Linear Algebra Quiz: Linear Independence

17 questions · exam conditions

0:00

Linear IndependenceQuestion 1 of 17

In the vector space $\mathbb{R}^n$ , suppose $S = \{v_1, v_2, \ldots, v_k\}$ is a linearly independent set with $k < n$ . A new vector $v_{k+1}$ is chosen randomly from $\mathbb{R}^n$ according to a continuous probability distribution. What is the probability that $S \cup \{v_{k+1}\}$ remains linearly independent?

The probability is

\frac{n-k}{n}

, representing the fraction of dimensions not yet occupied by the span of

S

The probability cannot be determined without specifying the exact probability distribution used to select

v_{k+1}

The probability is

1 - \frac{k}{n}

, since there are

k

directions that would create dependence out of

n

possible dimensions

The probability is 1, because the set of vectors that would make

S \cup \{v_{k+1}\}

linearly dependent has measure zero in

\mathbb{R}^n

← Back to quizzes

Linear Algebra Quiz

Linear Algebra Quiz: Linear Independence

Practice Linear Independence 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 Linear Independence, 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 probability is $\frac{n-k}{n}$ , representing the fraction of dimensions not yet occupied by the span of $S$
The probability cannot be determined without specifying the exact probability distribution used to select $v_{k+1}$
The probability is $1 - \frac{k}{n}$ , since there are $k$ directions that would create dependence out of $n$ possible dimensions
The probability is 1, because the set of vectors that would make $S \cup \{v_{k+1}\}$ linearly dependent has measure zero in $\mathbb{R}^n$ (correct answer)

Explanation: When you encounter probability questions involving linear independence in vector spaces, the key insight is understanding what it means for vectors to be linearly dependent and how this translates to geometric probability. The set

S \cup \{v_{k+1}\}

becomes linearly dependent if and only if

v_{k+1}

lies in the span of

S

. Since

S = \{v_1, v_2, \ldots, v_k\}

is linearly independent with

k < n

, the span of

S

forms a

k

-dimensional subspace of

\mathbb{R}^n

. This subspace has dimension strictly less than

n

, which means it has measure zero in

\mathbb{R}^n

. Since

v_{k+1}

is chosen from a continuous probability distribution, the probability of landing exactly in this lower-dimensional subspace is zero. Therefore, the probability that

S \cup \{v_{k+1}\}

remains linearly independent is 1. Option A incorrectly treats this as a discrete counting problem, suggesting

\frac{n-k}{n}

. Option C makes a similar error with

1 - \frac{k}{n}

. Both fail to recognize that we're dealing with continuous probability over infinite-dimensional spaces, not discrete counting. Option B incorrectly assumes the specific distribution matters, but any continuous distribution will assign zero probability to lower-dimensional subspaces. The correct answer is D because lower-dimensional subspaces always have measure zero in higher-dimensional spaces under continuous distributions. Study tip: Remember that "almost surely" results in probability often involve measure zero sets. When dealing with continuous distributions in vector spaces, lower-dimensional subspaces have probability zero.

Question 2

Consider the set of polynomials $S = \{1+x^2, x-x^2, k+2x\}$ in the vector space $P_2$ of polynomials of degree at most 2. For what value of $k$ is the set $S$ linearly dependent?

$k = -2$
$k = 0$
$k = 1$
$k = 2$ (correct answer)

Explanation: To check for linear dependence, we seek scalars

c_1, c_2, c_3

, not all zero, such that

c_1(1+x^2) + c_2(x-x^2) + c_3(k+2x) = 0

. Grouping terms by powers of

x

gives:

(c_1 - c_2)x^2 + (c_2 + 2c_3)x + (c_1 + kc_3) = 0x^2 + 0x + 0

This yields a system of linear equations:

$c_1 - c_2 = 0 \implies c_1 = c_2$
$c_2 + 2c_3 = 0$
$c_1 + kc_3 = 0$ Substituting (1) into (2) gives $c_1 + 2c_3 = 0$ , so $c_1 = -2c_3$ . Substituting this into (3) gives $(-2c_3) + kc_3 = 0$ , which simplifies to $(k-2)c_3 = 0$ . For a non-trivial solution to exist (where $c_3 \neq 0$ ), we must have $k-2=0$ . Therefore, $k=2$ .

Question 3

Let $A$ be a $5 \times 4$ matrix whose columns are the vectors $\mathbf{a}_1, \mathbf{a}_2, \mathbf{a}_3, \mathbf{a}_4$ . If the set $\{\mathbf{a}_1, \mathbf{a}_2, \mathbf{a}_3, \mathbf{a}_4\}$ is linearly independent, what is the rank of matrix $A$ ?

3
4 (correct answer)
5
The rank cannot be determined from the given information.

Explanation: The rank of a matrix is defined as the dimension of its column space, which is equal to the maximum number of linearly independent columns. The problem states that the four columns of the

5 \times 4

matrix

A

form a linearly independent set. Therefore, the column space is spanned by these four linearly independent vectors, and its dimension is 4. Thus, the rank of

A

is 4.

Question 4

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

$\mathrm{span}(S) = \mathbb{R}^4$ .
$\mathrm{span}(S)$ is a three-dimensional subspace of $\mathbb{R}^4$ . (correct answer)
$\mathrm{span}(S) = \mathbb{R}^3$ .
Any vector in $\mathbb{R}^4$ can be written as a unique linear combination of vectors in $S$ .

Explanation: The dimension of the span of a set of vectors is the maximum number of linearly independent vectors in the set. Since the set

S

contains three linearly independent vectors, they form a basis for the subspace they span. Therefore, the dimension of

\mathrm{span}(S)

is 3. As the vectors belong to

\mathbb{R}^4

, their span is a three-dimensional subspace of

\mathbb{R}^4

. (A) is false because you need at least 4 linearly independent vectors to span

\mathbb{R}^4

. (C) is false because

\mathrm{span}(S)

is a subset of

\mathbb{R}^4

, not the separate space

\mathbb{R}^3

. (D) is false because it implies

S

is a basis for

\mathbb{R}^4

, which is not true.

Question 5

Consider the functions $f(x) = \cos(2x)$ , $g(x) = \cos^2(x)$ , and $h(x) = \sin^2(x)$ in the vector space of continuous functions on $\mathbb{R}$ . Which statement correctly describes the linear dependence of the set $S=\{f, g, h\}$ ?

The set is linearly dependent because $\cos(2x) = \cos^2(x) - \sin^2(x)$ . (correct answer)
The set is linearly dependent because $\cos^2(x) + \sin^2(x) = 1$ .
The set is linearly independent because no function is a scalar multiple of another.
The set is linearly independent because their Wronskian is non-zero.

Explanation: The set of functions is linearly dependent if one function can be expressed as a linear combination of the others. Using the double-angle trigonometric identity, we know that

\cos(2x) = \cos^2(x) - \sin^2(x)

. This means

f(x) = 1 \cdot g(x) - 1 \cdot h(x)

. This can be rewritten as

1 \cdot f(x) - 1 \cdot g(x) + 1 \cdot h(x) = 0

. Since we have found a non-trivial linear combination of the functions that equals the zero function, the set

S

is linearly dependent. The identity in choice (B) is true but does not involve

f(x)=\cos(2x)

and thus does not show the dependence of this specific set. Choice (C) presents a common fallacy about linear independence.

Question 6

Let $S = \{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\}$ be a set of vectors in a vector space $V$ . If $\mathbf{v}_2 = \mathbf{0}$ (the zero vector), which linear combination demonstrates that $S$ must be linearly dependent?

$1\mathbf{v}_1 + 1\mathbf{v}_2 + 1\mathbf{v}_3$
$1\mathbf{v}_1 + 0\mathbf{v}_2 - 1\mathbf{v}_3$
$0\mathbf{v}_1 + 1\mathbf{v}_2 + 0\mathbf{v}_3$ (correct answer)
$c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + c_3\mathbf{v}_3$ only if $c_1=c_2=c_3=0$

Explanation: A set of vectors is linearly dependent if there exists a linear combination

c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + c_3\mathbf{v}_3 = \mathbf{0}

where at least one of the scalars

c_i

is non-zero. Since

\mathbf{v}_2 = \mathbf{0}

, we can choose

c_1=0

c_2=1

(which is non-zero), and

c_3=0

. The combination becomes

0\mathbf{v}_1 + 1\mathbf{v}_2 + 0\mathbf{v}_3 = 0\mathbf{v}_1 + 1(\mathbf{0}) + 0\mathbf{v}_3 = \mathbf{0} + \mathbf{0} + \mathbf{0} = \mathbf{0}

. Because we found a set of scalars that are not all zero and result in the zero vector, the set is linearly dependent. Choice (C) represents this non-trivial combination.

Question 7

In the vector space $M_{2 \times 2}$ of $2 \times 2$ matrices, consider the set $S = \{A, B, C\}$ where $A=\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ , $B=\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ , and $C=\begin{pmatrix} 1 & -2 \\ -2 & 1 \end{pmatrix}$ . Which statement is true?

$S$ is linearly independent.
$S$ is linearly dependent because $C = A - 2B$ . (correct answer)
$S$ is linearly dependent because $A = B + C$ .
$S$ is linearly dependent because all matrices in $S$ are invertible.

Explanation: To determine linear dependence, we check if one vector can be written as a linear combination of the others. Let's test the relationship in choice B:

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

Since

C = A - 2B

, which can be rewritten as

A - 2B - C = O

(where

O

is the zero matrix), we have found a non-trivial linear combination of the matrices that equals the zero matrix. Therefore, the set

S

is linearly dependent.

Question 8

Let $\{\mathbf{u}, \mathbf{v}, \mathbf{w}\}$ be a linearly independent set of vectors in a vector space $V$ over $\mathbb{R}$ . Which of the following correctly describes the set $S = \{\mathbf{u}+\mathbf{v}, \mathbf{v}+\mathbf{w}, \mathbf{u}+\mathbf{w}\}$ ?

It is linearly independent. (correct answer)
It is linearly dependent.
It is linearly independent if and only if $\mathbf{u}, \mathbf{v}, \mathbf{w}$ are orthogonal.
Its dependence cannot be determined without more information.

Explanation: To determine if

S

is linearly independent, we look for solutions to the equation

c_1(\mathbf{u}+\mathbf{v}) + c_2(\mathbf{v}+\mathbf{w}) + c_3(\mathbf{u}+\mathbf{w}) = \mathbf{0}

. We can rearrange this equation by grouping the vectors

\mathbf{u}, \mathbf{v}, \mathbf{w}

(c_1+c_3)\mathbf{u} + (c_1+c_2)\mathbf{v} + (c_2+c_3)\mathbf{w} = \mathbf{0}

Since

\{\mathbf{u}, \mathbf{v}, \mathbf{w}\}

is linearly independent, the only way this equation can hold is if all scalar coefficients are zero:

$c_1 + c_3 = 0$
$c_1 + c_2 = 0$
$c_2 + c_3 = 0$ From (1), $c_1 = -c_3$ . From (2), $c_2 = -c_1$ . Substituting gives $c_2 = -(-c_3) = c_3$ . Substituting this into (3) gives $c_3 + c_3 = 0$ , so $2c_3=0$ , which implies $c_3=0$ . This then implies $c_1=0$ and $c_2=0$ . Since the only solution is the trivial solution $c_1=c_2=c_3=0$ , the set $S$ is linearly independent.

Question 9

In $\mathbb{R}^n$ , suppose vectors $v_1, v_2, \ldots, v_k$ are linearly independent. A new vector $w$ is added to form the set $\{v_1, v_2, \ldots, v_k, w\}$ . Which condition is both necessary and sufficient for the expanded set to remain linearly independent?

Vector $w$ must not be the zero vector, since adding the zero vector always creates linear dependence in any set
Vector $w$ must not be expressible as $w = a_1v_1 + a_2v_2 + \cdots + a_kv_k$ for any scalars $a_1, a_2, \ldots, a_k$ (correct answer)
Vector $w$ must have at least one component that differs from the corresponding components of all vectors $v_1, \ldots, v_k$
Vector $w$ must satisfy $w \cdot v_i \neq 0$ for at least one $i \in \{1, 2, \ldots, k\}$ to ensure geometric independence

Explanation: The set

\{v_1, v_2, \ldots, v_k, w\}

is linearly independent if and only if no vector in the set can be written as a linear combination of the others. Since

v_1, \ldots, v_k

are already linearly independent, none of them can be expressed as combinations of the others. The only way dependence can arise is if

w

can be written as a linear combination of

v_1, \ldots, v_k

, or equivalently, if

w

lies in

\text{span}\{v_1, \ldots, v_k\}

. Choice A is insufficient -

w

being non-zero doesn't guarantee it's not in the span of the

v_i

's. Choice C is irrelevant - component-wise differences don't determine linear independence. Choice D confuses orthogonality with linear independence - vectors can be linearly independent without being orthogonal.

Question 10

Consider the polynomial space $P_3$ of polynomials of degree at most 3. Let $S = \{p_1(x), p_2(x), p_3(x)\}$ where $p_1(x) = 1 + x$ , $p_2(x) = x + x^2$ , and $p_3(x) = 1 + x^2$ . To determine if $S$ is linearly independent, a student sets up the equation $c_1p_1(x) + c_2p_2(x) + c_3p_3(x) = 0$ and substitutes three values: $x = 0, 1, 2$ . What can be concluded about this approach?

This approach is valid and sufficient, since three equations in three unknowns will determine if $c_1 = c_2 = c_3 = 0$ is the only solution
This approach is insufficient because polynomials require evaluation at four points to account for the four-dimensional basis of $P_3$
This approach is valid because the polynomial equation $c_1p_1(x) + c_2p_2(x) + c_3p_3(x) = 0$ holds for all $x$ if and only if all coefficients are zero (correct answer)
This approach may fail because the chosen values $x = 0, 1, 2$ might not provide linearly independent constraint equations for the unknowns $c_1, c_2, c_3$

Explanation: For polynomial functions, if

c_1p_1(x) + c_2p_2(x) + c_3p_3(x) = 0

as a polynomial (meaning the zero polynomial), then this equation must hold for ALL values of

x

, which occurs if and only if the coefficient of each power of

x

is zero. The student's approach of checking three specific values would work in this case because:

c_1(1 + x) + c_2(x + x^2) + c_3(1 + x^2) = (c_1 + c_3) + (c_1 + c_2)x + (c_2 + c_3)x^2

. Setting this equal to zero gives the system:

c_1 + c_3 = 0

c_1 + c_2 = 0

c_2 + c_3 = 0

. This has only the trivial solution, so

S

is linearly independent. Choice A is partially correct but misses the key insight. Choice B incorrectly focuses on dimension rather than the polynomial property. Choice D is wrong because the approach is actually valid for polynomials.

Question 11

Consider the vector space $\mathbb{R}^{2 \times 2}$ of $2 \times 2$ real matrices. Let $M_1 = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}$ , $M_2 = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ , and $M_3 = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ . To test linear independence of $\{M_1, M_2, M_3\}$ , which equation must be solved?

$$$ to find if only the trivial solution exists for the coefficients$ (correct answer)
$c_1 \det(M_1) + c_2 \det(M_2) + c_3 \det(M_3) = 0$ to check if the determinants are linearly independent
$\det(c_1 M_1 + c_2 M_2 + c_3 M_3) = 0$ to determine when the linear combination becomes singular
$$$ to check multiplicative relationships between matrices$

Explanation: When testing linear independence in any vector space, you're asking: "Can we write the zero vector as a linear combination of these vectors using only trivial (all zero) coefficients?" This fundamental definition applies whether you're working with vectors in

\mathbb{R}^n

or matrices in

\mathbb{R}^{2 \times 2}

. The correct approach is answer A: set up

c_1 M_1 + c_2 M_2 + c_3 M_3 = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}

and solve for the coefficients. If only

c_1 = c_2 = c_3 = 0

works, the matrices are linearly independent. If you find non-zero coefficients that satisfy this equation, they're linearly dependent. Answer B incorrectly focuses on the determinants of the individual matrices. Linear independence of matrices has nothing to do with whether their determinants form a linearly independent set of scalars. This confuses properties of the matrices with the matrices themselves. Answer C tests when a linear combination becomes singular (non-invertible), which is unrelated to linear independence. A linear combination could be invertible or singular regardless of whether the original matrices are linearly independent. Answer D involves matrix multiplication, creating an entirely different problem. Linear independence is about addition and scalar multiplication in the vector space, not about products between the matrices. Remember: linear independence always comes down to the same equation format across all vector spaces: "linear combination equals zero vector." Don't let the context (matrices, functions, polynomials) distract you from this core definition.

Question 12

Let $f_1(x) = \sin(x)$ , $f_2(x) = \cos(x)$ , and $f_3(x) = \sin(x + \pi/4)$ be functions in the vector space of continuous functions on $\mathbb{R}$ . A student wants to determine if $\{f_1, f_2, f_3\}$ is linearly independent by checking if $c_1 f_1(x) + c_2 f_2(x) + c_3 f_3(x) = 0$ has only the trivial solution. Using the identity $\sin(x + \pi/4) = \frac{\sqrt{2}}{2}(\sin(x) + \cos(x))$ , what should the student conclude?

The set is linearly independent because the functions $\sin(x)$ , $\cos(x)$ , and $\sin(x + \pi/4)$ have different periods and phase shifts
Linear independence cannot be determined from the given identity because trigonometric functions require integration methods to test dependence relationships
The set is linearly independent because $\sin(x + \pi/4)$ cannot be written as a rational linear combination of $\sin(x)$ and $\cos(x)$
The set is linearly dependent because $f_3(x) = \frac{\sqrt{2}}{2}f_1(x) + \frac{\sqrt{2}}{2}f_2(x)$ , so $f_3$ is in the span of $\{f_1, f_2\}$ (correct answer)

Explanation: When testing linear independence in any vector space, including function spaces, you need to determine whether one vector (or function) can be written as a linear combination of the others. The given trigonometric identity is the key to solving this problem. The identity

\sin(x + \pi/4) = \frac{\sqrt{2}}{2}(\sin(x) + \cos(x))

directly tells us that

f_3(x) = \frac{\sqrt{2}}{2}f_1(x) + \frac{\sqrt{2}}{2}f_2(x)

. This means

f_3

can be expressed as a linear combination of

f_1

and

f_2

, so

f_3

lies in the span of

\{f_1, f_2\}

. Therefore, the set

\{f_1, f_2, f_3\}

is linearly dependent, making answer D correct. Looking at the wrong answers: Choice A incorrectly focuses on periods and phase shifts, which don't determine linear independence—functions can have different characteristics yet still be linearly dependent. Choice B is false because the trigonometric identity gives us exactly the information needed; no integration is required to test linear dependence when you have an explicit linear combination. Choice C contains a fundamental error—

\sin(x + \pi/4)

absolutely can be written as a linear combination of

\sin(x)

and

\cos(x)

, as the identity demonstrates (the coefficients

\frac{\sqrt{2}}{2}

are rational numbers). Study tip: When checking linear independence of functions, look for trigonometric identities, polynomial relationships, or other algebraic connections that express one function in terms of others. If such a relationship exists, the set is automatically linearly dependent.

Question 13

Let $A$ be an $n \times n$ matrix, and let $v_1, v_2, \ldots, v_k$ be eigenvectors of $A$ corresponding to distinct eigenvalues $\lambda_1, \lambda_2, \ldots, \lambda_k$ respectively. A student claims that since eigenvectors corresponding to distinct eigenvalues are linearly independent, any subset of $\{Av_1, Av_2, \ldots, Av_k\}$ must also be linearly independent. How should this claim be evaluated?

The claim is correct because $Av_i = \lambda_i v_i$ , and scaling linearly independent vectors by non-zero constants preserves linear independence
The claim is incorrect because $Av_i = \lambda_i v_i$ , and if any $\lambda_i = 0$ , then $Av_i = 0$ , which makes any subset containing it linearly dependent (correct answer)
The claim is correct because linear transformations preserve linear independence when applied to eigenvectors with distinct eigenvalues
The claim is incorrect because the vectors $Av_1, Av_2, \ldots, Av_k$ lie in the column space of $A$ , which may have dimension less than $k$

Explanation: The key insight is that

Av_i = \lambda_i v_i

for each eigenvector. If any eigenvalue

\lambda_i = 0

, then

Av_i = 0 \cdot v_i = 0

. Since the zero vector is always linearly dependent (any non-zero scalar multiple of it gives zero), any set containing the zero vector is linearly dependent. Therefore, if

A

has eigenvalue 0, then the corresponding

Av_i = 0

, making any subset of

\{Av_1, \ldots, Av_k\}

that includes this vector linearly dependent. Choice A fails to consider the case where some

\lambda_i = 0

. Choice C makes an incorrect general statement about linear transformations preserving independence. Choice D mentions a true fact about column spaces but doesn't identify the specific issue with zero eigenvalues.

Question 14

Let $T: \mathbb{R}^4 \to \mathbb{R}^3$ be a linear transformation with matrix representation $A$ . If the vectors $u_1, u_2, u_3 \in \mathbb{R}^4$ are linearly independent and $T(u_1), T(u_2), T(u_3)$ are linearly dependent, what is the minimum possible value of $\text{nullity}(T)$ ?

The nullity must be at least 1, since linear transformations always reduce linear independence when the codomain has lower dimension
The nullity must be at least 2, since three independent vectors cannot map to three dependent vectors unless the kernel is non-trivial
The nullity must be at least 1, since there exists a non-trivial linear combination $c_1u_1 + c_2u_2 + c_3u_3$ that maps to zero (correct answer)
The nullity could be 0, since the dependence of $T(u_1), T(u_2), T(u_3)$ might result from the dimension constraint of the codomain

Explanation: Since

T(u_1), T(u_2), T(u_3)

are linearly dependent, there exist constants

c_1, c_2, c_3

(not all zero) such that

c_1T(u_1) + c_2T(u_2) + c_3T(u_3) = 0

. By linearity of T, this means

T(c_1u_1 + c_2u_2 + c_3u_3) = 0

. Since

u_1, u_2, u_3

are linearly independent and not all

c_i

are zero, the vector

c_1u_1 + c_2u_2 + c_3u_3

is non-zero. Therefore, this non-zero vector is in the kernel of T, so nullity(T) ≥ 1. Choice A gives the right answer but wrong reasoning - dimension constraints don't automatically create non-trivial kernels. Choice B is incorrect because we only need one non-zero vector in the kernel. Choice D is wrong because we've shown a non-zero vector must be in the kernel.

Question 15

Suppose $\{u_1, u_2, u_3\}$ is a linearly independent set in $\mathbb{R}^4$ , and $\{w_1, w_2\}$ is a linearly independent set in $\mathbb{R}^4$ such that $\text{span}\{w_1, w_2\} \cap \text{span}\{u_1, u_2, u_3\} = \{0\}$ . What can be concluded about the set $\{u_1, u_2, u_3, w_1, w_2\}$ ?

The set must be linearly dependent because it contains 5 vectors in $\mathbb{R}^4$ , and any set with more than 4 vectors in $\mathbb{R}^4$ is linearly dependent (correct answer)
The set must be linearly independent because the intersection condition ensures no vector from one span can be written using vectors from the other span
The linear independence cannot be determined without additional information about the specific vectors and their relationships
The set is linearly independent if and only if the combined span has dimension 4, which requires additional verification beyond the given intersection condition

Explanation: This is a direct application of a fundamental theorem: in an n-dimensional vector space, any set containing more than n vectors must be linearly dependent. Since

\mathbb{R}^4

has dimension 4, any set of 5 or more vectors in

\mathbb{R}^4

is automatically linearly dependent, regardless of any other properties the vectors might have. The intersection condition

\text{span}\{w_1, w_2\} \cap \text{span}\{u_1, u_2, u_3\} = \{0\}

tells us that the two subspaces intersect only at the origin, but this doesn't override the dimension constraint. Choice B incorrectly suggests that the intersection condition is sufficient for independence. Choice C is wrong because the dimension argument is definitive. Choice D is incorrect because no additional verification can make 5 vectors in

\mathbb{R}^4

linearly independent.

Question 16

Let $S = \{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3, \mathbf{v}_4\}$ be a set of four distinct, non-zero vectors in $\mathbb{R}^3$ . Which statement about the linear dependence of $S$ must be true?

The set $S$ is linearly independent.
The set $S$ is linearly dependent. (correct answer)
The set $S$ is linearly independent if and only if the vectors are mutually orthogonal.
The linear dependence of $S$ cannot be determined without knowing the specific vector components.

Explanation: The dimension of

\mathbb{R}^3

is 3. A fundamental theorem in linear algebra states that any set of

m

vectors in an

n

-dimensional space is linearly dependent if

m > n

. In this case, we have a set of

m=4

vectors in

\mathbb{R}^3

, where

n=3

. Since

4 > 3

, the set

S

must be linearly dependent, regardless of the specific vectors (as long as they are in

\mathbb{R}^3

). The other options are therefore incorrect.

Question 17

Let $S = \{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\}$ be a linearly independent set of vectors in $\mathbb{R}^5$ . Which of the following sets is also guaranteed to be linearly independent?

$\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3, \mathbf{0}\}$
$\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_1 + \mathbf{v}_2\}$
$\{\mathbf{v}_1 - \mathbf{v}_2, \mathbf{v}_2 - \mathbf{v}_3, \mathbf{v}_3 - \mathbf{v}_1\}$
$\{\mathbf{v}_1, \mathbf{v}_2\}$ (correct answer)

Explanation: The correct answer is D. A fundamental property of linear independence is that any subset of a linearly independent set is also linearly independent. Since

S

is linearly independent, its subset

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

must also be linearly independent. (A) is incorrect because any set containing the zero vector is linearly dependent. (B) is incorrect because one vector is a linear combination of the others:

1\cdot\mathbf{v}_1 + 1\cdot\mathbf{v}_2 - 1\cdot(\mathbf{v}_1 + \mathbf{v}_2) = \mathbf{0}

. (C) is incorrect because there is a non-trivial linear combination that sums to the zero vector:

1(\mathbf{v}_1 - \mathbf{v}_2) + 1(\mathbf{v}_2 - \mathbf{v}_3) + 1(\mathbf{v}_3 - \mathbf{v}_1) = \mathbf{0}