Chapter 8 Exercise A


Exercise 1

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 $T$.


Exercise 2

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


Exercise 3

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

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

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

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


Exercise 4

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 $0$.


Exercise 5

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.


Exercise 6

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 $V \subset \operatorname{null} T^2$.


Exercise 7

This follows directly from 8.19 and 5.32.


Exercise 8

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).


Exercise 9

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.


Exercise 10

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.


Exercise 12

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.


Exercise 13

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 $\mathcal{N} = 0$.


Exercise 14

This follows directly from 8.19 and 6.37.


Exercise 15

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 $j$.


Exercise 16

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 $v \in \operatorname{range} T^k$.


Exercise 17

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 $\operatorname{range} T^{m+k} = \operatorname{range} T^m$.


Exercise 18

This follows directly from the previous exercise and 8.4.


Exercise 19

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.


Exercise 20

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


Exercise 21

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

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.

You may also like...

Leave a Reply

Your email address will not be published. Required fields are marked *