Linear Algebra Quiz: Subspaces And Tests

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Subspaces And TestsQuestion 1 of 20

Let $A$ be a fixed $m \times n$ matrix and $\mathbf{b}$ be a fixed non-zero vector in $\mathbb{R}^m$ . Consider the sets $S_1 = \{\mathbf{x} \in \mathbb{R}^n \mid A\mathbf{x} = \mathbf{0}\}$ and $S_2 = \{\mathbf{x} \in \mathbb{R}^n \mid A\mathbf{x} = \mathbf{b}\}$ . Which statement is correct?

S_1

is a subspace of

\mathbb{R}^n

, but

S_2

is not a subspace because the zero vector is not in

S_2

S_2

is a subspace of

\mathbb{R}^n

, but

S_1

is not a subspace because it may only contain the zero vector.

Both

S_1

and

S_2

are subspaces of

\mathbb{R}^n

because they are solution sets to systems of linear equations.

Neither

S_1

nor

S_2

is a subspace of

\mathbb{R}^n

unless the matrix

A

is invertible.

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Linear Algebra Quiz

Linear Algebra Quiz: Subspaces And Tests

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

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

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All questions

Question 1

$S_1$ is a subspace of $\mathbb{R}^n$ , but $S_2$ is not a subspace because the zero vector is not in $S_2$ . (correct answer)
$S_2$ is a subspace of $\mathbb{R}^n$ , but $S_1$ is not a subspace because it may only contain the zero vector.
Both $S_1$ and $S_2$ are subspaces of $\mathbb{R}^n$ because they are solution sets to systems of linear equations.
Neither $S_1$ nor $S_2$ is a subspace of $\mathbb{R}^n$ unless the matrix $A$ is invertible.

Explanation:

S_1

is the null space of matrix

A

, which is always a subspace of

\mathbb{R}^n

S_2

is the solution set to a non-homogeneous linear system because

\mathbf{b} \neq \mathbf{0}

. For

\mathbf{x} = \mathbf{0}

, we have

A\mathbf{0} = \mathbf{0}

. Since

\mathbf{b} \neq \mathbf{0}

A\mathbf{0} \neq \mathbf{b}

, which means the zero vector

\mathbf{0}

is not in

S_2

. Therefore,

S_2

cannot be a subspace.

Question 2

Let $\mathbf{p}=(1,2,0)$ , $\mathbf{u}=(1,0,1)$ , and $\mathbf{v}=(0,1,-1)$ be vectors in $\mathbb{R}^3$ . Let $W = \{\mathbf{p} + s\mathbf{u} + t\mathbf{v} \mid s, t \in \mathbb{R}\}$ . Why is $W$ not a subspace of $\mathbb{R}^3$ ?

$W$ does not contain the zero vector.
The vectors $\mathbf{u}$ and $\mathbf{v}$ are linearly dependent.
$W$ is not closed under scalar multiplication. (correct answer)
The set $W$ is a line, and a line is never a subspace.

Explanation: The set

W

describes a plane in

\mathbb{R}^3

. To check if it is a subspace, we must check the three axioms. First, note that for

s=-1, t=-2

, we have

\mathbf{p} + s\mathbf{u} + t\mathbf{v} = (1,2,0) - (1,0,1) - 2(0,1,-1) = (0,0,1) \neq \mathbf{0}

. It can be shown

\mathbf{0} \notin W

. However, a more fundamental failure is closure. Let

\mathbf{w}_1 = \mathbf{p} + s_1\mathbf{u} + t_1\mathbf{v} \in W

. Consider the scalar multiple

2\mathbf{w}_1 = 2\mathbf{p} + 2s_1\mathbf{u} + 2t_1\mathbf{v}

. This vector is not in the form

\mathbf{p} + s\mathbf{u} + t\mathbf{v}

because of the

2\mathbf{p}

term. Thus,

W

is not closed under scalar multiplication. It is also not closed under addition.

Question 3

Let $V$ be the vector space of all $2 \times 2$ real matrices, and define $W = \{A \in V : A^T = -A \text{ and } \text{tr}(A) = 0\}$ . What can be concluded about the relationship between these two conditions?

The conditions are incompatible except for the zero matrix, since skew-symmetric matrices cannot have zero trace
Both conditions are necessary and independent, giving $W$ dimension 0 with only the zero matrix
The trace condition is redundant because skew-symmetric matrices automatically have trace zero, so $W$ has dimension 3
The trace condition is redundant because skew-symmetric matrices automatically have trace zero, so $W$ has dimension 1 (correct answer)

Explanation: When analyzing subspaces defined by multiple conditions, you need to determine whether those conditions are independent or if one implies the other. This question tests your understanding of skew-symmetric matrices and their fundamental properties. A

2 \times 2

skew-symmetric matrix has the form

A^T = -A

, which means

A = \begin{pmatrix} 0 & a \\ -a & 0 \end{pmatrix}

for some real number

a

. Notice that the diagonal entries must be zero (since

a_{ii} = -a_{ii}

implies

a_{ii} = 0

), and off-diagonal entries are negatives of each other. The trace of this matrix is

\text{tr}(A) = 0 + 0 = 0

. This isn't coincidental—every skew-symmetric matrix automatically has zero trace because its diagonal entries must all be zero. Therefore, the condition

\text{tr}(A) = 0

is redundant when combined with

A^T = -A

. Since

W

is effectively just the space of

2 \times 2

skew-symmetric matrices, and these have the form

\begin{pmatrix} 0 & a \\ -a & 0 \end{pmatrix}

, the space is one-dimensional (parameterized by the single variable

a

). Choice A is wrong because skew-symmetric matrices do have zero trace—they're perfectly compatible conditions. Choice B incorrectly claims both conditions are independent and that

W

has dimension 0. Choice C correctly identifies that the trace condition is redundant but incorrectly states the dimension as 3 (confusing this with

3 \times 3

skew-symmetric matrices). Remember: for any

n \times n

skew-symmetric matrix, the trace is automatically zero due to the zero diagonal entries.

Question 4

Let $W_1 = \text{span}\{(1, 2, 0, 1), (2, 1, 1, 0)\}$ and $W_2 = \{(x, y, z, w) \in \mathbb{R}^4 : x - y + z = 0\}$ . To determine whether $W_1 \subseteq W_2$ , what is the most efficient approach?

Compute the intersection $W_1 \cap W_2$ and verify it equals $W_1$ by comparing dimensions
Find a basis for $W_1$ , then verify each basis element satisfies the constraint equation for $W_2$
Check whether both spanning vectors of $W_1$ satisfy the constraint defining $W_2$ , since this is necessary and sufficient (correct answer)
Parameterize all vectors in $W_1$ and substitute into the constraint equation to verify it holds for all parameter values

Explanation: When you encounter subset questions involving a span and a constraint-defined subspace, you need to understand that every vector in the span can be written as a linear combination of the spanning vectors. This means if all spanning vectors satisfy the constraint, then every vector in the span will too. Option C is correct because it leverages this fundamental property. Since

W_1 = \text{span}\{(1, 2, 0, 1), (2, 1, 1, 0)\}

, every vector in

W_1

has the form

a(1, 2, 0, 1) + b(2, 1, 1, 0)

for scalars

a, b

. If both spanning vectors satisfy

x - y + z = 0

, then any linear combination will too. Let's verify: for

(1, 2, 0, 1)

1 - 2 + 0 = -1 \neq 0

. This vector doesn't satisfy the constraint, so

W_1 \not\subseteq W_2

. Option A is inefficient because computing intersections and comparing dimensions requires more work than simply checking the spanning vectors. Option B contains a subtle error: finding "a basis" suggests you might need to reduce the spanning set, but you can check the given spanning vectors directly without basis reduction. Option D is unnecessarily complicated—parameterizing all vectors means writing

a(1, 2, 0, 1) + b(2, 1, 1, 0) = (a + 2b, 2a + b, b, a)

and substituting into the constraint, which is algebraically more involved than checking the two spanning vectors individually. Strategy tip: For span-subset questions, always check if the spanning vectors satisfy the constraint first—it's the most direct path to your answer.

Question 5

Consider the set $T = \{f \in C^\infty(\mathbb{R}) : f''(x) + 4f'(x) + 4f(x) = 0\}$ where $C^\infty(\mathbb{R})$ denotes the vector space of infinitely differentiable functions on $\mathbb{R}$ . Which property of $T$ requires the most sophisticated verification to confirm it is a subspace?

Verifying that the zero function satisfies the differential equation, since this involves confirming that derivatives of zero are zero
Confirming closure under addition requires showing that if two functions satisfy the differential equation, their sum does too
Establishing closure under scalar multiplication, since scaling affects the derivatives in the differential equation
Proving that $T$ is non-empty, since the existence of non-trivial solutions to the differential equation is not immediately obvious (correct answer)

Explanation: While options A, B, and C involve routine verification using linearity of differentiation, option D requires solving the differential equation f'' + 4f' + 4f = 0 to prove solutions exist. The characteristic equation is r² + 4r + 4 = 0, which factors as (r + 2)² = 0, giving a repeated root r = -2. This yields the general solution f(x) = (c₁ + c₂x)e^(-2x) for constants c₁, c₂. Verifying this solution exists requires knowledge of differential equations theory, whereas the other properties follow directly from the linearity of the differential operator L[f] = f'' + 4f' + 4f.

Question 6

Let $S = \{A \in M_{2 \times 2}(\mathbb{R}) : A \begin{pmatrix} 1 \\ 2 \end{pmatrix} = 3 \begin{pmatrix} 1 \\ 2 \end{pmatrix}\}$ . A student argues that $S$ is not a subspace because "eigenvalue conditions are not linear constraints." What is wrong with this reasoning?

The reasoning is correct; eigenvalue conditions involve products $Av = \lambda v$ which are not linear in the matrix entries
The condition $A \begin{pmatrix} 1 \\ 2 \end{pmatrix} = 3 \begin{pmatrix} 1 \\ 2 \end{pmatrix}$ is actually a system of linear equations in the entries of $A$ (correct answer)
Eigenvalue conditions can be nonlinear, but this particular condition happens to define a subspace by coincidence
The student's reasoning fails because eigenvalues are always real numbers, making the constraints automatically linear

Explanation: The condition A[[1],[2]] = 3[[1],[2]] translates to [[a,b],[c,d]][[1],[2]] = [[3],[6]], which gives the system: a + 2b = 3 and c + 2d = 6. These are linear equations in the entries a, b, c, d of matrix A. While eigenvalue problems in general can involve nonlinear relationships (like det(A - λI) = 0), this specific condition with a fixed eigenvalue and eigenvector gives linear constraints on the matrix entries. Therefore S is indeed a subspace, and the student's reasoning is incorrect because they failed to recognize that fixing λ and v makes the constraint Av = λv linear in A.

Question 7

Consider the sets $A = \{(x, y, z) \in \mathbb{R}^3 : 2x + y - z = 0\}$ and $B = \{(x, y, z) \in \mathbb{R}^3 : x + y = 1, z = 0\}$ . A student claims that $A \cap B$ must be a subspace since "the intersection of any two subspaces is always a subspace." What error does this reasoning contain?

The intersection of subspaces is not always a subspace; this is only true under special conditions
Set $B$ is not a subspace, so the general theorem about intersections of subspaces does not apply (correct answer)
Set $A$ is not a subspace, so we cannot use properties of subspace intersections
The reasoning is correct, but $A \cap B = \emptyset$ , so the intersection is not a subspace

Explanation: The general theorem states that the intersection of subspaces is always a subspace, but this requires both sets to actually be subspaces. Set A = {(x,y,z) : 2x + y - z = 0} is indeed a subspace (it's the null space of the linear functional 2x + y - z). However, set B = {(x,y,z) : x + y = 1, z = 0} is not a subspace because it doesn't contain the zero vector (since 0 + 0 ≠ 1). Therefore, the theorem about intersections of subspaces cannot be applied here. The student's error is assuming that B is a subspace when it is not.

Question 8

Define $U = \{(a, b, c, d) \in \mathbb{R}^4 : a + b + c + d = 1\}$ and $V = \{(a, b, c, d) \in \mathbb{R}^4 : a + b + c + d = 0\}$ . A student claims that $U \cup V$ forms a subspace of $\mathbb{R}^4$ . What is the most direct way to show this claim is false?

Show that $U \cup V$ fails closure under addition by finding vectors $u_1, u_2 \in U$ such that $u_1 + u_2 \notin U \cup V$ (correct answer)
Show that $U \cup V$ fails closure under scalar multiplication by finding $u \in U$ and scalar $c$ such that $cu \notin U \cup V$
Show that $U$ is not a subspace because it does not contain the zero vector
Show that both $U$ and $V$ are themselves not subspaces, so their union cannot be a subspace either

Explanation: The most direct approach is to show closure fails under addition. Take u₁ = (1,0,0,0) ∈ U and u₂ = (0,1,0,0) ∈ U. Then u₁ + u₂ = (1,1,0,0), and 1+1+0+0 = 2. Since 2 ≠ 1, we have (1,1,0,0) ∉ U, and since 2 ≠ 0, we have (1,1,0,0) ∉ V. Therefore u₁ + u₂ ∉ U ∪ V, showing U ∪ V fails closure under addition. Note that V is actually a subspace (the zero vector is in V), but U is not since it doesn't contain the zero vector.

Question 9

Which of the following subsets of $\mathbb{R}^3$ is a subspace of $\mathbb{R}^3$ ?

$W_1 = \{(x, y, z) \in \mathbb{R}^3 \mid x + y - z = 4\}$
$W_2 = \{(x, y, z) \in \mathbb{R}^3 \mid x \ge 0, y \ge 0, z \ge 0\}$
$W_3 = \{(x, y, z) \in \mathbb{R}^3 \mid 2x - y = 0 \text{ and } 3y + z = 0\}$ (correct answer)
$W_4 = \{(x, y, z) \in \mathbb{R}^3 \mid x, y, \text{ and } z \text{ are rational numbers}\}$

Explanation:

W_3

is a subspace. It represents the intersection of two planes through the origin, which is a line through the origin. It contains the zero vector

(0,0,0)

. It is closed under addition and scalar multiplication because the defining equations are linear and homogeneous.

W_1

is not a subspace because it does not contain the zero vector.

W_2

is not a subspace because it is not closed under multiplication by negative scalars.

W_4

is not a subspace because it is not closed under multiplication by irrational scalars.

Question 10

Let $W$ be the subset of $\mathbb{R}^2$ defined by $W = \{(x, y) \mid xy = 0\}$ . Which statement correctly explains why $W$ is not a subspace of $\mathbb{R}^2$ ?

$W$ does not contain the zero vector $\mathbf{0}=(0,0)$ .
$W$ is not closed under scalar multiplication.
$W$ is not closed under vector addition. (correct answer)
The condition $xy=0$ is nonlinear, and subspaces must be defined by linear equations.

Explanation: To be a subspace, a set must be closed under vector addition. Let

\mathbf{u} = (1, 0)

and

\mathbf{v} = (0, 1)

. Since

1 \cdot 0 = 0

\mathbf{u} \in W

. Since

0 \cdot 1 = 0

\mathbf{v} \in W

. However, their sum is

\mathbf{u} + \mathbf{v} = (1, 1)

. For this vector,

x=1

and

y=1

, so

xy = 1 \neq 0

. Therefore,

\mathbf{u} + \mathbf{v} \notin W

, and

W

is not closed under addition.

Question 11

Let $P_3$ be the vector space of polynomials of degree at most 3. Let $S$ be the subset of $P_3$ defined by $S = \{ p(x) \in P_3 \mid p(0) = 2p(1) \}$ . Which statement about $S$ is correct?

$S$ is not a subspace because the zero polynomial is not in $S$ .
$S$ is not a subspace because it is not closed under scalar multiplication.
$S$ is not a subspace because it is not closed under addition.
$S$ is a subspace of $P_3$ . (correct answer)

Explanation:

S

is a subspace. We check the three conditions. (1) Zero vector: The zero polynomial is

z(x)=0

z(0)=0

and

2z(1)=2(0)=0

, so

z(0)=2z(1)

and

z(x) \in S

. (2) Closure under addition: Let

p, q \in S

. Then

p(0)=2p(1)

and

q(0)=2q(1)

. For the sum

p+q

, we have

(p+q)(0) = p(0)+q(0) = 2p(1)+2q(1) = 2(p(1)+q(1)) = 2(p+q)(1)

. Thus,

p+q \in S

. (3) Closure under scalar multiplication: Let

p \in S

and

c

be a scalar. Then

(cp)(0)=c \cdot p(0) = c \cdot (2p(1)) = 2(c \cdot p(1)) = 2(cp)(1)

. Thus,

cp \in S

. Since all conditions hold,

S

is a subspace.

Question 12

Let $V = C(\mathbb{R})$ be the vector space of all continuous functions from $\mathbb{R}$ to $\mathbb{R}$ . Which of the following subsets of $V$ is a subspace?

The set of all functions $f$ such that $f(x) \ge 0$ for all $x$ .
The set of all functions $f$ such that $f(7) = f(1)$ . (correct answer)
The set of all polynomial functions of degree exactly 4.
The set of all functions $f$ such that $\int_0^1 f(x) dx = 3$ .

Explanation: Let

S = \{ f \in V \mid f(7) = f(1) \}

. (1) The zero function satisfies

0=0

. (2) If

f,g \in S

, then

f(7)=f(1)

and

g(7)=g(1)

. Then

(f+g)(7) = f(7)+g(7) = f(1)+g(1) = (f+g)(1)

, so

f+g \in S

. (3) If

f \in S

, then

(cf)(7) = c f(7) = c f(1) = (cf)(1)

, so

cf \in S

. Thus,

S

is a subspace. Choice A fails closure under multiplication by negative scalars. Choice C does not contain the zero polynomial and fails closure. Choice D does not contain the zero function.

Question 13

Let $W$ be the set of all vectors in $\mathbb{R}^4$ whose components sum to zero. Which statement is true?

$W$ is a subspace of $\mathbb{R}^4$ . (correct answer)
$W$ is not a subspace because it is not closed under scalar multiplication.
$W$ is not a subspace because it is not closed under vector addition.
$W$ is not a subspace because it does not contain the zero vector.

Explanation:

W = \{(x_1, x_2, x_3, x_4) \in \mathbb{R}^4 \mid x_1+x_2+x_3+x_4=0\}

. This is a subspace. We check the axioms. (1) Zero vector: The components of

\mathbf{0}=(0,0,0,0)

sum to 0, so

\mathbf{0} \in W

. (2) Addition: Let

\mathbf{u}, \mathbf{v} \in W

. Then

\sum u_i = 0

and

\sum v_i = 0

. For the sum

\mathbf{u}+\mathbf{v}

, the sum of components is

\sum (u_i+v_i) = \sum u_i + \sum v_i = 0+0=0

. So

\mathbf{u}+\mathbf{v} \in W

. (3) Scalar multiplication: Let

\mathbf{u} \in W

and

c

be a scalar. The sum of components for

c\mathbf{u}

\sum (cu_i) = c (\sum u_i) = c(0)=0

. So

c\mathbf{u} \in W

. All three conditions hold.

Question 14

Let $V$ be the vector space of $2 \times 3$ matrices. Which of the following subsets of $V$ is NOT a subspace?

The set of matrices where the first row is all zeros.
The set of matrices $A$ such that $A \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$ .
The set of matrices where the sum of all entries is 0.
The set of matrices with at least one zero entry. (correct answer)

Explanation: The set of matrices with at least one zero entry is not a subspace because it is not closed under addition. For example, let

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

and

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

. Both

A

and

B

have a zero entry. Their sum is

A+B = \begin{pmatrix} 2 & 2 & 1 \\ 2 & 2 & 1 \end{pmatrix}

, which has no zero entries. Therefore, the set is not closed under addition. The other three options describe valid subspaces.

Question 15

Consider the set $S = \{(x,y) \in \mathbb{R}^2 \mid x^2 - y^2 = 0\}$ . Which of the following claims about $S$ is true?

$S$ is a subspace of $\mathbb{R}^2$ .
$S$ contains the zero vector and is closed under scalar multiplication, but is not closed under addition. (correct answer)
$S$ is closed under addition and scalar multiplication, but does not contain the zero vector.
$S$ contains the zero vector and is closed under addition, but is not closed under scalar multiplication.

Explanation: The condition

x^2 - y^2 = 0

is equivalent to

(x-y)(x+y)=0

, which means

y=x

y=-x

. So

S

is the union of two lines through the origin. (1) Zero vector:

(0,0)

is in

S

since

0^2-0^2=0

. (2) Scalar multiplication: If

(x,y) \in S

, then

x^2-y^2=0

. For any scalar

c

(cx)^2 - (cy)^2 = c^2(x^2-y^2) = c^2(0)=0

. So

(cx,cy) \in S

S

is closed under scalar multiplication. (3) Addition: Let

\mathbf{u}=(1,1) \in S

and

\mathbf{v}=(1,-1) \in S

. Their sum is

\mathbf{u}+\mathbf{v}=(2,0)

. For this vector,

2^2-0^2 = 4 \neq 0

, so

\mathbf{u}+\mathbf{v} \notin S

. Thus,

S

is not closed under addition.

Question 16

Let $U$ and $W$ be subspaces of a vector space $V$ . Which of the following sets is guaranteed to be a subspace of $V$ ?

The union $U \cup W$ .
The intersection $U \cap W$ . (correct answer)
The set difference $U \setminus W = \{\mathbf{v} \in U \mid \mathbf{v} \notin W\}$ .
The set of vectors not in $U$ , denoted $V \setminus U$ .

Explanation: The intersection of two subspaces is always a subspace. If

\mathbf{u}, \mathbf{v} \in U \cap W

, then

\mathbf{u}, \mathbf{v}

are in both

U

and

W

. Since

U

and

W

are subspaces,

\mathbf{u}+\mathbf{v}

and

c\mathbf{u}

are in both

U

and

W

, and thus in their intersection. Also,

\mathbf{0}

is in both

U

and

W

. The union

U \cup W

is not generally a subspace. The set differences

U \setminus W

and

V \setminus U

are not subspaces as they do not contain the zero vector.

Question 17

Let $M_{2\times2}$ be the vector space of $2 \times 2$ matrices. Let $S$ be the subset of $M_{2\times2}$ consisting of all singular matrices. Why is $S$ not a subspace of $M_{2\times2}$ ?

$S$ does not contain the zero matrix.
$S$ is not closed under scalar multiplication.
$S$ is not closed under matrix multiplication.
$S$ is not closed under matrix addition. (correct answer)

Explanation: A matrix is singular if its determinant is zero. Let

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

and

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

. We have

\det(A) = 0

and

\det(B) = 0

, so both

A

and

B

are in

S

. Their sum is

A+B = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I

. The determinant of the identity matrix

I

1

, so

A+B

is not singular. Therefore,

A+B \notin S

, which shows

S

is not closed under matrix addition.

Question 18

Let $W = \{p(x) \in P_4(\mathbb{R}) : p(-1) = 0, p(1) = 0, \text{ and } p'(0) = 0\}$ where $P_4(\mathbb{R})$ is the space of polynomials of degree at most 4. What is the dimension of $W$ ?

$\dim(W) = 1$ , since the three constraints eliminate 4 degrees of freedom from the 5-dimensional space
$\dim(W) = 3$ , since the constraints are independent and reduce the dimension from 5 to 2
$\dim(W) = 2$ , since we have 3 linear constraints on a 5-dimensional space (correct answer)
$\dim(W) = 4$ , since polynomial evaluation constraints don't affect the underlying polynomial degree

Explanation: When you encounter a subspace defined by constraints, you need to count how many independent linear conditions restrict the space and subtract from the original dimension. The space

P_4(\mathbb{R})

has dimension 5, with basis

\{1, x, x^2, x^3, x^4\}

. Any polynomial can be written as

p(x) = a_0 + a_1x + a_2x^2 + a_3x^3 + a_4x^4

. The three constraints translate to:

$p(-1) = 0$ : $a_0 - a_1 + a_2 - a_3 + a_4 = 0$
$p(1) = 0$ : $a_0 + a_1 + a_2 + a_3 + a_4 = 0$
$p'(0) = 0$ : Since $p'(x) = a_1 + 2a_2x + 3a_3x^2 + 4a_4x^3$ , we get $a_1 = 0$

These give us a system of three linear equations in five unknowns. The key insight is that these constraints are linearly independent—none can be derived from the others. Therefore, they reduce the dimension by exactly 3, giving us

\dim(W) = 5 - 3 = 2

. Option A incorrectly states the constraints eliminate 4 degrees of freedom and claims the result is dimension 1. Option B has the calculation backwards, saying 3 independent constraints somehow leave dimension 3. Option D ignores that evaluation constraints absolutely do restrict the space—they're linear conditions on the coefficients. Study tip: For constraint problems, write out what each condition means for the coefficients, verify the constraints are independent, then subtract the number of independent constraints from the original dimension. The dimension always equals (original dimension) - (number of independent linear constraints).

Question 19

Define $H = \{(x, y, z) \in \mathbb{R}^3 : x^2 + y^2 \leq 1, z = 0\}$ and $K = \{(x, y, z) \in \mathbb{R}^3 : z = 0\}$ . Which statement correctly analyzes these sets as potential subspaces?

Both $H$ and $K$ are subspaces, since they are both subsets of the $xy$ -plane with $z = 0$
$H$ is not a subspace because it fails closure under scalar multiplication, while $K$ is a subspace (correct answer)
$H$ is not a subspace because it fails closure under addition, while $K$ fails closure under scalar multiplication
Neither $H$ nor $K$ is a subspace because both involve quadratic constraints on the coordinates

Explanation: Set K = {(x,y,z) ∈ ℝ³ : z = 0} is the xy-plane, which is clearly a subspace (it's the null space of the linear functional z). Set H is the closed unit disk in the xy-plane. While H contains the zero vector (0,0,0) and is closed under addition of vectors within the unit disk, it fails closure under scalar multiplication. For example, (1,0,0) ∈ H since 1² + 0² = 1 ≤ 1, but 2·(1,0,0) = (2,0,0) ∉ H since 2² + 0² = 4 > 1. Therefore H is not a subspace due to failure of closure under scalar multiplication.

Question 20

Let $W = \{(x, y, z, w) \in \mathbb{R}^4 : x + 2y - z = 0 \text{ and } 2x + 4y - 2z + w = 0\}$ . Which of the following statements about $W$ is correct?

$W$ is a subspace of $\mathbb{R}^4$ with dimension 2, and the second constraint is redundant with the first (correct answer)
$W$ is a subspace of $\mathbb{R}^4$ with dimension 3, and both constraints provide independent information
$W$ is not a subspace because the system of constraints is inconsistent for some values
$W$ is a subspace of $\mathbb{R}^4$ with dimension 1, since both constraints must be satisfied simultaneously

Explanation: First, we check if W is a subspace by verifying it contains the zero vector, is closed under addition, and closed under scalar multiplication. The zero vector (0,0,0,0) satisfies both equations. For closure under addition and scalar multiplication, since both constraints are homogeneous linear equations, W is indeed a subspace. Next, we analyze the constraints: from the first equation x + 2y - z = 0, we get z = x + 2y. Substituting into the second equation: 2x + 4y - 2(x + 2y) + w = 0, which simplifies to 2x + 4y - 2x - 4y + w = 0, so w = 0. Thus W = {(x, y, x + 2y, 0) : x, y ∈ ℝ}, which has dimension 2 with basis vectors (1, 0, 1, 0) and (0, 1, 2, 0). The second constraint forces w = 0 but is otherwise redundant.