$.

More generally, without restricting $\mathbb{F}$ to $\mathbb{C}$, we will prove $N^{\operatorname{dim} V – 1} = 0$ and this fact can be used to show $N^{\operatorname{dim} V – 2} = 0$, which then can be used to show… and so on until $N^1$.

Let $\mathcal{N} = N^{\operatorname{dim} V – 1}$. Note that $\mathcal{N}$ is also normal and that $\mathcal{N}^2 = 0$. Then, for all $v \in V$,
$$||\mathcal{N}^*\mathcal{N}v||^2 = ||\mathcal{N}\mathcal{N}v|| = 0,$$ where the first equality comes from 7.20. Thus $\mathcal{N}^*\mathcal{N} = 0$. Therefore
$$||\mathcal{N}v||^2 = \langle \mathcal{N}v, \mathcal{N}v \rangle = \langle v, \mathcal{N}^*\mathcal{N}v \rangle = 0,$$ which shows that $\mathcal{N} = 0$.

14. Solution: This follows directly from 8.19 and 6.37.

15. Solution: By the same reasoning used in the proof of 8.4, it follows that $\operatorname{dim} \operatorname{null} N^{\operatorname{dim} V} \ge \operatorname{dim} V$. Thus $\operatorname{dim} \operatorname{null} N^{\operatorname{dim} V} = \operatorname{dim} V$ and so $N$ is nilpotent. We have $\operatorname{dim} V + 1$ null spaces each of different dimension. Since the sequence
$$\operatorname{dim} \operatorname{null} N^0, \operatorname{dim} \operatorname{null} N^1, \dots, \operatorname{dim} \operatorname{null} N^{\operatorname{dim} V}$$ must be sorted in strictly increasing order, the only way this can fit is if $\operatorname{dim} \operatorname{null} N^j = j$ for each $j$.

16. Solution: Obviously $V = \operatorname{range} T^0 = \operatorname{range} I$. Let $k$ be a nonnegative integer. Suppose $v \in \operatorname{range} T^{k+1}$. Then $v = T^{k+1}u$ for some $u \in V$. But then $v = T^k(Tu)$. This implies that $v \in \operatorname{range} T^k$.

17. Solution: By the Fundamental Theorem of Linear Maps (3.22), we have
$$\operatorname{dim} \operatorname{null} T^m + \operatorname{dim} \operatorname{range} T^m = \operatorname{dim} \operatorname{null} T^{m+1} + \operatorname{dim} \operatorname{range} T^{m+1},$$ which implies that $\operatorname{dim} \operatorname{null} T^m = \operatorname{dim} \operatorname{null} T^{m+1}$ (because $\operatorname{dim} \operatorname{range} T^m = \operatorname{dim} \operatorname{range} T^{m+1}$). Thus, by 8.3, for all $k > m$, we have
$$\operatorname{dim} \operatorname{null} T^m = \operatorname{dim} \operatorname{null} T^{m+k}.$$ Applying the Fundamental Theorem of Linear Maps again to $T^m$ and $T^{m+k}$ we see that
$$\operatorname{dim} \operatorname{range} T^m = \operatorname{dim} \operatorname{range} T^{m+k}.$$ Since $\operatorname{range} T^{m+k} \subset \operatorname{range} T^m$, it follows that $\operatorname{range} T^{m+k} = \operatorname{range} T^m$.

18. Solution: This follows directly from the previous exercise and 8.4.

19. Solution: This is just a matter of realizing that $\operatorname{null} T^m \subset \operatorname{null} T^{m+1}$ and $\operatorname{range} T^{m+1} \subset \operatorname{range} T^m$ and applying the Fundamental Theorem of Linear Maps.

20. Solution: By Exercise 19, $\operatorname{null} T^4 \neq \operatorname{null} T^5$. By Exercise 15, this implies that $T$ is nilpotent.

21. Solution: Let $W = \mathbb{F}^\infty \times \mathbb{F}^\infty$ and define $T \in \mathcal{L}(W)$ by
$$T\bigr((x_1, x_2, x_3, \dots), (y_1, y_2, y_3, \dots)\bigl) = \bigr((x_2, x_3, \dots), (0, y_1, y_2, y_3, \dots)\bigl),$$ that is, $T$ applies the backward shift operator (call it $B$) on the first slot and forward shift operator (call it $F$) on the second slot. Thus, for each positive integer $k$, we have $$\operatorname{null} B^k = \{(x_1, x_2, x_3, \dots) \in \mathbb{F}^\infty: x_j = 0 \text{ for all } j > k\}$$ and $$\operatorname{range} F^k = \{(x_1, x_2, x_3, \dots) \in \mathbb{F}^\infty: x_1 = x_2 = \dots = x_k = 0\}.$$ Moreover $\operatorname{range} B^k = \mathbb{F}^\infty$ and $\operatorname{null} F^k = \{0\}$. Note that $\operatorname{null} B^k \subsetneq \operatorname{null} B^{k+1}$ and $\operatorname{range} F^k \supsetneq \operatorname{range} F^{k+1}$. Thus $$\operatorname{null} T^k = \{(x, 0) \in \mathbb{F}^\infty \times \mathbb{F}^\infty: x \in \operatorname{null} B^k\}$$ and
$$\operatorname{range} T^k = \{(x, y) \in \mathbb{F}^\infty \times \mathbb{F}^\infty: y \in \operatorname{range} T^k\}.$$ Hence $\operatorname{null} T^k \subsetneq \operatorname{null} T^{k+1}$ and $\operatorname{range} T^k \supsetneq \operatorname{range} T^{k+1}$.