Linear Algebra Quiz: Orthogonality And Complements
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Orthogonality And ComplementsQuestion 1 of 20

Let WW be the subspace of R3\mathbb{R}^3 spanned by the vector v=(1−23)\mathbf{v} = \begin{pmatrix} 1 \\ -2 \\ 3 \end{pmatrix}. Which of the following sets is a basis for the orthogonal complement W⊥W^\perp?

{(210)}\left\{ \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix} \right\}
{(210),(−301)}\left\{ \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} -3 \\ 0 \\ 1 \end{pmatrix} \right\}
{(1−23),(210)}\left\{ \begin{pmatrix} 1 \\ -2 \\ 3 \end{pmatrix}, \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix} \right\}
{(420),(−602)}\left\{ \begin{pmatrix} 4 \\ 2 \\ 0 \end{pmatrix}, \begin{pmatrix} -6 \\ 0 \\ 2 \end{pmatrix} \right\}
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Linear Algebra Quiz

Linear Algebra Quiz: Orthogonality And Complements

Practice Orthogonality And Complements 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 Orthogonality And Complements, 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.

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Question 1

Let WW be the subspace of R3\mathbb{R}^3 spanned by the vector v=(1−23)\mathbf{v} = \begin{pmatrix} 1 \\ -2 \\ 3 \end{pmatrix}. Which of the following sets is a basis for the orthogonal complement W⊥W^\perp?

  1. {(210)}\left\{ \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix} \right\}
  2. {(210),(−301)}\left\{ \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} -3 \\ 0 \\ 1 \end{pmatrix} \right\} (correct answer)
  3. {(1−23),(210)}\left\{ \begin{pmatrix} 1 \\ -2 \\ 3 \end{pmatrix}, \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix} \right\}
  4. {(420),(−602)}\left\{ \begin{pmatrix} 4 \\ 2 \\ 0 \end{pmatrix}, \begin{pmatrix} -6 \\ 0 \\ 2 \end{pmatrix} \right\}
Explanation: The orthogonal complement W⊥W^\perp consists of all vectors x=(x1,x2,x3)\mathbf{x} = (x_1, x_2, x_3) such that v⋅x=0\mathbf{v} \cdot \mathbf{x} = 0. This gives the equation x1−2x2+3x3=0x_1 - 2x_2 + 3x_3 = 0. This is the equation of a plane in R3\mathbb{R}^3. Since dim⁡(W)=1\dim(W) = 1 and dim⁡(R3)=3\dim(\mathbb{R}^3) = 3, the dimension of W⊥W^\perp must be dim⁡(R3)−dim⁡(W)=3−1=2\dim(\mathbb{R}^3) - \dim(W) = 3 - 1 = 2. Therefore, a basis for W⊥W^\perp must contain two linearly independent vectors. Both vectors in choice B satisfy the equation: (2)−2(1)+3(0)=0(2) - 2(1) + 3(0) = 0 and (−3)−2(0)+3(1)=0(-3) - 2(0) + 3(1) = 0. They are also linearly independent. Thus, they form a basis for W⊥W^\perp.

Question 2

Let WW be the column space of the matrix A=(10012−1)A = \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 2 & -1 \end{pmatrix}. Which of the following vectors forms a basis for the orthogonal complement W⊥W^\perp?

  1. (−211)\begin{pmatrix} -2 \\ 1 \\ 1 \end{pmatrix} (correct answer)
  2. (102)\begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix}
  3. (20−1)\begin{pmatrix} 2 \\ 0 \\ -1 \end{pmatrix}
  4. (011)\begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}
Explanation: The orthogonal complement of the column space of AA, denoted (ColA)⊥(\mathrm{Col } A)^\perp, is the null space of ATA^T, i.e., Nul(AT)\mathrm{Nul}(A^T). First, we find the transpose of AA: AT=(10201−1)A^T = \begin{pmatrix} 1 & 0 & 2 \\ 0 & 1 & -1 \end{pmatrix}. To find the null space, we solve ATx=0A^T\mathbf{x} = \mathbf{0}. This corresponds to the system of equations x1+2x3=0x_1 + 2x_3 = 0 and x2−x3=0x_2 - x_3 = 0. From these equations, we get x1=−2x3x_1 = -2x_3 and x2=x3x_2 = x_3. A vector in the null space has the form (−2x3x3x3)=x3(−211)\begin{pmatrix} -2x_3 \\ x_3 \\ x_3 \end{pmatrix} = x_3 \begin{pmatrix} -2 \\ 1 \\ 1 \end{pmatrix}. Thus, a basis for W⊥W^\perp is {(−211)}\left\{ \begin{pmatrix} -2 \\ 1 \\ 1 \end{pmatrix} \right\}.

Question 3

In the inner product space C[0,1]C[0, 1] of continuous functions on the interval [0,1][0, 1] with the inner product ⟨f,g⟩=∫01f(x)g(x) dx\langle f, g \rangle = \int_0^1 f(x)g(x) \,dx, let f(x)=1f(x) = 1 and g(x)=ax−1g(x) = ax - 1. For what value of aa are f(x)f(x) and g(x)g(x) orthogonal?

  1. a=0a = 0
  2. a=1a = 1
  3. a=2a = 2 (correct answer)
  4. a=−2a = -2
Explanation: For f(x)f(x) and g(x)g(x) to be orthogonal, their inner product must be zero. We set up the integral: ⟨f,g⟩=∫01(1)(ax−1) dx=0\langle f, g \rangle = \int_0^1 (1)(ax - 1) \,dx = 0 We evaluate the integral: ∫01(ax−1) dx=[ax22−x]01=(a(1)22−1)−(a(0)22−0)=a2−1\int_0^1 (ax - 1) \,dx = \left[ \frac{ax^2}{2} - x \right]_0^1 = \left( \frac{a(1)^2}{2} - 1 \right) - \left( \frac{a(0)^2}{2} - 0 \right) = \frac{a}{2} - 1 Setting the result to zero gives a2−1=0\frac{a}{2} - 1 = 0, which solves to a2=1\frac{a}{2} = 1, so a=2a = 2.

Question 4

Let WW be the subspace of R4\mathbb{R}^4 consisting of all vectors (x1,x2,x3,x4)(x_1, x_2, x_3, x_4) that satisfy the equations x1−x2+x3=0x_1 - x_2 + x_3 = 0 and x2−x4=0x_2 - x_4 = 0. Which of the following vectors lies in W⊥W^\perp?

  1. (−1,0,1,0)(-1, 0, 1, 0)
  2. (1,0,1,−1)(1, 0, 1, -1) (correct answer)
  3. (1,1,0,1)(1, 1, 0, 1)
  4. (0,1,1,1)(0, 1, 1, 1)
Explanation: The subspace WW is the set of solutions to a homogeneous system of linear equations, which means WW is the null space of the matrix A=(1−110010−1)A = \begin{pmatrix} 1 & -1 & 1 & 0 \\ 0 & 1 & 0 & -1 \end{pmatrix}. The orthogonal complement of the null space of a matrix is its row space, i.e., W⊥=(NulA)⊥=RowAW^\perp = (\mathrm{Nul } A)^\perp = \mathrm{Row } A. Therefore, we are looking for a vector that is a linear combination of the rows of AA. The rows are r1=(1,−1,1,0)\mathbf{r}_1 = (1, -1, 1, 0) and r2=(0,1,0,−1)\mathbf{r}_2 = (0, 1, 0, -1). Let's check the options:
  • Choice A: (−1,0,1,0)(-1, 0, 1, 0) is in WW because (−1)−(0)+(1)=0(-1) - (0) + (1) = 0 and (0)−(0)=0(0) - (0) = 0. It cannot be in W⊥W^\perp unless it is the zero vector.
  • Choice B: (1,0,1,−1)(1, 0, 1, -1) is the sum of the two rows: r1+r2\mathbf{r}_1 + \mathbf{r}_2. Since it is a linear combination of the rows, it is in the row space, and thus in W⊥W^\perp.
  • Choices C and D are not linear combinations of the rows of AA.

Question 5

In R3\mathbb{R}^3, let WW be the plane defined by the equation 2x−y+3z=02x - y + 3z = 0. Which of the following provides a geometric description of the orthogonal complement W⊥W^\perp?

  1. The plane with equation x+2y+0z=0x + 2y + 0z = 0.
  2. The line passing through the origin with direction vector (2−13)\begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}. (correct answer)
  3. The line passing through the origin with direction vector (120)\begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix}.
  4. The plane with equation 2x−y+3z=12x - y + 3z = 1.
Explanation: The equation of the plane WW can be written as a dot product: (2−13)⋅(xyz)=0\begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} x \\ y \\ z \end{pmatrix} = 0. This shows that every vector in the plane WW is orthogonal to the normal vector n=(2,−1,3)\mathbf{n} = (2, -1, 3). The orthogonal complement W⊥W^\perp is the set of all vectors that are orthogonal to every vector in WW. This set is precisely the line spanned by the normal vector n\mathbf{n}. Therefore, W⊥W^\perp is the line through the origin with direction vector (2,−1,3)(2, -1, 3).

Question 6

The set of vectors S={(22−1),(2−12),(x22)}S = \left\{ \begin{pmatrix} 2 \\ 2 \\ -1 \end{pmatrix}, \begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix}, \begin{pmatrix} x \\ 2 \\ 2 \end{pmatrix} \right\} is given to be an orthogonal set. What must be the value of xx?

  1. x=2x = 2
  2. x=1x = 1
  3. x=−1x = -1 (correct answer)
  4. x=−2x = -2
Explanation: For a set of vectors to be orthogonal, every pair of distinct vectors in the set must have a dot product of zero. Let the vectors be v1,v2,v3\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3. We must have v1⋅v3=0\mathbf{v}_1 \cdot \mathbf{v}_3 = 0 and v2⋅v3=0\mathbf{v}_2 \cdot \mathbf{v}_3 = 0. Let's use the first condition: v1⋅v3=0\mathbf{v}_1 \cdot \mathbf{v}_3 = 0. (22−1)⋅(x22)=(2)(x)+(2)(2)+(−1)(2)=2x+4−2=2x+2\begin{pmatrix} 2 \\ 2 \\ -1 \end{pmatrix} \cdot \begin{pmatrix} x \\ 2 \\ 2 \end{pmatrix} = (2)(x) + (2)(2) + (-1)(2) = 2x + 4 - 2 = 2x + 2. Setting the dot product to zero: 2x+2=02x + 2 = 0, which gives 2x=−22x = -2, so x=−1x = -1. We can verify this with the second condition: v2⋅v3=0\mathbf{v}_2 \cdot \mathbf{v}_3 = 0. If x=−1x=-1, then (2−12)⋅(−122)=(2)(−1)+(−1)(2)+(2)(2)=−2−2+4=0\begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} -1 \\ 2 \\ 2 \end{pmatrix} = (2)(-1) + (-1)(2) + (2)(2) = -2 - 2 + 4 = 0. Both conditions are satisfied.

Question 7

Let y=(71)\mathbf{y} = \begin{pmatrix} 7 \\ 1 \end{pmatrix} and let WW be the subspace of R2\mathbb{R}^2 spanned by u=(42)\mathbf{u} = \begin{pmatrix} 4 \\ 2 \end{pmatrix}. The vector y\mathbf{y} can be written as y=w+z\mathbf{y} = \mathbf{w} + \mathbf{z}, where w∈W\mathbf{w} \in W and z∈W⊥\mathbf{z} \in W^\perp. What is the vector z\mathbf{z}?

  1. (63)\begin{pmatrix} 6 \\ 3 \end{pmatrix}
  2. (1−2)\begin{pmatrix} 1 \\ -2 \end{pmatrix} (correct answer)
  3. (3/23/2)\begin{pmatrix} 3/2 \\ 3/2 \end{pmatrix}
  4. (−12)\begin{pmatrix} -1 \\ 2 \end{pmatrix}
Explanation: The vector w\mathbf{w} is the orthogonal projection of y\mathbf{y} onto the line spanned by u\mathbf{u}. The formula is w=proju(y)=y⋅uu⋅uu\mathbf{w} = \mathrm{proj}_{\mathbf{u}}(\mathbf{y}) = \frac{\mathbf{y} \cdot \mathbf{u}}{\mathbf{u} \cdot \mathbf{u}} \mathbf{u}. First, calculate the dot products: y⋅u=(7)(4)+(1)(2)=28+2=30\mathbf{y} \cdot \mathbf{u} = (7)(4) + (1)(2) = 28 + 2 = 30. u⋅u=(4)(4)+(2)(2)=16+4=20\mathbf{u} \cdot \mathbf{u} = (4)(4) + (2)(2) = 16 + 4 = 20. Now find w\mathbf{w}: w=3020u=32(42)=(63)\mathbf{w} = \frac{30}{20} \mathbf{u} = \frac{3}{2} \begin{pmatrix} 4 \\ 2 \end{pmatrix} = \begin{pmatrix} 6 \\ 3 \end{pmatrix}. The vector z\mathbf{z} is the component of y\mathbf{y} orthogonal to WW, which is found by z=y−w\mathbf{z} = \mathbf{y} - \mathbf{w}. z=(71)−(63)=(1−2)\mathbf{z} = \begin{pmatrix} 7 \\ 1 \end{pmatrix} - \begin{pmatrix} 6 \\ 3 \end{pmatrix} = \begin{pmatrix} 1 \\ -2 \end{pmatrix}. Choice A is w\mathbf{w}, the projection onto WW, a common error.

Question 8

Let WW be the subspace of R4\mathbb{R}^4 spanned by the vectors v1=(1,2,−1,3)\mathbf{v}_1 = (1, 2, -1, 3) and v2=(2,1,0,−1)\mathbf{v}_2 = (2, 1, 0, -1). If u=(a,b,c,d)∈W⊥\mathbf{u} = (a, b, c, d) \in W^{\perp}, which of the following systems of equations must u\mathbf{u} satisfy?

  1. a+2b−c+3d=0a + 2b - c + 3d = 0 and 2a+b+d=02a + b + d = 0
  2. a+2b−c+3d=0a + 2b - c + 3d = 0 and 2a+b−d=02a + b - d = 0 (correct answer)
  3. a+2b−c+3d=0a + 2b - c + 3d = 0 and 2a+b=02a + b = 0
  4. a+2b−c+3d=0a + 2b - c + 3d = 0 and 2a+b+c−d=02a + b + c - d = 0
Explanation: For u\mathbf{u} to be in W⊥W^{\perp}, it must be orthogonal to every vector in WW. Since W=span{v1,v2}W = \text{span}\{\mathbf{v}_1, \mathbf{v}_2\}, we need u⋅v1=0\mathbf{u} \cdot \mathbf{v}_1 = 0 and u⋅v2=0\mathbf{u} \cdot \mathbf{v}_2 = 0. Computing: u⋅v1=a+2b−c+3d=0\mathbf{u} \cdot \mathbf{v}_1 = a + 2b - c + 3d = 0 and u⋅v2=2a+b+0c+(−1)d=2a+b−d=0\mathbf{u} \cdot \mathbf{v}_2 = 2a + b + 0c + (-1)d = 2a + b - d = 0. Choice A has the wrong sign for dd in the second equation. Choice C omits the dd term entirely. Choice D incorrectly includes cc in the second equation.

Question 9

Let VV be a finite-dimensional inner product space and WW a subspace of VV. If dim⁡(V)=5\dim(V) = 5 and dim⁡(W)=2\dim(W) = 2, and v∈V\mathbf{v} \in V can be written as v=w+u\mathbf{v} = \mathbf{w} + \mathbf{u} where w∈W\mathbf{w} \in W and u∈W⊥\mathbf{u} \in W^{\perp}, what is dim⁡(W⊥)\dim(W^{\perp})?

  1. 22
  2. 33 (correct answer)
  3. 44
  4. 55
Explanation: By the fundamental theorem about orthogonal complements in finite-dimensional inner product spaces, dim⁡(W)+dim⁡(W⊥)=dim⁡(V)\dim(W) + \dim(W^{\perp}) = \dim(V). Given dim⁡(V)=5\dim(V) = 5 and dim⁡(W)=2\dim(W) = 2, we have dim⁡(W⊥)=5−2=3\dim(W^{\perp}) = 5 - 2 = 3. Choice A incorrectly assumes dim⁡(W⊥)=dim⁡(W)\dim(W^{\perp}) = \dim(W). Choice D would mean W⊥=VW^{\perp} = V, which only occurs when W={0}W = \{\mathbf{0}\}. Choice C results from the incorrect calculation 5−1=45 - 1 = 4.

Question 10

Let WW be a subspace of an inner product space VV, and let v∈V\mathbf{v} \in V. The orthogonal projection of v\mathbf{v} onto WW is denoted projW(v)\text{proj}_W(\mathbf{v}). Which statement about v−projW(v)\mathbf{v} - \text{proj}_W(\mathbf{v}) is always true?

  1. v−projW(v)∈W\mathbf{v} - \text{proj}_W(\mathbf{v}) \in W and minimizes the distance to v\mathbf{v}
  2. v−projW(v)∈W⊥\mathbf{v} - \text{proj}_W(\mathbf{v}) \in W^{\perp} and is the closest vector in W⊥W^{\perp} to v\mathbf{v}
  3. v−projW(v)∈W⊥\mathbf{v} - \text{proj}_W(\mathbf{v}) \in W^{\perp} and minimizes ∥v−w∥\|\mathbf{v} - \mathbf{w}\| over all w∈W\mathbf{w} \in W (correct answer)
  4. v−projW(v)∈W⊥\mathbf{v} - \text{proj}_W(\mathbf{v}) \in W^{\perp} and equals projW⊥(v)\text{proj}_{W^{\perp}}(\mathbf{v}) in general
Explanation: By definition of orthogonal projection, v−projW(v)∈W⊥\mathbf{v} - \text{proj}_W(\mathbf{v}) \in W^{\perp}, and projW(v)\text{proj}_W(\mathbf{v}) is the unique vector in WW closest to v\mathbf{v}. This means ∥v−projW(v)∥≤∥v−w∥\|\mathbf{v} - \text{proj}_W(\mathbf{v})\| \leq \|\mathbf{v} - \mathbf{w}\| for all w∈W\mathbf{w} \in W. Choice A incorrectly places the difference vector in WW. Choice B incorrectly describes it as closest among vectors in W⊥W^{\perp}. Choice D incorrectly equates it with projW⊥(v)\text{proj}_{W^{\perp}}(\mathbf{v}).

Question 11

Let T:R4→R3T: \mathbb{R}^4 \to \mathbb{R}^3 be a linear transformation with matrix representation A=(120101−12102−1)A = \begin{pmatrix} 1 & 2 & 0 & 1 \\ 0 & 1 & -1 & 2 \\ 1 & 0 & 2 & -1 \end{pmatrix} . How are ker⁡(T)\ker(T) and (Col(A))⊥(\text{Col}(A))^{\perp} related?

  1. ker⁡(T)=(Col(A))⊥\ker(T) = (\text{Col}(A))^{\perp} since both are orthogonal complements of the column space
  2. ker⁡(T)\ker(T) and (Col(A))⊥(\text{Col}(A))^{\perp} exist in different spaces and have no direct relationship
  3. ker⁡(T)=(Col(A))⊥\ker(T) = (\text{Col}(A))^{\perp} where both subspaces exist in R3\mathbb{R}^3
  4. ker⁡(T)⊥Row(A)\ker(T) \perp \text{Row}(A) and ker⁡(T)=(Row(A))⊥\ker(T) = (\text{Row}(A))^{\perp} where Row(A)⊂R4(A) \subset \mathbb{R}^4 (correct answer)
Explanation: When analyzing relationships between subspaces associated with a linear transformation, you need to understand where each subspace lives and apply the Fundamental Theorem of Linear Algebra, which reveals four interconnected subspaces. For the linear transformation T:R4→R3T: \mathbb{R}^4 \to \mathbb{R}^3 with matrix AA, the kernel ker⁡(T)\ker(T) consists of all vectors x∈R4\mathbf{x} \in \mathbb{R}^4 such that Ax=0A\mathbf{x} = \mathbf{0}. The Fundamental Theorem tells us that ker⁡(T)\ker(T) is the orthogonal complement of the row space of AA. Since the rows of AA are vectors in R4\mathbb{R}^4, we have Row(A)⊂R4\text{Row}(A) \subset \mathbb{R}^4 and ker⁡(T)=(Row(A))⊥\ker(T) = (\text{Row}(A))^{\perp} in R4\mathbb{R}^4. This makes answer D correct. A is wrong because it confuses the column space relationship. The kernel and column space orthogonal complement aren't equal. B is incorrect because there is a fundamental relationship between these subspaces, just not the one initially suggested. C contains two errors: ker⁡(T)\ker(T) exists in R4\mathbb{R}^4 (the domain), not R3\mathbb{R}^3, and the kernel isn't the orthogonal complement of the column space. Key strategy: Remember the four fundamental subspaces come in orthogonal pairs within the same vector space: ker⁡(A)⊥Row(A)\ker(A) \perp \text{Row}(A) in Rn\mathbb{R}^n, and ker⁡(AT)⊥Col(A)\ker(A^T) \perp \text{Col}(A) in Rm\mathbb{R}^m. Always check which space each subspace lives in before claiming relationships.

Question 12

Consider the inner product space C[0,1]C[0,1] of continuous functions on [0,1][0,1] with inner product ⟨f,g⟩=∫01f(x)g(x)dx\langle f, g \rangle = \int_0^1 f(x)g(x) dx. Let W={f∈C[0,1]:∫01f(x)dx=0}W = \{f \in C[0,1] : \int_0^1 f(x) dx = 0\}. Which function is in W⊥W^{\perp}?

  1. f(x)=x−12f(x) = x - \frac{1}{2}
  2. f(x)=x(1−x)f(x) = x(1-x)
  3. f(x)=sin⁡(πx)f(x) = \sin(\pi x)
  4. f(x)=1f(x) = 1 (the constant function) (correct answer)
Explanation: When you encounter orthogonal complement problems in function spaces, you need to find which functions are perpendicular to every function in the given subspace WW. The key insight is understanding what it means for two functions to be orthogonal under the given inner product. The subspace WW contains all continuous functions whose integral over [0,1][0,1] equals zero. For a function gg to be in W⊥W^{\perp}, it must satisfy ⟨f,g⟩=0\langle f, g \rangle = 0 for every f∈Wf \in W. This means ∫01f(x)g(x)dx=0\int_0^1 f(x)g(x) dx = 0 whenever ∫01f(x)dx=0\int_0^1 f(x) dx = 0. The constant function g(x)=1g(x) = 1 is in W⊥W^{\perp} because for any f∈Wf \in W: ⟨f,1⟩=∫01f(x)⋅1dx=∫01f(x)dx=0\langle f, 1 \rangle = \int_0^1 f(x) \cdot 1 dx = \int_0^1 f(x) dx = 0 by the definition of WW. This makes choice D correct. Let's check why the others fail. For choice A, f(x)=x−12f(x) = x - \frac{1}{2}: we can verify ∫01(x−12)dx=0\int_0^1 (x - \frac{1}{2}) dx = 0, so this function is actually in WW, not in W⊥W^{\perp}. Similarly, choice C gives ∫01sin⁡(πx)dx=0\int_0^1 \sin(\pi x) dx = 0, placing it in WW. Choice B also integrates to zero: ∫01x(1−x)dx=16\int_0^1 x(1-x) dx = \frac{1}{6}, wait—this actually gives 16≠0\frac{1}{6} \neq 0, but it's still not in W⊥W^{\perp} since it doesn't satisfy the orthogonality condition with all functions in WW. Remember: W⊥W^{\perp} consists of functions orthogonal to the entire subspace WW, which often includes the "simplest" functions that naturally pair to zero with the constraint defining WW.

Question 13

Consider the subspace U={(x,y,z,w)∈R4:x+y−z=0 and 2x−y+w=0}U = \{(x, y, z, w) \in \mathbb{R}^4 : x + y - z = 0 \text{ and } 2x - y + w = 0\} of R4\mathbb{R}^4 with the standard inner product. Which of the following vectors spans U⊥U^{\perp}?

  1. {(1,1,−1,0),(2,−1,0,1)}\{(1, 1, -1, 0), (2, -1, 0, 1)\} (correct answer)
  2. {(1,1,−1,0),(−2,1,0,1)}\{(1, 1, -1, 0), (-2, 1, 0, 1)\}
  3. {(1,−1,1,0),(2,−1,0,1)}\{(1, -1, 1, 0), (2, -1, 0, 1)\}
  4. {(1,1,1,0),(2,−1,0,−1)}\{(1, 1, 1, 0), (2, -1, 0, -1)\}
Explanation: The subspace UU is defined by the system of linear equations. The orthogonal complement U⊥U^{\perp} is spanned by the normal vectors to these hyperplanes. From x+y−z=0x + y - z = 0, we get normal vector (1,1,−1,0)(1, 1, -1, 0). From 2x−y+w=02x - y + w = 0, we get normal vector (2,−1,0,1)(2, -1, 0, 1). Choice B has the wrong sign in the second vector. Choice C has the wrong signs in the first vector. Choice D has wrong signs in both vectors' components.

Question 14

In Rn\mathbb{R}^n with the standard inner product, let S={v1,v2,…,vk}S = \{\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_k\} be a set of vectors. If (span(S))⊥={0}(\text{span}(S))^{\perp} = \{\mathbf{0}\}, which statement must be true?

  1. The vectors in SS are orthonormal and k=nk = n
  2. The vectors in SS span Rn\mathbb{R}^n and are linearly independent
  3. The vectors in SS span Rn\mathbb{R}^n but may be linearly dependent (correct answer)
  4. The vectors in SS are linearly independent but may not span Rn\mathbb{R}^n
Explanation: If (span(S))⊥={0}(\text{span}(S))^{\perp} = \{\mathbf{0}\}, then span(S)(S) has no non-zero orthogonal vectors, which means span(S)=Rn(S) = \mathbb{R}^n (since dim⁡(span(S))+dim⁡((span(S))⊥)=n\dim(\text{span}(S)) + \dim((\text{span}(S))^{\perp}) = n and dim⁡({0})=0\dim(\{\mathbf{0}\}) = 0). However, the vectors in SS need not be linearly independent; we could have redundant vectors. Choice A incorrectly requires orthonormality and specific cardinality. Choice B incorrectly requires linear independence. Choice D contradicts the given condition since if span(S)≠Rn(S) \neq \mathbb{R}^n, then (span(S))⊥≠{0}(\text{span}(S))^{\perp} \neq \{\mathbf{0}\}.

Question 15

Let UU and VV be subspaces of a finite-dimensional inner product space WW. If U⊥VU \perp V (meaning every vector in UU is orthogonal to every vector in VV), which relationship between (U+V)⊥(U + V)^{\perp} and U⊥∩V⊥U^{\perp} \cap V^{\perp} is correct?

  1. (U+V)⊥=U⊥∩V⊥(U + V)^{\perp} = U^{\perp} \cap V^{\perp} regardless of the orthogonality condition (correct answer)
  2. (U+V)⊥=U⊥∩V⊥(U + V)^{\perp} = U^{\perp} \cap V^{\perp} and this equality depends on U⊥VU \perp V
  3. (U+V)⊥=U⊥∩V⊥(U + V)^{\perp} = U^{\perp} \cap V^{\perp} and dim⁡((U+V)⊥)=dim⁡(U⊥)+dim⁡(V⊥)−dim⁡(W)\dim((U + V)^{\perp}) = \dim(U^{\perp}) + \dim(V^{\perp}) - \dim(W)
  4. (U+V)⊥⊂U⊥∩V⊥(U + V)^{\perp} \subset U^{\perp} \cap V^{\perp} but equality holds only when U∩V={0}U \cap V = \{\mathbf{0}\}
Explanation: The identity (U+V)⊥=U⊥∩V⊥(U + V)^{\perp} = U^{\perp} \cap V^{\perp} is a fundamental property of orthogonal complements that holds for any subspaces UU and VV, regardless of whether U⊥VU \perp V. A vector is orthogonal to U+VU + V if and only if it's orthogonal to both UU and VV. Choice B incorrectly suggests the equality depends on orthogonality. Choice C provides an incorrect dimension formula. Choice D incorrectly suggests only inclusion holds generally, when equality always holds.

Question 16

Let AA be a 4×64 \times 6 matrix with rank(A)=3\mathrm{rank}(A) = 3. What is the dimension of the orthogonal complement of the row space of AA?

  1. 1
  2. 2
  3. 3 (correct answer)
  4. 4
Explanation: The orthogonal complement of the row space of AA, (RowA)⊥(\mathrm{Row } A)^\perp, is the null space of AA, Nul(A)\mathrm{Nul}(A). The question asks for the dimension of Nul(A)\mathrm{Nul}(A). According to the Rank-Nullity Theorem, for a matrix with nn columns, rank(A)+dim⁡(Nul(A))=n\mathrm{rank}(A) + \dim(\mathrm{Nul}(A)) = n. In this case, AA has n=6n=6 columns and its rank is given as 3. Therefore, 3+dim⁡(Nul(A))=63 + \dim(\mathrm{Nul}(A)) = 6, which implies dim⁡(Nul(A))=3\dim(\mathrm{Nul}(A)) = 3.

Question 17

Let WW be a subspace of Rn\mathbb{R}^n. Which of the following statements is NOT always true?

  1. W⊥W^\perp is a subspace of Rn\mathbb{R}^n.
  2. W∩W⊥={0}W \cap W^\perp = \{\mathbf{0}\}.
  3. dim⁡(W)+dim⁡(W⊥)=n\dim(W) + \dim(W^\perp) = n.
  4. W∪W⊥=RnW \cup W^\perp = \mathbb{R}^n. (correct answer)
Explanation: The union of two subspaces is not generally a subspace, and it does not typically equal the entire vector space. For example, in R3\mathbb{R}^3, if WW is a line through the origin (a 1D subspace), then W⊥W^\perp is a plane through the origin (a 2D subspace). The union W∪W⊥W \cup W^\perp consists of only the points on that line and on that plane, which does not include all points in R3\mathbb{R}^3. The other three statements are fundamental properties of orthogonal complements in Rn\mathbb{R}^n.

Question 18

Let UU and WW be subspaces of Rn\mathbb{R}^n. Which of the following expressions is always equivalent to (U⊥+W⊥)⊥(U^\perp + W^\perp)^\perp?

  1. U+WU + W
  2. U∪WU \cup W
  3. U⊥∩W⊥U^\perp \cap W^\perp
  4. U∩WU \cap W (correct answer)
Explanation: When you encounter questions about orthogonal complements and subspace operations, the key is understanding how these operations interact with each other. This requires familiarity with fundamental properties of orthogonal complements. To find (U⊥+W⊥)⊥(U^\perp + W^\perp)^\perp, we need to apply the crucial property that for any subspaces V1V_1 and V2V_2: (V1+V2)⊥=V1⊥∩V2⊥(V_1 + V_2)^\perp = V_1^\perp \cap V_2^\perp. Applying this property with V1=U⊥V_1 = U^\perp and V2=W⊥V_2 = W^\perp: (U⊥+W⊥)⊥=(U⊥)⊥∩(W⊥)⊥(U^\perp + W^\perp)^\perp = (U^\perp)^\perp \cap (W^\perp)^\perp Since taking the orthogonal complement twice returns you to the original subspace (i.e., (V⊥)⊥=V(V^\perp)^\perp = V for any subspace VV), we have: (U⊥)⊥=U(U^\perp)^\perp = U and (W⊥)⊥=W(W^\perp)^\perp = W Therefore: (U⊥+W⊥)⊥=U∩W(U^\perp + W^\perp)^\perp = U \cap W Now for the incorrect options: A confuses the relationship—U+WU + W would actually equal (U⊥∩W⊥)⊥(U^\perp \cap W^\perp)^\perp, not our expression. B represents set union, which isn't even a subspace operation (the union of subspaces is generally not a subspace). C gives us U⊥∩W⊥U^\perp \cap W^\perp, which equals (U+W)⊥(U + W)^\perp—essentially the "opposite" of what we want. Strategy tip: Memorize the key duality: (V1+V2)⊥=V1⊥∩V2⊥(V_1 + V_2)^\perp = V_1^\perp \cap V_2^\perp and (V1∩V2)⊥=V1⊥+V2⊥(V_1 \cap V_2)^\perp = V_1^\perp + V_2^\perp. These relationships flip sums to intersections and vice versa when taking orthogonal complements.

Question 19

Let WW be the subspace of R4\mathbb{R}^4 spanned by the vectors u1=(1,1,0,0)\mathbf{u}_1 = (1, 1, 0, 0) and u2=(0,1,1,0)\mathbf{u}_2 = (0, 1, 1, 0). Which of the following vectors is in W⊥W^\perp?

  1. (1,−1,1,5)(1, -1, 1, 5) (correct answer)
  2. (1,−1,−5,5)(1, -1, -5, 5)
  3. (1,2,1,0)(1, 2, 1, 0)
  4. (1,1,−1,−1)(1, 1, -1, -1)
Explanation: A vector v\mathbf{v} is in W⊥W^\perp if and only if it is orthogonal to every vector in a spanning set for WW. Therefore, we need to find a vector v\mathbf{v} such that v⋅u1=0\mathbf{v} \cdot \mathbf{u}_1 = 0 and v⋅u2=0\mathbf{v} \cdot \mathbf{u}_2 = 0. Let's test choice A: v=(1,−1,1,5)\mathbf{v} = (1, -1, 1, 5). v⋅u1=(1)(1)+(−1)(1)+(1)(0)+(5)(0)=1−1=0\mathbf{v} \cdot \mathbf{u}_1 = (1)(1) + (-1)(1) + (1)(0) + (5)(0) = 1 - 1 = 0. v⋅u2=(1)(0)+(−1)(1)+(1)(1)+(5)(0)=−1+1=0\mathbf{v} \cdot \mathbf{u}_2 = (1)(0) + (-1)(1) + (1)(1) + (5)(0) = -1 + 1 = 0. Since the vector in choice A is orthogonal to both spanning vectors, it is in W⊥W^\perp. Choice B is orthogonal to u1\mathbf{u}_1 but not u2\mathbf{u}_2. Choice C is in WW itself (it's u1+u2\mathbf{u}_1 + \mathbf{u}_2). Choice D is orthogonal to u2\mathbf{u}_2 but not u1\mathbf{u}_1.

Question 20

Let S={v1,v2,…,vk}S = \{\mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_k\} be a set of non-zero vectors in Rn\mathbb{R}^n. If SS is an orthogonal set, which of the following statements must be true?

  1. SS is a basis for Rn\mathbb{R}^n.
  2. For any vector v∈Rn\mathbf{v} \in \mathbb{R}^n, v⋅vi=0\mathbf{v} \cdot \mathbf{v}_i = 0 for some ii.
  3. The span of SS is equal to Rn\mathbb{R}^n.
  4. SS is a linearly independent set. (correct answer)
Explanation: When you encounter questions about orthogonal sets, focus on what orthogonality guarantees versus what it doesn't. An orthogonal set means every pair of distinct vectors has a dot product of zero: vi⋅vj=0\mathbf{v}_i \cdot \mathbf{v}_j = 0 for all i≠ji \neq j. The key insight is that orthogonal sets of non-zero vectors are automatically linearly independent. Here's why: suppose we have a linear combination c1v1+c2v2+⋯+ckvk=0c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + \cdots + c_k\mathbf{v}_k = \mathbf{0}. Taking the dot product of both sides with v1\mathbf{v}_1 gives us c1(v1⋅v1)+c2(v2⋅v1)+⋯=0c_1(\mathbf{v}_1 \cdot \mathbf{v}_1) + c_2(\mathbf{v}_2 \cdot \mathbf{v}_1) + \cdots = 0. Since the vectors are orthogonal, all terms except the first vanish, leaving c1∥v1∥2=0c_1\|\mathbf{v}_1\|^2 = 0. Since v1\mathbf{v}_1 is non-zero, we must have c1=0c_1 = 0. Repeating this argument for each vector shows all coefficients are zero, proving linear independence. Therefore, D is correct. A is wrong because we need exactly nn linearly independent vectors to form a basis for Rn\mathbb{R}^n, but kk could be less than nn. B is incorrect—there's no reason why every vector in Rn\mathbb{R}^n must be orthogonal to some vector in SS. C fails for the same reason as A: if k<nk < n, the span of SS is only a kk-dimensional subspace, not all of Rn\mathbb{R}^n. Remember: orthogonality always implies linear independence for non-zero vectors, but it doesn't guarantee the set spans the entire space or forms a basis.