Linear Algebra Quiz: Span And Generating Sets

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Span And Generating SetsQuestion 1 of 15

Let $S = \{\vec{v}_1, \dots, \vec{v}_k\}$ be a set of vectors in a vector space $V$ . If $\mathrm{span}(S) = V$ and the dimension of $V$ is $n$ , which of the following statements is a necessary condition?

k = n

k \ge n

S

is a linearly independent set.

Every vector in

V

has a unique representation as a linear combination of vectors in

S

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Linear Algebra Quiz

Linear Algebra Quiz: Span And Generating Sets

Practice Span And Generating Sets 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 Span And Generating Sets, giving you a quick way to practice the rules, question types, and explanations that matter most for Linear Algebra.

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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

$k = n$
$k \ge n$ (correct answer)
$S$ is a linearly independent set.
Every vector in $V$ has a unique representation as a linear combination of vectors in $S$ .

Explanation: A fundamental theorem of linear algebra states that any generating set for a vector space

V

must contain at least as many vectors as the dimension of

V

. Therefore, if

\mathrm{dim}(V)=n

and

\mathrm{span}(S)=V

, it must be true that

k \ge n

. Choice A is only true if

S

is a basis. Choice C is not necessary; a generating set can be linearly dependent. Choice D is only true if

S

is a basis (linearly independent), as linear dependence implies non-unique representations.

Question 2

Let $S_1 = \{\begin{pmatrix} 1 \\ 2 \end{pmatrix}, \begin{pmatrix} 2 \\ 4 \end{pmatrix}\}$ and $S_2 = \{\begin{pmatrix} 3 \\ 6 \end{pmatrix}\}$ be sets of vectors in $\mathbb{R}^2$ . Which statement correctly describes the relationship between their spans?

$\mathrm{span}(S_1) = \mathrm{span}(S_2)$ (correct answer)
$\mathrm{span}(S_1)$ is a proper superset of $\mathrm{span}(S_2)$
$\mathrm{span}(S_2)$ is a proper superset of $\mathrm{span}(S_1)$
$\mathrm{span}(S_1)$ and $\mathrm{span}(S_2)$ intersect only at the origin

Explanation: In set

S_1

, the vector

\begin{pmatrix} 2 \\ 4 \end{pmatrix}

is a scalar multiple of

\begin{pmatrix} 1 \\ 2 \end{pmatrix}

(specifically, 2 times the first vector). Therefore, the second vector is redundant, and

\mathrm{span}(S_1) = \mathrm{span}(\{\begin{pmatrix} 1 \\ 2 \end{pmatrix}\})

. The vector in

S_2

\begin{pmatrix} 3 \\ 6 \end{pmatrix}

, is also a scalar multiple of

\begin{pmatrix} 1 \\ 2 \end{pmatrix}

(3 times). Thus,

\mathrm{span}(S_2)

is the same set of vectors. Both sets span the same line through the origin in

\mathbb{R}^2

Question 3

Consider the set of vectors $S = \{\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}, \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}, \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}\}$ in $\mathbb{R}^3$ . Which of the following is a minimal generating set for $\mathrm{span}(S)$ ?

$\{\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}\}$
$\{\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}\}$ (correct answer)
$\{\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}, \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}\}$
$\{\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}\}$

Explanation: First, we check for linear dependencies in

S

. Let the vectors be

\vec{v}_1, \vec{v}_2, \vec{v}_3, \vec{v}_4

. We can see that

\vec{v}_3 = \vec{v}_1 - \vec{v}_2

and

\vec{v}_4 = \vec{v}_1 + \vec{v}_2

. This means

\vec{v}_3

and

\vec{v}_4

are redundant and can be removed without changing the span. The set

\{\vec{v}_1, \vec{v}_2\}

generates

\mathrm{span}(S)

. Since

\vec{v}_1

and

\vec{v}_2

are not scalar multiples of each other, they are linearly independent. A minimal generating set is a basis, which must be linearly independent. Therefore,

\{\vec{v}_1, \vec{v}_2\}

is a minimal generating set. Choice A is too small (spans a line, not a plane). Choice C is not minimal because it's linearly dependent. Choice D spans the xy-plane, which is a different subspace from the one spanned by

S

Question 4

In the vector space $M_{2\times2}$ of $2 \times 2$ matrices, consider the subspace $W$ spanned by $S = \{\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\}$ . Which of the following matrices is not in $W$ ?

$\begin{pmatrix} 3 & 5 \\ 5 & 3 \end{pmatrix}$
$\begin{pmatrix} 1 & -1 \\ -1 & 1 \end{pmatrix}$
$\begin{pmatrix} 2 & 2 \\ 2 & 0 \end{pmatrix}$ (correct answer)
$\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$

Explanation: Any matrix in the subspace

W

must be a linear combination of the two generating matrices. A general matrix in

W

can be written as:

a\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} + b\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} a & b \\ b & a \end{pmatrix}

. This means any matrix in

W

must be symmetric (which is true for all options except D, which is trivially symmetric) and its diagonal elements must be equal. We check each option: A) Diagonals are both 3. In W. B) Diagonals are both 1. In W. C) Diagonals are 2 and 0. They are not equal. This matrix is not in W. D) The zero matrix is in every subspace. Here,

a=0

and

b=0

Question 5

The set $S = \{\begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}\}$ generates a plane $W$ in $\mathbb{R}^3$ . Which of the following vectors $\vec{v}$ , if added to $S$ , would create a new set that generates all of $\mathbb{R}^3$ ?

$\vec{v} = \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}$
$\vec{v} = \begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix}$
$\vec{v} = \begin{pmatrix} -1 \\ -1 \\ -2 \end{pmatrix}$
$\vec{v} = \begin{pmatrix} 3 \\ 2 \\ 4 \end{pmatrix}$ (correct answer)

Explanation: To generate all of

\mathbb{R}^3

, the new vector

\vec{v}

must not lie in the plane

W

spanned by

S

. A vector

(x,y,z)

W

can be written as

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

. So, any vector in

W

must satisfy the condition

z = x+y

. We test each option: A) For

(1,1,2)

2 = 1+1

. It is in

W

. B) For

(2,-1,1)

1 = 2+(-1)

. It is in

W

. C) For

(-1,-1,-2)

-2 = -1+(-1)

. It is in

W

. D) For

(3,2,4)

4 \neq 3+2

. This vector is not in

W

. Adding it to the set will span a 3-dimensional space, which is all of

\mathbb{R}^3

Question 6

Let $S = \{\vec{u}, \vec{v}\}$ and $T = \{\vec{u}, \vec{v}, \vec{u}+\vec{v}\}$ be two sets of vectors in a vector space $V$ . Let $W_S = \mathrm{span}(S)$ and $W_T = \mathrm{span}(T)$ . Which statement accurately describes the relationship between $W_S$ and $W_T$ ?

$W_S = W_T$ (correct answer)
$W_S$ is a proper subspace of $W_T$
$W_T$ is a proper subspace of $W_S$
The relationship cannot be determined without knowing if $\vec{u}$ and $\vec{v}$ are linearly independent.

Explanation: The span of a set of vectors is the set of all possible linear combinations of those vectors. The vector

\vec{u}+\vec{v}

is, by its form, a linear combination of

\vec{u}

and

\vec{v}

. This means

\vec{u}+\vec{v}

is already in

\mathrm{span}(\{\vec{u}, \vec{v}\})

. Adding a vector that is already in the span to a generating set does not change the span. Therefore,

\mathrm{span}(T) = \mathrm{span}(S)

, so

W_S = W_T

. This holds true whether or not

\vec{u}

and

\vec{v}

are linearly independent.

Question 7

Let $\{\vec{v}_1, \dots, \vec{v}_n\}$ be a generating set for a non-trivial vector space $V$ . Suppose $\vec{u}$ is a vector in $V$ . What must be true about the set $S' = \{\vec{v}_1, \dots, \vec{v}_n, \vec{u}\}$ ?

$S'$ is a basis for $V$ .
$S'$ is linearly independent.
$\mathrm{span}(S')$ is a larger space than $V$ .
$S'$ is a linearly dependent generating set for $V$ . (correct answer)

Explanation: Since

\{\vec{v}_1, \dots, \vec{v}_n\}

is a generating set for

V

, its span is

V

. Adding any vector

\vec{u}

from

V

to this set will not change the span, so

\mathrm{span}(S') = V

. Thus,

S'

is also a generating set for

V

. Furthermore, because

\vec{u}

is in

V = \mathrm{span}(\{\vec{v}_1, \dots, \vec{v}_n\})

\vec{u}

can be written as a linear combination of

\{\vec{v}_1, \dots, \vec{v}_n\}

. By definition, this makes the set

S'

linearly dependent. Therefore,

S'

is a linearly dependent generating set for

V

Question 8

For which value of the scalar $k$ does the set of vectors $S = \{\begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}, \begin{pmatrix} 2 \\ -2 \\ k \end{pmatrix}\}$ not form a generating set for $\mathbb{R}^3$ ?

$k = -2$
$k = 0$
$k = 2$ (correct answer)
$k = 4$

Explanation: A set of three vectors in

\mathbb{R}^3

fails to be a generating set (i.e., does not span

\mathbb{R}^3

) if and only if the vectors are linearly dependent. This occurs when the matrix formed by these vectors as columns has a determinant of zero. The determinant of the matrix

A = \begin{pmatrix} 1 & 0 & 2 \\ 0 & 1 & -2 \\ 2 & 1 & k \end{pmatrix}

is calculated as

1(k - (-2)) - 0(...) + 2(0 - 2) = k + 2 - 4 = k - 2

. Setting the determinant to zero, we get

k - 2 = 0

, which gives

k = 2

. When

k=2

, the third vector is a linear combination of the first two (

2\vec{v}_1 - 2\vec{v}_2

), so the set is linearly dependent and spans only a plane.

Question 9

Let $\vec{u}$ and $\vec{v}$ be two non-zero, non-parallel vectors in $\mathbb{R}^3$ . Let $\vec{w} = 2\vec{u} - 3\vec{v}$ . Which geometric object best describes the set of all linear combinations of the vectors in $S = \{\vec{u}, \vec{v}, \vec{w}\}$ ?

A line through the origin
A plane through the origin (correct answer)
The entire space $\mathbb{R}^3$
A single point at the origin

Explanation: The set of all linear combinations is the span of

S

, denoted

\mathrm{span}(S)

. Since

\vec{w}

is a linear combination of

\vec{u}

and

\vec{v}

, it is a redundant vector. Adding a redundant vector to a set does not change its span. Therefore,

\mathrm{span}(\{\vec{u}, \vec{v}, \vec{w}\}) = \mathrm{span}(\{\vec{u}, \vec{v}\})

. The span of two non-zero, non-parallel vectors is a plane through the origin.

Question 10

Let $S = \{\vec{v}_1, \vec{v}_2, \vec{v}_3\}$ be a set of vectors in $\mathbb{R}^3$ . You are given that the system $A\vec{x} = \vec{b}$ , where $A = [\vec{v}_1 \ \vec{v}_2 \ \vec{v}_3]$ , has a unique solution for at least one vector $\vec{b} \in \mathbb{R}^3$ . What can be concluded about $\mathrm{span}(S)$ ?

$\mathrm{span}(S)$ is a plane in $\mathbb{R}^3$ .
$\mathrm{span}(S)$ contains $\vec{b}$ but may not be all of $\mathbb{R}^3$ .
$\mathrm{span}(S)$ is all of $\mathbb{R}^3$ . (correct answer)
The relationship between $\mathrm{span}(S)$ and $\mathbb{R}^3$ depends on the specific vector $\vec{b}$ .

Explanation: The existence of a unique solution to

A\vec{x} = \vec{b}

implies that the null space of matrix

A

contains only the zero vector. For a square matrix

A

, this is equivalent to saying that

A

is invertible. An invertible matrix has linearly independent columns. Therefore, the vectors

\vec{v}_1, \vec{v}_2, \vec{v}_3

are linearly independent. A set of three linearly independent vectors in

\mathbb{R}^3

must form a basis for

\mathbb{R}^3

. By definition, a basis for a space spans that entire space. Thus,

\mathrm{span}(S) = \mathbb{R}^3

Question 11

Let $S = \{1+x^2, x-x^2\}$ be a set of polynomials in $\mathcal{P}_2$ , the space of polynomials of degree at most 2. Which of the following polynomials $p(x)$ , when added to $S$ , results in a set that does not generate $\mathcal{P}_2$ ?

$p(x) = 1$
$p(x) = x^2$
$p(x) = 1+2x-x^2$ (correct answer)
$p(x) = 2+x+3x^2$

Explanation: The set will fail to generate

\mathcal{P}_2

if the new polynomial

p(x)

is in the span of the original set

S

. Using the standard basis

\{1, x, x^2\}

, the vectors for

S

are

(1, 0, 1)

and

(0, 1, -1)

. Any vector in their span has the form

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

. We test each option:\nA) For

p(x)=1

, the vector is

(1,0,0)

. This requires

a=1, b=0

, but then

a-b=1 \neq 0

. Not in the span.\nB) For

p(x)=x^2

, the vector is

(0,0,1)

. This requires

a=0, b=0

, but then

a-b=0 \neq 1

. Not in the span.\nC) For

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

, the vector is

(1,2,-1)

. This requires

a=1, b=2

. Then

a-b = 1-2 = -1

. This matches. Since

p(x)

is a linear combination of the other two (

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

), the set is linearly dependent and cannot span the 3-dimensional space

\mathcal{P}_2

.\nD) For

p(x)=2+x+3x^2

, the vector is

(2,1,3)

. This requires

a=2, b=1

, but then

a-b=1 \neq 3

. Not in the span.

Question 12

Consider the set $T = \{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\}$ where $\mathbf{v}_1 = (1,2,0)$ , $\mathbf{v}_2 = (0,1,1)$ , and $\mathbf{v}_3 = (2,3,-1)$ . For which vector $\mathbf{b}$ is the equation $\mathbf{b} = c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + c_3\mathbf{v}_3$ inconsistent?

$(1,4,1)$
$(3,7,1)$
$(2,5,0)$
$(0,1,2)$ (correct answer)

Explanation: We need to find which vector is not in span(T). First, let's check if the vectors in T are linearly independent by row reducing the matrix

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

. Applying

R_3 \leftarrow R_3 - 2R_1

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

. Then

R_3 \leftarrow R_3 + R_2

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

. Since the rank is 2, the vectors span a 2-dimensional subspace. From the reduced form, we see that

\mathbf{v}_3 = 2\mathbf{v}_1 - \mathbf{v}_2

. So span(T) = span{

\mathbf{v}_1, \mathbf{v}_2

}. For

(0,1,2)

to be in the span, we need

(0,1,2) = c_1(1,2,0) + c_2(0,1,1)

, giving

c_1 = 0

2c_1 + c_2 = 1 \Rightarrow c_2 = 1

c_2 = 2

. This is inconsistent since

c_2

cannot be both 1 and 2.

Question 13

Let $A = \{(1,1,0), (1,0,1), (0,1,1)\}$ and $B = \{(2,1,1), (1,2,1), (1,1,2)\}$ . Which statement about $\text{span}(A)$ and $\text{span}(B)$ is correct?

$\text{span}(A) = \text{span}(B)$ and both equal $\mathbb{R}^3$ (correct answer)
$\text{span}(A) = \text{span}(B)$ but neither equals $\mathbb{R}^3$
$\text{span}(A) \neq \text{span}(B)$ but both equal $\mathbb{R}^3$
$\text{span}(A) \neq \text{span}(B)$ and neither equals $\mathbb{R}^3$

Explanation: First, let's check if the vectors in A are linearly independent. The matrix

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

has determinant

1(0-1) - 1(1-0) + 0 = -1-1 = -2 ≠ 0

, so A spans

\mathbb{R}^3

. For B, the matrix

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

has determinant

2(4-1) - 1(2-1) + 1(1-2) = 6-1-1 = 4 ≠ 0

, so B also spans

\mathbb{R}^3

. Since both A and B span

\mathbb{R}^3

, we have

\text{span}(A) = \text{span}(B) = \mathbb{R}^3

Question 14

Let $S = \{\vec{v}_1, \vec{v}_2, \vec{v}_3\}$ be a set of vectors in $\mathbb{R}^4$ . Let $V = \mathrm{span}(S)$ . Which of the following statements about $V$ must be true?

$V$ is a 3-dimensional subspace of $\mathbb{R}^4$ .
The set $S$ forms a basis for the subspace $V$ .
$V$ cannot be all of $\mathbb{R}^4$ . (correct answer)
$V$ cannot be spanned by any set with only two vectors.

Explanation: To span the entire space

\mathbb{R}^4

, a set must contain at least 4 vectors. Since the set

S

only contains 3 vectors, its span,

V

, can have a dimension of at most 3. Therefore,

V

cannot be the entire 4-dimensional space

\mathbb{R}^4

. The other options are not guaranteed: A) The dimension could be less than 3 if the vectors are linearly dependent. B)

S

is a generating set for

V

by definition, but it is only a basis if it is also linearly independent. D) If the three vectors are collinear,

V

would be a line and could be spanned by a single non-zero vector.

Question 15

Consider the vector space $\mathbb{P}_2$ of polynomials of degree at most 2. Let $W = \text{span}\{1 + x^2, x - x^2, 2 + x + x^2\}$ . Which polynomial is NOT in $W$ ?

$3 - 2x + 4x^2$
$1 + 2x - x^2$
$4 + 3x + 2x^2$ (correct answer)
$-1 + x + 3x^2$

Explanation: To determine if a polynomial is in W, we need to check if it can be written as

a(1+x^2) + b(x-x^2) + c(2+x+x^2) = (a+2c) + (b+c)x + (a-b+c)x^2

. For option C:

4+3x+2x^2

, we need

a+2c=4

b+c=3

a-b+c=2

. From the first two equations:

a=4-2c

and

b=3-c

. Substituting into the third:

(4-2c)-(3-c)+c = 4-2c-3+c+c = 1 ≠ 2

. This is inconsistent, so

4+3x+2x^2 ∉ W

. We can verify the others work: A gives

a=1, b=-2, c=1

; B gives

a=-1, b=2, c=1

; D gives

a=0, b=1, c=2