Subspaces & Tests - Linear Algebra

1. 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?
   Answer: $W$ is a subspace of $\mathbb{R}^4$ with dimension 2, and the second constraint is redundant with the first 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.
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 $\mathbb{R}^3$?
   Answer: $W$ is not closed under scalar multiplication. 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.
3. 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 $M_{2\times2}$?
   Answer: $S$ is not closed under matrix addition. 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 $I$ is $1$, so $A+B$ is not singular. Therefore, $A+B \notin S$, which shows $S$ is not closed under matrix addition.
4. 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?
   Answer: Set $B$ is not a subspace, so the general theorem about intersections of subspaces does not apply 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.
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?
   Answer: Proving that $T$ is non-empty, since the existence of non-trivial solutions to the differential equation is not immediately obvious 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.
6. Let $U$ and $W$ be subspaces of a vector space $V$. Which of the following sets is guaranteed to be a subspace of $V$?
   Answer: The intersection $U \cap W$. 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.
7. Let $W$ be the set of all vectors in $\mathbb{R}^4$ whose components sum to zero. Which statement is true?
   Answer: $W$ is a subspace of $\mathbb{R}^4$. $W = \{(x_1, x_2, x_3, x_4) \in \mathbb{R}^4 \mid x_1+x_2+x_3+x_4=0\}$. 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 $c\mathbf{u}$ is $\sum (cu_i) = c (\sum u_i) = c(0)=0$. So $c\mathbf{u} \in W$. All three conditions hold.
8. Which of the following subsets of $\mathbb{R}^3$ is a subspace of $\mathbb{R}^3$?
   Answer: $W_3 = \{(x, y, z) \in \mathbb{R}^3 \mid 2x - y = 0 \text{ and } 3y + z = 0\}$ $W_3$ 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.
9. 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?
   Answer: Show that $U \cup V$ fails closure under addition by finding vectors $u_1, u_2 \in U$ such that $u_1 + u_2 \notin U \cup V$ 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.
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 $\mathbb{R}^2$?
    Answer: $W$ is not closed under vector addition. 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.
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?
    Answer: $S$ is a subspace of $P_3$. $S$ 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.
12. 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?
    Answer: The trace condition is redundant because skew-symmetric matrices automatically have trace zero, so $W$ has dimension 1 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.
13. 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 $W$?
    Answer: $\dim(W) = 2$, since we have 3 linear constraints on a 5-dimensional space 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).
14. 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?
    Answer: Check whether both spanning vectors of $W_1$ satisfy the constraint defining $W_2$, since this is necessary and sufficient 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.
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?
    Answer: $S$ contains the zero vector and is closed under scalar multiplication, but is not closed under addition. The condition $x^2 - y^2 = 0$ is equivalent to $(x-y)(x+y)=0$, which means $y=x$ or $y=-x$. 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.
16. 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?
    Answer: 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 $A$ 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.
17. 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?
    Answer: $S_1$ is a subspace of $\mathbb{R}^n$, but $S_2$ is not a subspace because the zero vector is not in $S_2$. $S_1$ 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.
18. 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?
    Answer: The set of all functions $f$ such that $f(7) = f(1)$. 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.
19. Let $V$ be the vector space of $2 \times 3$ matrices. Which of the following subsets of $V$ is NOT a subspace?
    Answer: The set of matrices with at least one zero entry. 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.
20. 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?
    Answer: $H$ is not a subspace because it fails closure under scalar multiplication, while $K$ is a subspace 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.

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