Linear Independence - Linear Algebra

1. 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?
   Answer: 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 $\{f_1, f_2\}$ 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.
2. 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 $A$?
   Answer: 4 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.
3. 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?
   Answer: Vector $w$ must not be expressible as $w = a_1v_1 + a_2v_2 + \cdots + a_kv_k$ for any scalars $a_1, a_2, \ldots, a_k$ 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.
4. 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 $\text{nullity}(T)$?
   Answer: 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 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.
5. 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?
   Answer: 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 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.
6. 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?
   Answer: The set $S$ is linearly dependent. 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.
7. 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?
   Answer: The probability is 1, because the set of vectors that would make $S \cup \{v_{k+1}\}$ linearly dependent has measure zero in $\mathbb{R}^n$ 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.
8. 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 $\{u_1, u_2, u_3, w_1, w_2\}$?
   Answer: 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 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.
9. 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?
   Answer: $S$ is linearly dependent because $C = A - 2B$. 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.
10. 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?
    Answer: $c_1 M_1 + c_2 M_2 + c_3 M_3 = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$ to find if only the trivial solution exists for the coefficients 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.
11. 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?
    Answer: 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 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.
12. 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 $S=\{f, g, h\}$?
    Answer: The set is linearly dependent because $\cos(2x) = \cos^2(x) - \sin^2(x)$. 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.
13. 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?
    Answer: $\{\mathbf{v}_1, \mathbf{v}_2\}$ 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: $1(\mathbf{v}_1 - \mathbf{v}_2) + 1(\mathbf{v}_2 - \mathbf{v}_3) + 1(\mathbf{v}_3 - \mathbf{v}_1) = \mathbf{0}$.
14. 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?
    Answer: $0\mathbf{v}_1 + 1\mathbf{v}_2 + 0\mathbf{v}_3$ 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.
15. 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?
    Answer: $\mathrm{span}(S)$ is a three-dimensional subspace of $\mathbb{R}^4$. 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.
16. 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 $S = \{\mathbf{u}+\mathbf{v}, \mathbf{v}+\mathbf{w}, \mathbf{u}+\mathbf{w}\}$?
    Answer: It is linearly independent. 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: 1) $c_1 + c_3 = 0$ 2) $c_1 + c_2 = 0$ 3) $c_2 + c_3 = 0$ 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.
17. 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?
    Answer: $k = 2$ 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: 1) $c_1 - c_2 = 0 \implies c_1 = c_2$ 2) $c_2 + 2c_3 = 0$ 3) $c_1 + kc_3 = 0$ 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, $k=2$.

Start this quiz