Linear Algebra Quiz: Vector Space Definition

20 questions · exam conditions

0:00

Vector Space DefinitionQuestion 1 of 20

A student attempts to prove that $W = \{(x, y, z) \in \mathbb{R}^3 : x^2 + y^2 = z^2\}$ is a vector space by showing it's closed under scalar multiplication. They argue: "If $(x, y, z) \in W$ , then $x^2 + y^2 = z^2$ . For any scalar $k$ , we have $(kx)^2 + (ky)^2 = k^2x^2 + k^2y^2 = k^2(x^2 + y^2) = k^2z^2 = (kz)^2$ , so $(kx, ky, kz) \in W$ ." Which statement best evaluates this argument?

The argument is correct and sufficient to prove

W

is a vector space, since closure under scalar multiplication is the most restrictive vector space axiom

The argument is incorrect because scalar multiplication of vectors in

W

can produce vectors outside

W

when the scalar is negative

The argument correctly proves closure under scalar multiplication, but this alone is insufficient since

W

fails other vector space axioms, particularly closure under addition

The argument is correct about closure under scalar multiplication, but irrelevant since

W

obviously fails to be a vector space because it's not a linear subspace

← Back to quizzes

Linear Algebra Quiz

Linear Algebra Quiz: Vector Space Definition

Practice Vector Space Definition in Linear Algebra with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Vector Space Definition, giving you a quick way to practice the rules, question types, and explanations that matter most for Linear Algebra.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

The argument is correct and sufficient to prove $W$ is a vector space, since closure under scalar multiplication is the most restrictive vector space axiom
The argument is incorrect because scalar multiplication of vectors in $W$ can produce vectors outside $W$ when the scalar is negative
The argument correctly proves closure under scalar multiplication, but this alone is insufficient since $W$ fails other vector space axioms, particularly closure under addition (correct answer)
The argument is correct about closure under scalar multiplication, but irrelevant since $W$ obviously fails to be a vector space because it's not a linear subspace

Explanation: When determining if a set is a vector space, you must verify that it satisfies all vector space axioms, including closure under addition, closure under scalar multiplication, existence of a zero vector, and existence of additive inverses. The student's mathematical work on scalar multiplication is actually correct. If

(x, y, z) \in W

with

x^2 + y^2 = z^2

, then for any scalar

k

, we have

(kx)^2 + (ky)^2 = k^2(x^2 + y^2) = k^2z^2 = (kz)^2

, so

(kx, ky, kz) \in W

. This proves closure under scalar multiplication. However, this single property is insufficient to conclude that

W

is a vector space. Most critically,

W

fails closure under addition. Consider two vectors in

W

(1, 0, 1)

since

1^2 + 0^2 = 1^2

, and

(0, 1, 1)

since

0^2 + 1^2 = 1^2

. Their sum is

(1, 1, 2)

, but

1^2 + 1^2 = 2 \neq 4 = 2^2

, so the sum is not in

W

. Option A is wrong because one axiom cannot prove vector space status, and closure under scalar multiplication isn't necessarily the "most restrictive." Option B incorrectly claims the scalar multiplication argument fails—it doesn't. Option D dismissively calls the work "irrelevant" when systematic verification of axioms is the proper approach. Study tip: When checking if a set forms a vector space, systematically verify each axiom. Even if some properties hold, a single failed axiom means the set isn't a vector space. Always test closure under addition with concrete examples.

Question 2

Let $T = \{(x, y, z, w) \in \mathbb{R}^4 : x - y + 2z = 0 \text{ and } 2x + z - w = 0\}$ with standard operations from $\mathbb{R}^4$ . A student correctly identifies that $T$ is a vector space and claims it has dimension 2. Which reasoning best supports this dimensional analysis?

The system has 2 equations in 4 unknowns, so by the rank-nullity theorem applied to homogeneous systems, the solution space has dimension $4 - 2 = 2$
The two constraint equations are linearly independent, and each equation reduces the dimension by exactly 1, giving dimension $4 - 2 = 2$ (correct answer)
Solving the system yields 2 free parameters, and since each free parameter contributes 1 to the dimension, the space has dimension 2
The system can be parameterized as $(y + 2z, y, z, 2y + 5z)$ with free variables $y, z$ , confirming that the solution set is 2-dimensional

Explanation: To determine dimension, we need to verify the constraint equations are linearly independent. The coefficient matrix is

\begin{pmatrix} 1 & -1 & 2 & 0 \\ 2 & 0 & 1 & -1 \end{pmatrix}

. Row reducing: the second row becomes

\begin{pmatrix} 0 & 2 & -3 & -1 \end{pmatrix}

, giving rank 2. Since the equations are linearly independent, each reduces the dimension by 1. Choice A assumes rank 2 without verification. Choice C describes the result but not the reasoning. Choice D gives a parameterization (actually

(y + 2z, y, z, 2(y + 2z) + z) = (y + 2z, y, z, 2y + 5z)

) but doesn't explain why this confirms dimension 2.

Question 3

Consider the set $V = \mathbb{R}^2$ with standard vector addition, but with scalar multiplication defined as $c \odot \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} cx \\ y \end{pmatrix}$ for any scalar $c \in \mathbb{R}$ . Which vector space axiom is not satisfied by this structure?

$1\vec{u} = \vec{u}$
$c(\vec{u} + \vec{v}) = c\vec{u} + c\vec{v}$
$(c+d)\vec{u} = c\vec{u} + d\vec{u}$ (correct answer)
Existence of an additive inverse for every vector.

Explanation: To check axiom

(c+d)\vec{u} = c\vec{u} + d\vec{u}

, let

\vec{u} = \begin{pmatrix} x \\ y \end{pmatrix}

. The left side is

(c+d) \odot \vec{u} = \begin{pmatrix} (c+d)x \\ y \end{pmatrix}

. The right side is

c \odot \vec{u} + d \odot \vec{u} = \begin{pmatrix} cx \\ y \end{pmatrix} + \begin{pmatrix} dx \\ y \end{pmatrix} = \begin{pmatrix} (c+d)x \\ 2y \end{pmatrix}

. These are not equal unless

y=0

. Thus, this distributive property fails. The other listed properties hold:

1 \odot \vec{u} = \begin{pmatrix} 1x \\ y \end{pmatrix} = \vec{u}

;

c \odot (\vec{u} + \vec{v})

and

c \odot \vec{u} + c \odot \vec{v}

both equal

\begin{pmatrix} c(u_1+v_1) \\ u_2+v_2 \end{pmatrix}

; standard addition ensures additive inverses exist.

Question 4

Let $V$ be the set of rational numbers, $\mathbb{Q}$ . If we consider $V$ with standard addition and scalar multiplication by scalars from the field of real numbers, $\mathbb{R}$ , which axiom fails, proving that $V$ is not a vector space over $\mathbb{R}$ ?

Closure under scalar multiplication. (correct answer)
Closure under vector addition.
Existence of an additive identity (zero vector).
Commutativity of vector addition.

Explanation: For

V

to be a vector space over

\mathbb{R}

, the product of any scalar

c \in \mathbb{R}

and any vector

\vec{v} \in V

must be in

V

. Let the vector be

1 \in \mathbb{Q}

and the scalar be

\sqrt{2} \in \mathbb{R}

. Their product is

\sqrt{2} \cdot 1 = \sqrt{2}

, which is not a rational number, so it is not in

V

. Therefore, the set is not closed under scalar multiplication. The other axioms listed hold: the sum of two rationals is rational (closure under addition),

0

is a rational number (additive identity), and addition is commutative.

Question 5

Consider the set $V = \mathbb{R}^2$ with standard vector addition. If scalar multiplication is defined as $c \odot \vec{v} = c^2\vec{v}$ for any scalar $c \in \mathbb{R}$ and vector $\vec{v} \in \mathbb{R}^2$ , which vector space axiom fails?

$c(d\vec{v}) = (cd)\vec{v}$
$1\vec{v} = \vec{v}$
$c(\vec{u} + \vec{v}) = c\vec{u} + c\vec{v}$
$(c+d)\vec{v} = c\vec{v} + d\vec{v}$ (correct answer)

Explanation: The distributive property

(c+d)\vec{v} = c\vec{v} + d\vec{v}

fails. The left side is

(c+d) \odot \vec{v} = (c+d)^2\vec{v} = (c^2 + 2cd + d^2)\vec{v}

. The right side is

c \odot \vec{v} + d \odot \vec{v} = c^2\vec{v} + d^2\vec{v} = (c^2+d^2)\vec{v}

. These expressions are not equal in general. The other axioms listed are satisfied:

c \odot (d \odot \vec{v}) = c^2(d^2\vec{v}) = (cd)^2\vec{v} = (cd) \odot \vec{v}

;

1 \odot \vec{v} = 1^2\vec{v} = \vec{v}

; and

c \odot (\vec{u}+\vec{v}) = c^2(\vec{u}+\vec{v}) = c^2\vec{u}+c^2\vec{v} = c\odot\vec{u}+c\odot\vec{v}

Question 6

Let $S$ be the set of all $2 \times 2$ invertible matrices with real entries, equipped with standard matrix addition and scalar multiplication. Which of the following is a valid reason that $S$ is not a vector space over $\mathbb{R}$ ?

The sum of two invertible matrices is not necessarily invertible. (correct answer)
Scalar multiplication of an invertible matrix by a non-zero scalar may result in a non-invertible matrix.
Matrix addition for invertible matrices is not commutative.
The $2 \times 2$ identity matrix $I$ does not satisfy $\vec{A} + I = \vec{A}$ for all $\vec{A} \in S$ .

Explanation: The set

S

is not closed under addition. For example, the identity matrix

I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}

is invertible, and so is

-I = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}

. Their sum is

I + (-I) = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}

, the zero matrix, which has a determinant of 0 and is therefore not invertible. So the sum of two matrices in

S

is not always in

S

. Choice B is incorrect; if

c \neq 0

and

A

is invertible,

\det(cA) = c^2\det(A) \neq 0

, so

cA

is invertible. Choice C is incorrect; matrix addition is always commutative. Choice D is incorrect because the additive identity is the zero matrix, not the identity matrix.

Question 7

The definition of a vector space requires that scalars be taken from a field. A field is a set with addition and multiplication that satisfy certain properties, including the existence of multiplicative inverses for all non-zero elements. Based on this requirement, which of the following sets of scalars cannot be used to define a vector space?

$\mathbb{Q}$ (the rational numbers)
$\mathbb{R}$ (the real numbers)
$\mathbb{C}$ (the complex numbers)
$\mathbb{Z}$ (the integers) (correct answer)

Explanation: The set of integers,

\mathbb{Z}

, is not a field because not every non-zero element has a multiplicative inverse that is also an integer. For example, the multiplicative inverse of

2

1/2

, which is not in

\mathbb{Z}

. Since the scalars for a vector space must form a field,

\mathbb{Z}

cannot be used. The sets of rational, real, and complex numbers are all fields.

Question 8

In any vector space $V$ , several properties can be proven to be true based on the fundamental vector space axioms. Which of the following properties is a theorem derived from the axioms, rather than being an axiom itself?

For any $\vec{u}, \vec{v} \in V$ , $\vec{u} + \vec{v} = \vec{v} + \vec{u}$ .
For any scalar $c$ , $c\vec{0} = \vec{0}$ . (correct answer)
For any $\vec{u} \in V$ , there exists an element $-\vec{u} \in V$ such that $\vec{u} + (-\vec{u}) = \vec{0}$ .
There exists a zero vector $\vec{0} \in V$ such that for all $\vec{u} \in V$ , $\vec{u} + \vec{0} = \vec{u}$ .

Explanation: The property

c\vec{0} = \vec{0}

can be proven from the other axioms. One proof is:

c\vec{0} = c(\vec{0} + \vec{0}) = c\vec{0} + c\vec{0}

. Adding the additive inverse of

c\vec{0}

to both sides gives

\vec{0} = c\vec{0}

. The other three choices are the axiom of commutativity of addition, the axiom of additive inverse, and the axiom of additive identity, respectively.

Question 9

A structure that satisfies all vector space axioms except for the existence of additive inverses is sometimes called a cone. Which of the following, using standard operations over non-negative real scalars ( $c \ge 0$ ), is an example of a cone that is not a vector space?

The set of points on the line $y=x+1$ in $\mathbb{R}^2$ .
The set of all polynomials of degree at most 5 with non-negative coefficients. (correct answer)
The set of all $2 \times 2$ matrices with a determinant of 0.
The set of all rational numbers $\mathbb{Q}$ with scalars from $\mathbb{R}_{\ge 0}$ .

Explanation: Let

V

be the set of polynomials of degree at most 5 with non-negative coefficients. The sum of two such polynomials also has non-negative coefficients. Multiplying by a non-negative scalar preserves non-negative coefficients. The zero polynomial is in this set. All associative and distributive laws hold. However, the additive inverse axiom fails. For example, the polynomial

p(x)=x^2+1

is in

V

, but its additive inverse,

-p(x)=-x^2-1

, does not have non-negative coefficients and is therefore not in

V

. The line in (A) lacks a zero vector. The set in (C) is not closed under addition. The set in (D) is not closed under scalar multiplication.

Question 10

Let $V$ be the set of positive real numbers, $\mathbb{R}^+$ . Define 'vector addition' as standard multiplication ( $x \oplus y = xy$ ) and 'scalar multiplication' as standard multiplication ( $c \odot x = cx$ ) for $c \in \mathbb{R}$ . Why is $V$ not a vector space with these operations?

An additive identity element does not exist in $V$ .
An additive inverse does not exist in $V$ for every element.
The set is not closed under 'vector addition'.
The set is not closed under 'scalar multiplication'. (correct answer)

Explanation: Closure under scalar multiplication requires that for any

c \in \mathbb{R}

and

x \in V

, the result

c \odot x

must be in

V

. Let

x=2 \in \mathbb{R}^+

and

c=-1 \in \mathbb{R}

. Then

c \odot x = (-1)(2) = -2

. Since

-2

is not a positive real number,

-2 \notin V

. Thus, the set is not closed under scalar multiplication. The other axioms do not fail: The additive identity is

1

since

x \cdot 1 = x

. The additive inverse of

x

1/x

since

x \cdot (1/x) = 1

. The product of two positive numbers is positive, so it's closed under vector addition.

Question 11

Let $S$ be the set of all points $(x, y)$ in $\mathbb{R}^2$ that lie on the line defined by the equation $y = 2x + 1$ . Using standard vector addition and scalar multiplication, $S$ is not a vector space. Which axiom failure is the most direct consequence of the line not passing through the origin?

The set does not contain the additive identity element. (correct answer)
Vector addition is not associative for points in the set.
The set is not closed under scalar multiplication.
An additive inverse does not exist for every element in the set.

Explanation: The additive identity (or zero vector) in

\mathbb{R}^2

(0,0)

. For this vector to be in the set

S

, it must satisfy the equation

y=2x+1

. Substituting

(0,0)

gives

0 = 2(0)+1

, which simplifies to

0=1

. This is false, so the zero vector is not in

S

. The failure to contain the additive identity is a primary reason why this set is not a vector space. While it also fails closure under addition and scalar multiplication, the absence of the zero vector is the key feature distinguishing this line from a subspace (like

y=2x

Question 12

Let $U$ be the x-axis and $W$ be the y-axis in $\mathbb{R}^2$ . Consider the set $V = U \cup W$ , which is the set of all points that are on either the x-axis or the y-axis. With standard vector operations, $V$ is not a vector space. Which axiom failure most directly demonstrates this?

Closure under vector addition. (correct answer)
Closure under scalar multiplication.
Existence of an additive identity.
Existence of additive inverses.

Explanation: The set

V

is not closed under vector addition. To show this, take one vector from

U

and one from

W

. For example, let

\vec{u} = (1, 0) \in U

and

\vec{w} = (0, 1) \in W

. Both vectors are in

V

. Their sum is

\vec{u} + \vec{w} = (1, 1)

. The vector

(1, 1)

is not on the x-axis (since its y-component is non-zero) and not on the y-axis (since its x-component is non-zero). Therefore, the sum is not in

V

. The other axioms listed hold: the zero vector

(0,0)

is in

V

; scalar multiples of vectors in

V

remain in

V

; and additive inverses of vectors in

V

are also in

V

Question 13

Consider the set $V$ of all continuous functions $f: [0, 1] \to \mathbb{R}$ such that $\int_{0}^{1} f(x) dx = 0$ . With the standard operations of function addition and scalar multiplication, which of the following statements is true?

$V$ is not a vector space because it is not closed under addition.
$V$ is not a vector space because it does not contain the zero vector.
$V$ is a vector space. (correct answer)
$V$ is not a vector space because it is not closed under scalar multiplication.

Explanation:

V

is a vector space because it satisfies all the required axioms. Specifically, it is a subspace of the vector space of all continuous functions on

[0,1]

. 1) If

f, g \in V

, then

\int (f+g) dx = \int f dx + \int g dx = 0+0=0

, so it's closed under addition. 2) If

f \in V

and

c \in \mathbb{R}

, then

\int (cf) dx = c \int f dx = c \cdot 0=0

, so it's closed under scalar multiplication. 3) The zero function,

z(x)=0

, has an integral of 0, so the zero vector is in

V

. All other axioms are inherited from the parent space of continuous functions.

Question 14

Let $V = \mathbb{R}^2$ with standard vector addition. Define a non-standard scalar multiplication by $c \odot \vec{v} = \vec{0}$ for all scalars $c \in \mathbb{R}$ and all vectors $\vec{v} \in V$ . Which vector space axiom is not satisfied by this structure?

Closure under scalar multiplication.
The distributive property $(c+d)\vec{v} = c\vec{v} + d\vec{v}$ .
The multiplicative identity property $1\vec{v} = \vec{v}$ . (correct answer)
The existence of an additive identity $\vec{0}$ .

Explanation: The multiplicative identity axiom requires that for any vector

\vec{v}

, multiplying by the scalar

1

returns

\vec{v}

. With the given operation,

1 \odot \vec{v} = \vec{0}

. This is only equal to

\vec{v}

\vec{v}

is the zero vector itself. For any non-zero vector, the axiom

1\vec{v} = \vec{v}

fails. The other listed axioms hold: the result of scalar multiplication is always

\vec{0}

, which is in

\mathbb{R}^2

, so it is closed. Both sides of the distributive law evaluate to

\vec{0}

. The additive identity

\vec{0}

exists in

\mathbb{R}^2

Question 15

Consider the set $V = \{(x, y, z) \in \mathbb{R}^3 : x + 2y - z = 1\}$ with the standard operations of vector addition and scalar multiplication from $\mathbb{R}^3$ . Which statement correctly identifies whether $V$ is a vector space and provides the most fundamental reason?

$V$ is not a vector space because it fails the closure property under vector addition, since adding two vectors in $V$ does not necessarily produce a vector in $V$ (correct answer)
$V$ is not a vector space because it lacks a zero vector, since $(0, 0, 0)$ does not satisfy the constraint $x + 2y - z = 1$
$V$ is a vector space because it is a subset of $\mathbb{R}^3$ and inherits all necessary properties from the ambient space
$V$ is a vector space because it forms a two-dimensional plane in $\mathbb{R}^3$ and contains infinitely many vectors

Explanation: For

V

to be a vector space, it must be closed under addition. Take vectors

(1, 0, 0)

and

(0, 0.5, 0)

, both in

V

since they satisfy

x + 2y - z = 1

. Their sum is

(1, 0.5, 0)

, but

1 + 2(0.5) - 0 = 2 \neq 1

, so the sum is not in

V

. Choice B is also correct but closure failure is more fundamental. Choice C is wrong because subsets don't automatically inherit vector space structure. Choice D confuses geometric properties with algebraic requirements.

Question 16

Consider the set $S = \{f: \mathbb{R} \to \mathbb{R} : f(x+1) = f(x) + 1 \text{ for all } x \in \mathbb{R}\}$ with pointwise addition and scalar multiplication of functions. Which statement best explains why $S$ fails to be a vector space?

The zero function is not in $S$ because $0(x+1) = 0 \neq 1 = 0(x) + 1$ , violating the required functional equation (correct answer)
The set $S$ fails closure under scalar multiplication because if $f \in S$ and $k \neq 1$ , then $kf$ does not satisfy the functional equation
The set $S$ fails closure under addition because the sum of two functions satisfying $f(x+1) = f(x) + 1$ produces $g(x+1) = g(x) + 2$
Functions in $S$ cannot have additive inverses because the functional equation forces all functions to be strictly increasing

Explanation: For

S

to be a vector space, it must contain the zero function. But the zero function

0(x) = 0

for all

x

would need to satisfy

0(x+1) = 0(x) + 1

, which gives

0 = 1

, a contradiction. Choice B is wrong because if

f(x+1) = f(x) + 1

, then

(kf)(x+1) = kf(x+1) = k(f(x) + 1) = kf(x) + k

, so

kf \in S

only when

k = 1

, but this isn't the fundamental issue. Choice C is wrong because

(f+g)(x+1) = f(x+1) + g(x+1) = f(x) + 1 + g(x) + 1 = (f+g)(x) + 2

. Choice D is incorrect about the increasing property.

Question 17

Let $V$ be the set of sequences $(a_1, a_2, a_3, \ldots)$ of real numbers such that $\sum_{n=1}^{\infty} |a_n|^2 < \infty$ , with componentwise addition and scalar multiplication. A student claims that $V$ is not a vector space because "infinite sequences cannot form vector spaces." Which response most accurately addresses this claim?

The student is correct because vector spaces must be finite-dimensional, and infinite sequences necessarily create infinite-dimensional structures that violate the definition
The student is incorrect because $V$ does form a vector space; the convergence condition ensures that all vector space axioms are satisfied, including closure under the defined operations (correct answer)
The student is partially correct because while infinite sequences can form vector spaces, this particular set fails closure under addition since the sum of two convergent series may diverge
The student is incorrect about infinite sequences in general, but correct about this specific set because scalar multiplication of a convergent sequence can produce a divergent sequence

Explanation: The set

V

is actually the Hilbert space

\ell^2

, which is indeed a vector space. The condition

\sum_{n=1}^{\infty} |a_n|^2 < \infty

ensures closure: if

\sum |a_n|^2

and

\sum |b_n|^2

converge, then

\sum |a_n + b_n|^2

converges by the triangle inequality, and

\sum |ka_n|^2 = |k|^2 \sum |a_n|^2

converges for scalar

k

. Choice A is wrong because vector spaces can be infinite-dimensional. Choice C is wrong because the sum of two

\ell^2

sequences is in

\ell^2

. Choice D is wrong because scalar multiplication preserves the

\ell^2

property.

Question 18

Consider the set $M = \left\{\begin{pmatrix} a & b \\ c & d \end{pmatrix} : a, b, c, d \in \mathbb{Q}, \text{ and } ad - bc = 1\right\}$ with standard matrix addition and scalar multiplication over $\mathbb{Q}$ . Which statement correctly explains why $M$ fails to be a vector space?

$M$ fails closure under scalar multiplication because multiplying a matrix with determinant 1 by a rational scalar $k \neq \pm 1$ produces a matrix with determinant $k^2 \neq 1$
$M$ fails closure under addition because the sum of two matrices with determinant 1 does not necessarily have determinant 1
$M$ does not contain the zero matrix, since the zero matrix has determinant 0, not 1, violating the constraint $ad - bc = 1$ (correct answer)
$M$ fails commutativity of addition because matrices with determinant constraints do not necessarily commute under addition

Explanation: For

M

to be a vector space, it must contain the additive identity (zero matrix). But the zero matrix has determinant 0, not 1, so

\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \notin M

. This immediately disqualifies

M

as a vector space. Choice A is also correct (if

\det(A) = 1

, then

\det(kA) = k^2 \det(A) = k^2 \neq 1

for

k \neq \pm 1

), and Choice B is also true, but the absence of the zero element is the most fundamental violation. Choice D is incorrect because matrix addition is always commutative regardless of constraints on entries.

Question 19

Consider the set $N = \{f \in C(\mathbb{R}) : f(x) = f(-x) \text{ for all } x \in \mathbb{R}\}$ where $C(\mathbb{R})$ denotes continuous functions from $\mathbb{R}$ to $\mathbb{R}$ . Under pointwise operations, $N$ forms a vector space of even functions. Which of the following sets is most analogous to $N$ in terms of vector space structure?

The set of all differentiable functions $g: \mathbb{R} \to \mathbb{R}$ such that $g'(0) = 1$ under pointwise operations
The set of all polynomials $p(x) \in \mathbb{R}[x]$ such that $p(1) = 0$ under standard polynomial operations
The set of all convergent sequences of real numbers under componentwise operations
The set of all $3 \times 3$ symmetric matrices with real entries under standard matrix operations (correct answer)

Explanation: When analyzing vector space analogies, you need to focus on the fundamental structural properties that define how elements behave under the given operations. The set

N

of even functions has a key characteristic: it's defined by a linear constraint (the symmetry condition

f(x) = f(-x)

) that's preserved under vector addition and scalar multiplication. The correct answer is D because symmetric matrices form a vector space through the same structural principle. If

A

and

B

are symmetric matrices, then

A + B

is symmetric, and

cA

is symmetric for any scalar

c

. The symmetry condition

A^T = A

is a linear constraint that's preserved under the vector space operations, just like the even function condition. Option A fails because

g'(0) = 1

is not preserved under addition—if two functions both have derivative 1 at zero, their sum has derivative 2 at zero. This violates the closure property required for vector spaces. Option B also fails closure: if

p(1) = 0

and

q(1) = 0

, then

(p + q)(1) = 0

, which works, but the zero polynomial doesn't satisfy

p(1) = 0

, violating the zero vector requirement. Option C describes a perfectly valid vector space, but it lacks the constraining structural feature that characterizes

N

—there's no special symmetry or constraint condition being imposed. Study tip: When comparing vector spaces, look for preserved structural constraints rather than just checking if something forms a vector space. The most analogous spaces share similar types of defining conditions.

Question 20

Consider the set $U = \{p(x) \in \mathbb{R}[x] : p(0) = 0 \text{ and } p'(0) = 0\}$ where $\mathbb{R}[x]$ denotes polynomials with real coefficients and $p'(x)$ denotes the derivative. Under standard polynomial addition and scalar multiplication, which statement correctly characterizes $U$ as a vector space?

$U$ is not a vector space because the constraint $p'(0) = 0$ is not preserved under polynomial addition when combined with $p(0) = 0$
$U$ is a vector space, and it consists precisely of polynomials of the form $a_2x^2 + a_3x^3 + a_4x^4 + \cdots$ where $a_i \in \mathbb{R}$ (correct answer)
$U$ is not a vector space because it lacks a multiplicative identity element, since the polynomial $1$ does not satisfy $p(0) = 0$
$U$ is a vector space, but only when restricted to polynomials of degree at most $n$ for some fixed $n$ , since infinite-degree constraints cannot be maintained

Explanation:

U

is indeed a vector space. A polynomial

p(x) = a_0 + a_1x + a_2x^2 + \cdots

satisfies

p(0) = 0

iff

a_0 = 0

, and

p'(0) = 0

iff

a_1 = 0

. So

U

consists of polynomials

a_2x^2 + a_3x^3 + \cdots

. This set is closed under addition and scalar multiplication, contains the zero polynomial, and satisfies all vector space axioms. Choice A is wrong because both constraints are preserved under addition. Choice C confuses vector spaces with algebras (vector spaces don't need multiplicative identity). Choice D is wrong because the constraint is maintained for polynomials of any degree.