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 $\operatorname{range} T^k \supsetneq \operatorname{range} T^{k+1}$.

About Linearity

This website is supposed to help you study Linear Algebras. Please only read these solutions after thinking about the problems carefully. Do not just copy these solutions.
This entry was posted in Chapter 8 and tagged .