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Functional Analysis Exercises 12 : Linear Operators

Avishek Sen Gupta on 13 December 2021

This post lists solutions to the exercises in the Linear Operators section 2.6 of Erwin Kreyszig’s Introductory Functional Analysis with Applications. This is a work in progress, and proofs may be refined over time.

2.6.1. Show that the identity, zero, and differentiation operators are linear.

Proof:

The identity operator is $$I_x x=x$$. We have:

$I_x(\alpha x + \beta y) = \alpha x + \beta y = \alpha I_x x + \beta I_x y$

The zero operator is $$0x=0$$. We have:

$0(\alpha x + \beta y) = \alpha 0 + \beta 0 = \alpha 0x + \beta 0y$

The differentiation operator is $$Tx=x'(t)$$. We have:

$T(\alpha x(t) + \beta y(t)) = (\alpha x(t) + \beta y(t))' = \alpha x'(t) + \beta y'(t) = \alpha Tx + \beta Ty$ $\blacksquare$

2.6.2. Show that the operators $$T_1, \cdots , T_4$$ from $$\mathbb{R}^2$$ into $$\mathbb{R}^2$$ defined by

$(\xi_1, \xi_2) \mapsto (\xi_1, 0) \\ (\xi_1, \xi_2) \mapsto (0, \xi_2) \\ (\xi_1, \xi_2) \mapsto (\xi_2, \xi_1) \\ (\xi_1, \xi_2) \mapsto (\gamma\xi_1, \gamma\xi_2)$

respectively, are linear, and interpret these operators geometrically.

Proof:

We take $$x=(x_1,x_2)$$ and $$y=(y_1,y_2)$$.

$T_1(x_1,x_2) = (x_1,0)$

We have:

$T_1(\alpha x + \beta y) = T[(\alpha x_1, \alpha x_2) + (\beta y_1, \beta y_2)] \\ = T_1(\alpha x_1 + \beta y_1, \alpha x_2 + \beta y_2) = (\alpha x_1 + \beta y_1, 0) \\ = (\alpha x_1, 0) + (\beta y_1, 0) = \alpha T_1 x + \beta T_1 y$ $\blacksquare$

We take $$x=(x_1,x_2)$$ and $$y=(y_1,y_2)$$.

$T_2(x_1,x_2) = (0,x_2)$

We have:

$T_2(\alpha x + \beta y) = T[(\alpha x_1, \alpha x_2) + (\beta y_1, \beta y_2)] \\ = T_2(\alpha x_1 + \beta y_1, \alpha x_2 + \beta y_2) = (0, \alpha x_2 + \beta y_2) \\ = (0, \alpha x_2) + (0, \beta y_2) = \alpha T_2 x + \beta T_2 y$ $\blacksquare$

We take $$x=(x_1,x_2)$$ and $$y=(y_1,y_2)$$.

$T_3(x_1,x_2) = (x_2,x_1)$

We have:

$T_3(\alpha x + \beta y) = T[(\alpha x_1, \alpha x_2) + (\beta y_1, \beta y_2)] \\ = T_3(\alpha x_1 + \beta y_1, \alpha x_2 + \beta y_2) = (\alpha x_2 + \beta y_2, \alpha x_1 + \beta y_1) = \alpha (x_2, x_1) + \beta (y_2, y_1) = \alpha T_3 x + \beta T_3 y$ $\blacksquare$

We take $$x=(x_1,x_2)$$ and $$y=(y_1,y_2)$$.

$T_4(x_1,x_2) = (\gamma x_1, \gamma x_2)$

We have:

$T_4(\alpha x + \beta y) = T[(\alpha x_1, \alpha x_2) + (\beta y_1, \beta y_2)] \\ = T_4(\alpha x_1 + \beta y_1, \alpha x_2 + \beta y_2) = (\gamma \alpha x_1 + \gamma \beta y_1, \gamma \alpha x_2 + \gamma \beta y_2) \\ = \alpha \gamma (x_1, x_2) + \beta \gamma (y_1, y_2) = \alpha T_4 x + \beta T_4 y$ $\blacksquare$

2.6.3. What are the domain, range and null space of $$T_1, T_2, T_3$$ in Prob. 2?

The operators are:

$T_1(x_1,x_2) = (x_1,0) \\ T_2(x_1,x_2) = (0, x_2) \\ T_3(x_1,x_2) = (x_2, x_1)$

For $$T_1$$, we have:

Domain is $$\mathbb{R}^2$$.
Range is $$((-\infty,0), (+\infty,0))$$.
Null Space is $$((0,-\infty), (0,+\infty))$$.

For $$T_2$$, we have:

Domain is $$\mathbb{R}^2$$.
Range is $$((0,-\infty), (0,+\infty))$$. Null Space is $$((-\infty,0), (+\infty,0))$$.

For $$T_2$$, we have:

Domain is $$\mathbb{R}^2$$.
Range is $$\mathbb{R}^2$$.
Null Space is $$(0,0)$$.

2.6.4. What is the null space of $$T_4$$ in Prob. 2? Of $$T_1$$ and $$T_2$$ in 2.6-7? Of $$T$$ in 2.6-4?

The null space of $$T_4(x_1,x_2)=(\gamma x_1, \gamma x_2)$$ is $$(0,0)$$.

The null space of $$T_1$$, which is the cross product $$T_1: \mathbb{R}^3 \rightarrow \mathbb{R}^3$$ with a fixed vector $$a=(a_1, a_2, a_3)$$ is $$(ka_1, ka_2, ka_3)$$, with $$k \in \mathbb{R}$$.

The null space of $$T_2=\langle x,a \rangle$$ with $$a=(a_1,a_2,a_3)$$ as a constant vector is the plane defined by $$a_1 x + a_2 y + a_3 z = 0$$

The null space of $$T: x(t) \mapsto x'(t)$$ is the space of functions $$x(t)=c$$, where $$c \in \mathbb{R}$$.

2.6.5. Let $$T: X \rightarrow Y$$ be a linear operator. Show that the image of a subspace $$V$$ of $$X$$ is a vector space, and so is the inverse image of a subspace $$W$$ of $$Y$$.

Proof:

Let $$x,y \in V$$, and $$x',y' \in \text{Im } V$$. Then we have:

$\alpha x' + \beta y' = \alpha Tx + \beta Ty = T(\alpha x + \beta y)$

We know that $$\alpha x + \beta y \in V$$, therefore $$T(\alpha x + \beta y) \in \text{Im } V$$. Thus $$\alpha x' + \beta y' \in \text{Im } V$$. Thus $$\text{Im } V$$ is a vector subspace of $$V$$ and is thus a vector space.

$\blacksquare$

Let $$x,y \in \text{Im } W$$, and $$x',y' \in W$$. Then we have:

$\alpha x + \beta y = \alpha T^{-1} x' + \beta T^{-1} y' = T^{-1}(\alpha x' + \beta y')$

We know that $$\alpha x' + \beta y' \in W$$, thus $$T(\alpha x' + \beta y') \in \text{Im } W$$. Thus $$\alpha x + \beta y \in \text{Im } W$$. Thus $$\text{Im } W$$ is a vector subspace of $$W$$ and is thus a vector space.

$\blacksquare$

2.6.6. If the product (the composite) of two linear operators exists, show that it is linear.

Proof:

$Tx = (T_1 \circ T_2) x = T_1(T_2 x)$

Thus for $$z=\alpha x + \beta y$$, we have:

$Tz = (T_1 \circ T_2) z = T_1(T_2 z) = T_1(T_2 (\alpha x + \beta y)) \\ T_1(\alpha T_2 x + \beta T_2 y) = T_1(\alpha T_2 x) + T_1(\beta T_2 y) \\ = \alpha (T_1 (T_2 x)) + \beta (T_1 (T_2 y)) \\ = \alpha (T_1 \circ T_2) x + \beta (T_1 \circ T_2) y = \alpha T x + \beta T y$ $\blacksquare$

2.6.7. (Commutativity) Let $$X$$ be any vector space and $$S: X \rightarrow X$$ and $$T: X \rightarrow X$$ any operators. $$S$$ and $$T$$ are said to commute if $$ST = TS$$, that is, $$(ST)x = (TS)x$$ for all $$x \in X$$. Do $$T_1$$ and $$T_3$$, in Prob. 2 commute?

Proof:

We have the following operators:

$T_1(x_1,x_2)=(x_1,0) \\ T_2(x_1,x_2)=(0,x_2)$

We have :

$(T_1 T_2) (x_1,x_2) = T_1(T_2(x_1,x_2)) = T_1(0,x_2) = (0,0) \\ (T_2 T_1) (x_1,x_2) = T_2(T_1(x_1,x_2)) = T_2(x_1,0) = (0,0)$

Thus $$T_1, T_2$$ commute.

$\blacksquare$

2.6.8. Write the operators in Prob. 2 using $$2 \times 2$$ matrices.

The operators are:

$(\xi_1, \xi_2) \mapsto (\xi_1, 0) \\ (\xi_1, \xi_2) \mapsto (0, \xi_2) \\ (\xi_1, \xi_2) \mapsto (\xi_2, \xi_1) \\ (\xi_1, \xi_2) \mapsto (\gamma\xi_1, \gamma\xi_2)$

$$(\xi_1, \xi_2) \mapsto (\xi_1, 0)$$ can be represented as:

$\begin{bmatrix} 1 && 0 \\ 0 && 0 \end{bmatrix}$

$$(\xi_1, \xi_2) \mapsto (0, \xi_2)$$ can be represented as:

$\begin{bmatrix} 0 && 0 \\ 0 && 1 \end{bmatrix}$

$$(\xi_1, \xi_2) \mapsto (\xi_2, \xi_1)$$ can be represented as:

$\begin{bmatrix} 0 && 1 \\ 1 && 0 \end{bmatrix}$

$$(\xi_1, \xi_2) \mapsto (\gamma\xi_1, \gamma\xi_2)$$ can be represented as:

$\begin{bmatrix} \gamma && 0 \\ 0 && \gamma \end{bmatrix}$

2.6.9. In 2.6-8, write $$y = Ax$$ in terms of components, show that $$T$$ is linear and give examples.

Proof:

Let $$A$$ be an arbitrary $$m \times n$$ matrix, and $$x$$ be an $$n \times 1$$ vector. Then we have $$T:x \mapsto Ax$$

Then we have $$y$$ as an $$m \times 1$$ vector, and $$y_k = \sum\limits_{i=1}^n A_{ki} x_i$$.

Let $$z=\alpha x + \beta y$$. Then we have: $$z_i=\alpha x_i + \beta y_i$$.

Then we have, for $$y=A(\alpha x + \beta y)$$:

$y_k = \sum\limits_{i=1}^n A_{ki} (\alpha x_i + \beta y_i) \\ = \sum\limits_{i=1}^n (\alpha A_{ki} x_i + \beta A_{ki} y_i) \\ = \alpha \sum\limits_{i=1}^n A_{ki} x_i + \beta \sum\limits_{i=1}^n A_{ki} y_i$

Thus $$y=\alpha Ax + \beta Ay = \alpha Tx + \beta Ty$$ and $$T$$ is linear.

$\blacksquare$

2.6.10. Formulate the condition in 2.6-10(a) in terms of the null space of $$T$$.

The condition is:

The inverse $$T^{-1}: \mathcal{R}(T) \rightarrow \mathcal{D}(T)$$ exists if and only if $$Tx=0 \Rightarrow x=0$$.

Reformulated in terms of the null space of $$T$$, it states the inverse $$T^{-1}: \mathcal{R}(T) \rightarrow \mathcal{D}(T)$$ exists if and only if the null space of $$T$$ consists of only the zero vector.

2.6.11. Let $$X$$ be the vector space of all complex $$2 \times 2$$ matrices and define $$T: X \rightarrow X$$ by $$Tx = bx$$, where $$b \in X$$ is fixed and $$bx$$ denotes the usual product of matrices. Show that $$T$$ is linear. Under what condition does $$T^{-1}$$ exist?

Proof:

We have $$Tx=bx=c$$, so we can write each individual element of the result $$c_{ij}$$ as:

$c_{ij}= \sum\limits_{k=1}^2 b_{ik} x_{kj}$

We can thus write, for $$z=\alpha x + \beta y$$:

${(Tz)}_{ij} = \sum\limits_{k=1}^2 b_{ik} (\alpha x_{kj} + \beta y_{kj}) \\ = \alpha \sum\limits_{k=1}^2 b_{ik} x_{kj} + \beta \sum\limits_{k=1}^2 b_{ik} y_{kj} \\ = \alpha {(Tx)}_{ij} + \beta {(Ty)}_{ij} \\ \Rightarrow T(\alpha x + \beta y) = \alpha Tx + \beta Ty$

$$T^{-1}$$ exists when the determinant $$ad-bc \neq 0$$.

Alternatively, the two column vectors in the $$2 \times 2$$ matrix should be linearly independent.

$\blacksquare$

2.6.12. Does the inverse of $$T$$ in 2.6-4 (differentiation operator) exist?

Proof:

For the inverse to exist, we need the operator to be injective, that is:

$x_1 \neq x_2 \Rightarrow Tx_1 \neq Tx_2 \\ \text{or, } Tx_1 = Tx_2 \Rightarrow x_1 = x_2$

As a counter-example, take:

$x_1(t) = x + 2 \\ x_2(t) = x + 3$

Then, applying the differentiation operator on them, we get:

$Tx_1(t) = x_1'(t) = 1 \\ Tx_2(t) = x_2'(t) = 1$

Thus $$Tx_1=Tx_2$$, but $$x_1 \neq x_2$$. Thus, the inverse of $$T$$ does not exist.

$\blacksquare$

2.6.13. Let $$T: \mathcal{D}(T) \rightarrow Y$$ be a linear operator whose inverse exists. If $$\{x_1, \cdots, x_n\}$$ is a linearly independent set in $$\mathcal{D}(T)$$, show that the set $$\{Tx_1, \cdots, Tx_n\}$$ is linearly independent.

Proof:

$$\{x_1, \cdots, x_n\}$$ is a linearly independent set. Thus, $$\sum\limits_{i=1}^n \alpha_i x_i = 0 \Rightarrow \alpha_i=0$$.

Assume then that $$\sum\limits_{i=1}^n \alpha_i Tx_i = 0$$. Then we have:

$\sum\limits_{i=1}^n \alpha_i Tx_i = T(\sum\limits_{i=1}^n \alpha_i x_i) = 0$

Because $$T0=0$$, we then have:

$T(\sum\limits_{i=1}^n \alpha_i x_i) = 0 = T0 \Rightarrow \sum\limits_{i=1}^n \alpha_i x_i = 0$

But this implies that all $$\alpha_i = 0$$.
Thus $$\sum\limits_{i=1}^n \alpha_i Tx_i = 0 \Rightarrow \alpha_i = 0$$

$\blacksquare$

2.6.14. Let $$T: X \rightarrow Y$$ be a linear operator and $$\dim X = \dim Y = n < \infty$$. Show that $$\mathcal{R}(T) = Y$$ if and only if $$T^{-1}$$ exists.

Proof:

$(\Rightarrow)$

Suppose that $$\mathcal{R}(T) = Y$$. For $$T^{-1}$$ to exist, we need to prove that the null space contains only the zero vector.

By the Rank-Nullity Theorem, we have:

$\dim \mathcal{R}(T) + \dim \mathcal{N}(T) = \dim X \\ \dim \mathcal{N}(T) = \dim X - \dim \mathcal{R}(T) = n - n = 0$

Thus the null space of $$T$$ consists of only the zero vector, and thus $$T^{-1}$$ exists.

$\blacksquare$ $(\Leftarrow)$

Conversely, assume that the inverse $$T^{-1}$$ exists. Then $$Tx=0 \Rightarrow x=0$$. Thus the null space of $$T$$ has only the zero vector, i.e., it has dimension zero.

By the Rank-Nullity Theorem, we have:

$\dim \mathcal{R}(T) + \dim \mathcal{N}(T) = \dim X \\ \dim \mathcal{R}(T) = \dim X - \dim \mathcal{N}(T) = n - 0 = n = \dim Y$

Thus, $$\mathcal{R}(T) = Y$$.

$\blacksquare$

2.6.15. Consider the vector space $$X$$ of all real-valued functions which are defined on $$\mathbb{R}$$ and have derivatives of all orders everywhere on $$\mathbb{R}$$. Define $$T: X \rightarrow X$$ by $$y(t) = Tx(t) = x'(t)$$. Show that $$\mathcal{R}(T)$$ is all of $$X$$ but $$T^{-1}$$ does not exist. Compare with Prob. 14 and comment.

Proof:

(TODO)

$\blacksquare$
tags: Mathematics - Proof - Functional Analysis - Pure Mathematics - Kreyszig