# Total Internal Reflection

Technology and Art

# 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:

• (VA1) Symmetric with respect to addition, i.e., $$x+y=y+x, x,y \in X$$
• (VA2) Existence of identity element, i.e., $$x+\theta=x, x,\theta \in X$$
• (VA3) Existence of inverse element, i.e., $$x+(-x)=\theta, x,\theta \in X$$
• (VA4) Associative with respect to addition, i.e., $$x+(y+z)=(x+y)+z, x,y,z \in X$$

• (VM1) Associative with respect to scalar multiplication, i.e., $$\alpha (\beta x) = (\alpha \beta) x, x \in X, \alpha, \beta \in \mathbb{R}$$
• (VM2) Existence of identity element, i.e., $$\alpha_0 x=x, x \in X, \alpha_0=1 \in \mathbb{R}$$
• (VM3) Distributive with respect to addition of scalars, i.e., $$(\alpha + \beta) x=\alpha x + \beta x, x \in X, \alpha, \beta \in \mathbb{R}$$
• (VM4) Distributive with respect to addition of vectors, i.e., $$\alpha (x+y)=\alpha x + \alpha y, x,y \in X, \alpha \in \mathbb{R}$$

#### 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 $$\mathbb{R}$$.

• (VA1) Symmetric with respect to addition, i.e., $$x+y=y+x, x,y \in \mathbb{R}$$
• (VA2) Existence of identity element, i.e., $$x+0=x, x,0 \in \mathbb{R}$$
• (VA3) Existence of inverse element, i.e., $$x+(-x)=0, x,0 \in \mathbb{R}$$
• (VA4) Associative with respect to addition, i.e., $$x+(y+z)=(x+y)+z, x,y,z \in \mathbb{R}$$

• (VM1) Associative with respect to scalar multiplication, i.e., $$\alpha (\beta x) = (\alpha \beta) x, x \in X, \alpha, \beta \in \mathbb{R}$$
• (VM2) Existence of identity element, i.e., $$1x=x, x \in X, 1 \in \mathbb{R}$$
• (VM3) Distributive with respect to addition of scalars, i.e., $$(\alpha + \beta) x=\alpha x + \beta x, x \in X, \alpha, \beta \in \mathbb{R}$$
• (VM4) Distributive with respect to addition of vectors, i.e., $$\alpha (x+y)=\alpha x + \alpha y, x,y \in X, \alpha \in \mathbb{R}$$
$\blacksquare$

Consider $$\mathbb{C}$$.

• (VA1) Symmetric with respect to addition, i.e., $$(a+ib)+(c+id)=(c+id)+(a+ib)=(a+c) + i(b+d), a+ib,c+id \in \mathbb{C}$$
• (VA2) Existence of identity element, i.e., $$(a+ib)+(0_0i)=a+ib, a+ib, c+id \in \mathbb{C}$$
• (VA3) Existence of inverse element, i.e., $$(a+ib)+(-a-ib)=0+0i, x,0 \in \mathbb{C}$$
• (VA4) Associative with respect to addition, i.e., $$x_1+ix_2+(y_1+iy_2+z_1+iz_2) \\ =(x_1+ix_2+y_1+iy_2)+z_1+iz_2 \\ =(x_1+y_1+z_1)+i(x_2+y_2+z_2), x_1+ix_2,y_1+iy_2,z_1+iz_2 \in \mathbb{C}$$

• (VM1) Associative with respect to scalar multiplication, i.e., $$\alpha (\beta x) = (\alpha \beta) x, x \in X, \alpha, \beta \in \mathbb{R}$$
• (VM2) Existence of identity element, i.e., $$(1+0i)(a+ib)=a+ib, a+ib \in \mathbb{C}$$
• (VM3) Distributive with respect to addition of scalars, i.e., $$(\alpha + \beta) x=\alpha x + \beta x, x \in X, \alpha, \beta \in \mathbb{R}$$
• (VM4) Distributive with respect to addition of vectors, i.e., $$\alpha (x+y)=\alpha x + \alpha y, x,y \in X, \alpha \in \mathbb{R}$$
$\blacksquare$

#### 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