Total Internal Reflection

Technology and Art


Code

Basis-Processing
Cataract
COMRADE
Duck-Angular
Exo
IRIS
MuchHeap
Snail-MapReduce
Underline
Lambda-Queuer
jQuery-Jenkins Radiator

Contact

Github
Twitter
LinkedIn

Site Feed

Functional Analysis Exercises 7 : Vector Spaces

Avishek Sen Gupta on 3 November 2021

This post lists solutions to the exercises in the Vector Space section 2.1 of Erwin Kreyszig’s Introductory Functional Analysis with Applications. This is a work in progress, and proofs may be refined over time.

Notes

The requirements for a space to be a vector space are:

2.1.1 Show that the set of all real numbers, with the usual addition and multiplication, constitutes a one-dimensional real vector space, and the set of all complex numbers constitutes a one-dimensional complex vector space.

Proof:

Consider \(\mathbb{R}\).

\[\blacksquare\]

Consider \(\mathbb{C}\).

\[\blacksquare\]

2.1.2 Prove (1) and (2).

Proof:

We have to prove:

(1a) \(0x=\theta\)
(1b) \(\alpha \theta=\theta\)
(1c) \((-1) x=-x\)

where \(\alpha \in \mathbb{R}, x, \theta \in X\)

(1a) We have:

\[\begin{array} {lr} (0x) + (0x) = (0+0)x = 0x && \mathbf{\text{ (by (VM3))}} \\ (0x) + (0x) + (-(0x))= (0x) + (-(0x)) && \mathbf{\text{ (adding (0x) on both sides)}} \\ (0x) + \theta = \theta && \mathbf{\text{ (by (VA3))}} \\ 0x = \theta && \mathbf{\text{ (by (VA2))}} \end{array}\] \[\blacksquare\]

(1b) We have:

\[\begin{array} {lr} \alpha(0x)=(\alpha 0)x && \mathbf{\text{ (by (VM1))}} \\ (\alpha 0)x=0x=\theta \end{array}\] \[\blacksquare\]

(1c) We have:

\[\begin{array} {lr} 0x = \theta && \mathbf{\text{ (already proved)}} \\ (1-1)x = \theta \\ 1x + (-1)x = \theta && \mathbf{\text{ (by (VM3))}} \\ x + (-1)x = \theta &&\mathbf{\text{ (by (VM2))}} \\ x + (-x) + (-1) x = \theta + (-x) &&\mathbf{\text{ (adding (-x) on both sides)}} \\ \theta + (-1) x = \theta + (-x) && \mathbf{\text{ (by (VA3))}} \\ (-1) x + \theta = (-x) + \theta && \mathbf{\text{ (by (VA1))}} \\ (-1) x = (-x) && \mathbf{\text{ (by (VA2))}} \end{array}\] \[\blacksquare\]

2.1.3 Describe the span of \(M = {(1,1,1), (0,0,2)}\) in \(\mathbb{R^1}\).

Answer:

The span of \(M\) is described by:

\[\begin{bmatrix} 1 && 0 \\ 1 && 0 \\ 1 && 2 \end{bmatrix} \bullet \begin{bmatrix} x \\ y \end{bmatrix}\]

Geometrically, this is a plane whose normal is perpendicular to both \((1,1,1)\) and \((0,0,2)\). Specifically, from the first perpendicularity with \((0,0,0)\):

\[0x+0y+2z=0 \\ z=0\]

From the second perpendicularity with \((1,1,1)\):

\[x+y+z=0 \\ x+y=0 \text{ (since z=0)} \\ x=-y\]

Choose \(x=1\), then \(y=-1\). Then, one choice of the normal vector is \((1,-1,0)\). The equation of the plane then becomes:

\[x-y=0\]

2.1.4 Which of the following subsets of \(\mathbb{R}^3\) constitute a subspace of \(\mathbb{R}^3\)? [Here, \(x = (\xi_1, \xi_2, \xi_3)\).]

(a) All \(x\) with \(\xi_1=\xi_2\) and \(\xi_3=0\).
(b) All \(x\) with \(\xi_1=\xi_2+1\).
(c) All \(x\) with positive \(\xi_1\), \(\xi_2\), \(\xi_3\).
(d) All \(x\) with \(\xi_1-\xi_2+\xi_3=k=\text{const}\).

Answer:

For a subset to be a subspace, it needs to satisfy the following criterion:

\[\alpha x + \beta y \in X, \alpha, \beta \in \mathbb{R}\]

(a) Consider two arbitrary members \(x=(\xi, \xi, 0)\) and \(y=(\eta, \eta, 0)\) of the given subset (call it \(X\)).

Then, we have:

\[\alpha x + \beta y=\alpha (\xi, \xi, 0) + \beta (\eta, \eta, 0) \\ = (\alpha\xi, \alpha\xi, \alpha 0) + (\beta\eta, \beta\eta, \beta 0) \\ = (\alpha\xi + \beta\eta, \alpha\xi + \beta\eta, \alpha 0 + \beta 0) \\ = (\alpha\xi + \beta\eta, \alpha\xi + \beta\eta, 0) \in X\]

Thus, \(X\) is a subspace of \(\mathbb{R}^3\).

(b) Consider two arbitrary members \(x=(\xi+1, \xi, 0)\) and \(y=(\eta+1, \eta, 0)\) of the given subset (call it \(X\)).

Then, we have:

\[\require{cancel} \alpha x + \beta y=\alpha (\xi+1, \xi, 0) + \beta (\eta+1, \eta, 0) \\ = (\alpha\xi + \alpha, \alpha\xi, \alpha 0) + (\beta\eta + \beta, \beta\eta, \beta 0) \\ = [(\alpha\xi + \beta\eta) + (\alpha + \beta), (\alpha\xi + \beta\eta), \alpha 0 + \beta 0] \\ = [(\alpha\xi + \beta\eta) + (\alpha + \beta), (\alpha\xi + \beta\eta), 0] \cancel{\in} X\\\]

Thus, \(X\) is not a subspace of \(\mathbb{R}^3\).

(c) Consider two arbitrary members \(x=(\xi_1, \xi_2, \xi_3)\) and \(y=(\eta_1, \eta_2, \eta_3)\) of the given subset (call it \(X\)), with \(\xi_i,\eta_i \geq 0\).

Then, we have:

\[\alpha x + \beta y=\alpha (\xi_1, \xi_2, \xi_3) + \beta (\eta_1, \eta_2, \eta_3) \\ = (\alpha\xi_1, \alpha\xi_2, \alpha\xi_3) + (\beta\eta_1, \beta\eta_2, \beta\eta_3) \\ = (\alpha\xi_1 + \beta\eta_1, \alpha\xi_2 + \beta\eta_2, \alpha\xi_3 + \beta\eta_3)\]

Choose any \(\xi_1>\eta_1\), and \(\alpha=-1\), \(\beta=1\). Then, we have:

\[\require{cancel} \alpha\xi_1 + \beta\eta_1 = -\xi_1 + \eta_1 < 0 \notin X\]

Thus, \(X\) is not a subspace of \(\mathbb{R}^3\).

(d) Consider two arbitrary members \(x=(\xi_1, \xi_2, k-\xi_1+\xi_2)\) and \(y=(\eta_1, \eta_2, k-\eta_1+\eta_2)\) of the given subset (call it \(X\)).

Then we have:

\[\require{cancel} \alpha x + \beta y=\alpha (\xi_1, \xi_2, k-\xi_1+\xi_2) + \beta (\eta_1, \eta_2, k-\eta_1+\eta_2) \\ = [\alpha\xi_1, \alpha\xi_2, \alpha(k-\xi_1+\xi_2)] + [\beta\eta_1, \beta\eta_2, \beta(k-\eta_1+\eta_2)] \\ = [\alpha\xi_1 + \beta\eta_1, \alpha\xi_2 + \beta\eta_2, (\alpha + \beta) k - (\alpha\xi_1 + \beta\eta_1) + (\alpha\xi_2 + \beta\eta_2)] \notin X\]

Thus, \(X\) is not a subspace of \(\mathbb{R}^3\).

\[\blacksquare\]

2.1.5 s. Show that \({x_1, \cdots, x_n}\), where \(x_j(t) = t^j\), is a linearly independent set in the space \(C[a,b]\).

Proof:

A set \(\{x_1, x_2, \cdots, x_n\}\) is linearly independent if:

\[\alpha_1 x_1 + \alpha_2 x_2 + \cdots \alpha_n x_n = x_0\]

only for all \(\alpha_i=0\).

Any linear combination of the given set, call it \(X\), is given by:

\[f(t)=\alpha_1 t^1 + \alpha_2 t^2 + \cdots + \alpha_n t^n\]

To prove that \(X\) is linearly independent, we need to prove that that the zero vector \(f(t)=0=f_0(t)\) is not possible for any combination of \(\alpha_i\).

This is a polynomial of degree \(n\). Fix all \(\alpha_i\), with not all of them zero.
By the Fundamental Theorem of Algebra, we know that it can have at most \(n\) roots of this polynomial. Thus, there are at most \(n\) values of \(t\) for which \(f(t)=0\). Thus, it is not zero for the remaining uncountable values of \(t \in [a,b]\), therefore for an arbitrary combination of \(\alpha_i\), \(f(t) \neq f_0(t)\).

Thus, \(X\) is a linearly independent set.

\[\blacksquare\]

2.1.6 Show that in an \(n\)-dimensional vector space \(X\), the representation of any \(x\) as a linear combination of given basis vectors \(e_1, \cdots, e_n\) is unique.

Proof:

Assume that \(x=\alpha_1 e_1 + \alpha_2 e_2 + \cdots + \alpha_n e_n\). Assume that \(x\) can also be represented by a different linear combination \(\beta_i\), such that:

\[x=\beta_1 e_1 + \beta_2 e_2 + \cdots + \beta_n e_n\]

Then we have:

\[\alpha_1 e_1 + \alpha_2 e_2 + \cdots + \alpha_n e_n=\beta_1 e_1 + \beta_2 e_2 + \cdots + \beta_n e_n \\ (\alpha_1 e_1 + \alpha_2 e_2 + \cdots + \alpha_n e_n)+(-\beta_1 e_1) +(-\beta_2 e_2) + \cdots + (-\beta_n e_n)=\beta_1 e_1 + \beta_2 e_2 + \cdots + \beta_n e_n + (-\beta_1 e_1) +(-\beta_2 e_2) + \cdots + (-\beta_n e_n) \\ (\alpha_1 - \beta_1) e_1 + (\alpha_2 - \beta_2) e_2 + \cdots + (\alpha_n -\beta_n) e_n =(\beta_1 - \beta_1) e_1 + (\beta_2 - \beta_2) e_2 + \cdots + (\beta_n - \beta_n) e_n \\ (\alpha_1 - \beta_1) e_1 + (\alpha_2 - \beta_2) e_2 + \cdots + (\alpha_n -\beta_n) e_n =0 e_1 + 0 e_2 + \cdots + 0 e_n \\\]

Then equating the coefficients on both sides, we get:

\[\alpha_i-\beta_i=0 \\ \alpha_i=\beta_i\]

Thus, the representation of any \(x\) as a linear combination of given basis vectors \(e_1, \cdots, e_n\) is unique.

\[\blacksquare\]

2.1.7 Let \(\{e_1, \cdots, e_n\}\) be a basis for a complex vector space \(X\). Find a basis for \(X\) regarded as a real vector space. What is the dimension of \(X\) in either case?

Answer:

A complex vector space is a vector space whose field of scalars is the complex numbers. Then any vector \(x\) in this complex vector space is representable as:

\[x=(a_1 + ib_1) e_1 + (a_2 + ib_2) e_2 + \cdots + (a_n + ib_n) e_n \\ =(a_1 e_1 + a_2 e_2 + \cdots + a_n e_n) + (b_1 ie_1 + b_2 ie_2 + \cdots + b_n ie_n)\]

The basis for \(X\) regarded as a real vector space are:

\[(e_1, e_2, \cdots, e_n, ie_1, ie_2, \cdots, ie_n)\]

The dimension of the complex space is \(n\).
The dimension of the complex space regarded as a real vector space is \(2n\).


2.1.8 If \(M\) is a linearly dependent set in a complex vector space \(X\), is \(M\) linearly dependent in \(X\), regarded as a real vector space?

Proof:

Consider the set \(\{u=i,v=-1\}\). This is a linearly dependent set because \(iv=u\), since \(i.i=-1\). But there is no \(\alpha \in \mathbb{C}\) which gives \(\alpha u=v\), i.e., \(\alpha i=-1\).

Thus, \(X\), regarded as a real vector space, is not necessarily dependent.

\[\blacksquare\]

2.1.9 On a fixed interval \([a,b] \subset \mathbb{R}\), consider the set \(X\) consisting of all polynomials with real coefficients and of degree not exceeding a given \(n\), and the polynomial \(x=0\) (for which a degree is not defined in the usual discussion of degree). Show that \(X\), with the usual addition and the usual multiplication by real numbers, is a real vector space of dimension \(n+1\). Find a basis for \(X\). Show that we can obtain a complex vector space \(\tilde{X}\) in a similar fashion if we let those coefficients be complex. Is \(X\) a subspace of \(\tilde{X}\)?

Proof:

A polynomial not exceeding degree \(n\) is given as: \(f(x)=\sum\limits_{i=0}^n \alpha_i x^i\).

\[\blacksquare\]

2.1.10 If \(Y\) and \(Z\) are subspaces of a vector space \(X\), show that \(Y\cap Z\) is a subspace of \(X\), but \(Y\cup Z\) need not be one. Give examples.

Proof:

We have, for all \(\alpha, \beta \in \mathbb{R}\):

\[\alpha y_1 + \beta y_2 \in Y \\ \alpha z_1 + \beta z_2 \in Z\]

Assume \(x \in Y \cap Z\).
Then \(x \in Y\) and \(x \in Z\).
Then \(\alpha x_1 + \beta x_2 \in Y\) and \(\alpha x_1 + \beta x_2 \in Z\).
Then \(\alpha x_1 + \beta x_2 \in Y \cap Z\)

Thus, \(Y \cap Z\) is a subspace of \(X\).

\[\blacksquare\]

In \(\mathbb{R}^2\), the vector space with the basis vector \(u=(1,0)\) and the vector space with the basis vector \(v=(0,1)\) gives two vector spaces \(Y\) and \(Z\). Choose \(\alpha=1\), \(\beta=1\), then \(\alpha u + \beta v = (1,1)\) does not belong to \(Y \cup Z\).

Thus, \(Y \cap Z\) is not necessarily a subspace of \(X\).

\[\blacksquare\]

2.1.11 If \(M \neq \emptyset\) is any subset of a vector space \(X\), show that span \(M\) is a subspace of \(X\).

Proof:

Let \(M \subset X\). Let \(e_1, e_2, \cdots, e_n \in M\). Then, the span of \(M\) is \(\sum\limits_{i=1}^n\alpha_i e_n\).

Then, every \(x=\sum\limits_{i=1}^n\alpha_i e_n, x \in M\). Pick any two arbitrary points in \(M\), so that:

\[x_1=\sum\limits_{i=1}^n\alpha_i e_i \\ x_2=\sum\limits_{i=1}^n\beta_i e_i \\\]

Then, we get:

\[ax_1+bx_2=\sum\limits_{i=1}^n a\alpha_i e_i + \sum\limits_{i=1}^n b\beta_i e_i \\ =\sum\limits_{i=1}^n (a\alpha_i+b\beta_i) e_i = \sum\limits_{i=1}^n k_i e_i \in M\]

Thus, span \(M\) is a subspace of \(X\).

\[\blacksquare\]

2.1.12 Show that the set of all real two-rowed square matrices forms a vector space \(X\). What is the zero vector in \(X\)? Determine dim \(X\). Find a basis for \(X\). Give examples of subspaces of X. Do the symmetric matrices \(x \in X\) form a subspace? The singular matrices?

Proof:

A real-valued \(2 \times 2\) matrix is of the form:

\[\begin{bmatrix} a && b \\ c && d \end{bmatrix}\]

Assume \(x_1\), \(x_2\) as below:

\[x_1=\begin{bmatrix} a && b \\ c && d \end{bmatrix}\\ x_2=\begin{bmatrix} p && q \\ q && s \end{bmatrix}\]

Then, we have:

\[\alpha x_1 + \beta x_2 = \alpha \begin{bmatrix} a && b \\ c && d \end{bmatrix} + \beta \begin{bmatrix} p && q \\ r && s \end{bmatrix} \\ = \begin{bmatrix} \alpha a && \alpha b \\ \alpha c && \alpha d \end{bmatrix} + \begin{bmatrix} \beta p && \beta q \\ \beta r && \beta s \end{bmatrix} \\ = \begin{bmatrix} \alpha a + \beta p && \alpha b + \beta q \\ \alpha c + \beta r && \alpha d + \beta s \end{bmatrix}\]

This is also a \(2 \times 2\) matrix, thus the set of all real two-rowed square matrices forms a vector space.

The zero vector of \(X\) is \(\begin{bmatrix} 0 && 0 \\ 0 && 0 \end{bmatrix}\)

The dimension of \(X\) is 4. A possible basis for \(X\) is:

\[\begin{bmatrix} 1 && 0 \\ 0 && 0 \end{bmatrix}, \begin{bmatrix} 0 && 1 \\ 0 && 0 \end{bmatrix}, \begin{bmatrix} 0 && 0 \\ 1 && 0 \end{bmatrix}, \begin{bmatrix} 0 && 0 \\ 0 && 1 \end{bmatrix}\]

An example of a subspace of \(X\) is \(\begin{bmatrix} k && 0 \\ 0 && 0 \end{bmatrix}, k \in \mathbb{R}\).

Yes, the symmetric matrices form a subspace. No, the singular matrices do not form a subspace. Here is a counter-example. Take \(x\) and \(y\) to be as follows:

\[x=\begin{bmatrix} 2 && 6 \\ 4 && 12 \end{bmatrix}, y=\begin{bmatrix} 1 && 1 \\ 1 && 1 \end{bmatrix}\]

The determinants for both \(x\) and \(y\) are zero, thus, they are singular.

Then, setting \(\alpha=1\), \(\beta=1\), we get:

\[\alpha x + \beta = \begin{bmatrix} 2 && 6 \\ 4 && 12 \end{bmatrix} + \begin{bmatrix} 1 && 1 \\ 1 && 1 \end{bmatrix}= \begin{bmatrix} 3 && 7 \\ 5 && 13 \end{bmatrix}\]

The determinant of the result is \(39-35=4 \neq 0\), thus it is not a singular matrix.

\[\blacksquare\]

2.1.13 (Product) Show that the Cartesian product \(X = X_1 \times X_2\) of two vector spaces over the same field becomes a vector space if. we define the two algebraic operations by

\((x_1. x_2) + (y_1, y_2) = (x_1 +y_1. x_2 + y_2)\),
\(\alpha(x_1, x_2) = (\alpha x_1, \alpha x_2)\).

Proof:

[Easy. TODO]

\[\blacksquare\]

2.1.14 (Quotient space, codimension) Let \(Y\) be a subspace of a vector space \(X\). The coset of an element \(x \in X\) with respect to \(Y\) is denoted by \(x + Y\) and is defined to be the set (see Fig. 12)

\(x+Y={v\vert v=x+y, \in Y}\).

Show that the distinct cosets form a partition of X. Show that under algebraic operations defined by (see Figs. 13, 14)

\[(w+Y)+(x+Y)=(w+x)+Y \alpha(x+Y)=\alpha x+Y\]

these cosets constitute the elements of a vector space. This space is called the quotient space (or sometimes factor space) of \(X\) by \(Y\) (or modulo \(Y\)) and is denoted by \(X/Y\). Its dimension is called the codimension of \(Y\) and is denoted by codim \(Y\), that is, \(\text{codim } Y=\text{dim }(X/Y)\).

Proof:

To prove that the cosets form a partition of \(X\), we need to prove that an arbitrary element \(x \in X\) belongs to one and only one coset.

Suppose \(x \in X\) belongs to two cosets \(u+Y\) and \(v+Y\). Then, we have from the definition:

\[x=v|v=u+y_1,y \in Y \\ x=v|v=v+y_2,y \in Y\]

Then, we have:

\[u+y_1=v+y_2 \\ u-v=y_2-y_1 \in Y\\\]

Then \(u-v \in Y\). Then \(u-v=y_0, y_0 \in Y\).
Then \(u=v+y_0\), where \(y_0 \in Y\). Then \(u \in v+Y\). Since \(u \in u+Y\), \(u+Y\) and \(v+Y\) are the same coset.

\[\blacksquare\]

[Easy to prove that it’s a vector space. TODO]


2.1.15 Let \(X=\mathbb{R}^3\) and \(Y=\{(\xi_1,0,0) \vert \xi \in \mathbb{R}\}\). Find \(X/Y\), \(X/X\), \(X/{0}\).

Answer:

Loosely, the quotient space is the set of points which can translate cosets to cover the entire vector space.

For \(X/Y\), we get the coset as the set of parallel subspaces along the vector \((1,0,0)\).

For \(X/X\), any translation of the coset \(X\) covers the entire vector space \(X\). No translation also covers the entire space. This implies that \(x=\{0\}\).

For \(X/\{0\}\), we have \(x+\{0\}=\{v:v=x+0\}\), i.e., each coset is the point itself. Thus the set of points required to partition \(X\) into cosets is \(X\) itself.


tags: Mathematics - Proof - Functional Analysis - Pure Mathematics - Kreyszig