Orthogonality & Complements - Linear Algebra

1. The set of vectors $S = \left\{ \begin{pmatrix} 2 \\ 2 \\ -1 \end{pmatrix}, \begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix}, \begin{pmatrix} x \\ 2 \\ 2 \end{pmatrix} \right\}$ is given to be an orthogonal set. What must be the value of $x$?
   Answer: $x = -1$ 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 $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$. We must have $\mathbf{v}_1 \cdot \mathbf{v}_3 = 0$ and $\mathbf{v}_2 \cdot \mathbf{v}_3 = 0$. Let's use the first condition: $\mathbf{v}_1 \cdot \mathbf{v}_3 = 0$. $\begin{pmatrix} 2 \\ 2 \\ -1 \end{pmatrix} \cdot \begin{pmatrix} x \\ 2 \\ 2 \end{pmatrix} = (2)(x) + (2)(2) + (-1)(2) = 2x + 4 - 2 = 2x + 2$. Setting the dot product to zero: $2x + 2 = 0$, which gives $2x = -2$, so $x = -1$. We can verify this with the second condition: $\mathbf{v}_2 \cdot \mathbf{v}_3 = 0$. If $x=-1$, then $\begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} -1 \\ 2 \\ 2 \end{pmatrix} = (2)(-1) + (-1)(2) + (2)(2) = -2 - 2 + 4 = 0$. Both conditions are satisfied.
2. Let $\mathbf{y} = \begin{pmatrix} 7 \\ 1 \end{pmatrix}$ and let $W$ be the subspace of $\mathbb{R}^2$ spanned by $\mathbf{u} = \begin{pmatrix} 4 \\ 2 \end{pmatrix}$. The vector $\mathbf{y}$ can be written as $\mathbf{y} = \mathbf{w} + \mathbf{z}$, where $\mathbf{w} \in W$ and $\mathbf{z} \in W^\perp$. What is the vector $\mathbf{z}$?
   Answer: $\begin{pmatrix} 1 \\ -2 \end{pmatrix}$ The vector $\mathbf{w}$ is the orthogonal projection of $\mathbf{y}$ onto the line spanned by $\mathbf{u}$. The formula is $\mathbf{w} = \mathrm{proj}_{\mathbf{u}}(\mathbf{y}) = \frac{\mathbf{y} \cdot \mathbf{u}}{\mathbf{u} \cdot \mathbf{u}} \mathbf{u}$. First, calculate the dot products: $\mathbf{y} \cdot \mathbf{u} = (7)(4) + (1)(2) = 28 + 2 = 30$. $\mathbf{u} \cdot \mathbf{u} = (4)(4) + (2)(2) = 16 + 4 = 20$. Now find $\mathbf{w}$: $\mathbf{w} = \frac{30}{20} \mathbf{u} = \frac{3}{2} \begin{pmatrix} 4 \\ 2 \end{pmatrix} = \begin{pmatrix} 6 \\ 3 \end{pmatrix}$. The vector $\mathbf{z}$ is the component of $\mathbf{y}$ orthogonal to $W$, which is found by $\mathbf{z} = \mathbf{y} - \mathbf{w}$. $\mathbf{z} = \begin{pmatrix} 7 \\ 1 \end{pmatrix} - \begin{pmatrix} 6 \\ 3 \end{pmatrix} = \begin{pmatrix} 1 \\ -2 \end{pmatrix}$. Choice A is $\mathbf{w}$, the projection onto $W$, a common error.
3. Let $W$ be the subspace of $\mathbb{R}^4$ spanned by the vectors $\mathbf{u}_1 = (1, 1, 0, 0)$ and $\mathbf{u}_2 = (0, 1, 1, 0)$. Which of the following vectors is in $W^\perp$?
   Answer: $(1, -1, 1, 5)$ A vector $\mathbf{v}$ is in $W^\perp$ if and only if it is orthogonal to every vector in a spanning set for $W$. Therefore, we need to find a vector $\mathbf{v}$ such that $\mathbf{v} \cdot \mathbf{u}_1 = 0$ and $\mathbf{v} \cdot \mathbf{u}_2 = 0$. Let's test choice A: $\mathbf{v} = (1, -1, 1, 5)$. $\mathbf{v} \cdot \mathbf{u}_1 = (1)(1) + (-1)(1) + (1)(0) + (5)(0) = 1 - 1 = 0$. $\mathbf{v} \cdot \mathbf{u}_2 = (1)(0) + (-1)(1) + (1)(1) + (5)(0) = -1 + 1 = 0$. Since the vector in choice A is orthogonal to both spanning vectors, it is in $W^\perp$. Choice B is orthogonal to $\mathbf{u}_1$ but not $\mathbf{u}_2$. Choice C is in $W$ itself (it's $\mathbf{u}_1 + \mathbf{u}_2$). Choice D is orthogonal to $\mathbf{u}_2$ but not $\mathbf{u}_1$.
4. Let $S = \{\mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_k\}$ be a set of non-zero vectors in $\mathbb{R}^n$. If $S$ is an orthogonal set, which of the following statements must be true?
   Answer: $S$ is a linearly independent set. 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: $\mathbf{v}_i \cdot \mathbf{v}_j = 0$ for all $i \neq j$. The key insight is that orthogonal sets of non-zero vectors are automatically linearly independent. Here's why: suppose we have a linear combination $c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + \cdots + c_k\mathbf{v}_k = \mathbf{0}$. Taking the dot product of both sides with $\mathbf{v}_1$ gives us $c_1(\mathbf{v}_1 \cdot \mathbf{v}_1) + c_2(\mathbf{v}_2 \cdot \mathbf{v}_1) + \cdots = 0$. Since the vectors are orthogonal, all terms except the first vanish, leaving $c_1\|\mathbf{v}_1\|^2 = 0$. Since $\mathbf{v}_1$ is non-zero, we must have $c_1 = 0$. 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 $n$ linearly independent vectors to form a basis for $\mathbb{R}^n$, but $k$ could be less than $n$. B is incorrect—there's no reason why every vector in $\mathbb{R}^n$ must be orthogonal to some vector in $S$. C fails for the same reason as A: if $k < n$, the span of $S$ is only a $k$-dimensional subspace, not all of $\mathbb{R}^n$. 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.
5. Consider the inner product space $C[0,1]$ of continuous functions on $[0,1]$ with inner product $\langle f, g \rangle = \int_0^1 f(x)g(x) dx$. Let $W = \{f \in C[0,1] : \int_0^1 f(x) dx = 0\}$. Which function is in $W^{\perp}$?
   Answer: $f(x) = 1$ (the constant function) When you encounter orthogonal complement problems in function spaces, you need to find which functions are perpendicular to every function in the given subspace $W$. The key insight is understanding what it means for two functions to be orthogonal under the given inner product. The subspace $W$ contains all continuous functions whose integral over $[0,1]$ equals zero. For a function $g$ to be in $W^{\perp}$, it must satisfy $\langle f, g \rangle = 0$ for every $f \in W$. This means $\int_0^1 f(x)g(x) dx = 0$ whenever $\int_0^1 f(x) dx = 0$. The constant function $g(x) = 1$ is in $W^{\perp}$ because for any $f \in W$: $\langle f, 1 \rangle = \int_0^1 f(x) \cdot 1 dx = \int_0^1 f(x) dx = 0$ by the definition of $W$. This makes choice D correct. Let's check why the others fail. For choice A, $f(x) = x - \frac{1}{2}$: we can verify $\int_0^1 (x - \frac{1}{2}) dx = 0$, so this function is actually in $W$, not in $W^{\perp}$. Similarly, choice C gives $\int_0^1 \sin(\pi x) dx = 0$, placing it in $W$. Choice B also integrates to zero: $\int_0^1 x(1-x) dx = \frac{1}{6}$, wait—this actually gives $\frac{1}{6} \neq 0$, but it's still not in $W^{\perp}$ since it doesn't satisfy the orthogonality condition with all functions in $W$. Remember: $W^{\perp}$ consists of functions orthogonal to the entire subspace $W$, which often includes the "simplest" functions that naturally pair to zero with the constraint defining $W$.
6. Let $W$ be a subspace of an inner product space $V$, and let $\mathbf{v} \in V$. The orthogonal projection of $\mathbf{v}$ onto $W$ is denoted $\text{proj}_W(\mathbf{v})$. Which statement about $\mathbf{v} - \text{proj}_W(\mathbf{v})$ is always true?
   Answer: $\mathbf{v} - \text{proj}_W(\mathbf{v}) \in W^{\perp}$ and minimizes $\|\mathbf{v} - \mathbf{w}\|$ over all $\mathbf{w} \in W$ By definition of orthogonal projection, $\mathbf{v} - \text{proj}_W(\mathbf{v}) \in W^{\perp}$, and $\text{proj}_W(\mathbf{v})$ is the unique vector in $W$ closest to $\mathbf{v}$. This means $\|\mathbf{v} - \text{proj}_W(\mathbf{v})\| \leq \|\mathbf{v} - \mathbf{w}\|$ for all $\mathbf{w} \in W$. Choice A incorrectly places the difference vector in $W$. Choice B incorrectly describes it as closest among vectors in $W^{\perp}$. Choice D incorrectly equates it with $\text{proj}_{W^{\perp}}(\mathbf{v})$.
7. Let $A$ be a $4 \times 6$ matrix with $\mathrm{rank}(A) = 3$. What is the dimension of the orthogonal complement of the row space of $A$?
   Answer: 3 The orthogonal complement of the row space of $A$, $(\mathrm{Row } A)^\perp$, is the null space of $A$, $\mathrm{Nul}(A)$. The question asks for the dimension of $\mathrm{Nul}(A)$. According to the Rank-Nullity Theorem, for a matrix with $n$ columns, $\mathrm{rank}(A) + \dim(\mathrm{Nul}(A)) = n$. In this case, $A$ has $n=6$ columns and its rank is given as 3. Therefore, $3 + \dim(\mathrm{Nul}(A)) = 6$, which implies $\dim(\mathrm{Nul}(A)) = 3$.
8. Let $V$ be a finite-dimensional inner product space and $W$ a subspace of $V$. If $\dim(V) = 5$ and $\dim(W) = 2$, and $\mathbf{v} \in V$ can be written as $\mathbf{v} = \mathbf{w} + \mathbf{u}$ where $\mathbf{w} \in W$ and $\mathbf{u} \in W^{\perp}$, what is $\dim(W^{\perp})$?
   Answer: $3$ By the fundamental theorem about orthogonal complements in finite-dimensional inner product spaces, $\dim(W) + \dim(W^{\perp}) = \dim(V)$. Given $\dim(V) = 5$ and $\dim(W) = 2$, we have $\dim(W^{\perp}) = 5 - 2 = 3$. Choice A incorrectly assumes $\dim(W^{\perp}) = \dim(W)$. Choice D would mean $W^{\perp} = V$, which only occurs when $W = \{\mathbf{0}\}$. Choice C results from the incorrect calculation $5 - 1 = 4$.
9. Let $W$ be the subspace of $\mathbb{R}^4$ consisting of all vectors $(x_1, x_2, x_3, x_4)$ that satisfy the equations $x_1 - x_2 + x_3 = 0$ and $x_2 - x_4 = 0$. Which of the following vectors lies in $W^\perp$?
   Answer: $(1, 0, 1, -1)$ The subspace $W$ is the set of solutions to a homogeneous system of linear equations, which means $W$ is the null space of the matrix $A = \begin{pmatrix} 1 & -1 & 1 & 0 \\ 0 & 1 & 0 & -1 \end{pmatrix}$. The orthogonal complement of the null space of a matrix is its row space, i.e., $W^\perp = (\mathrm{Nul } A)^\perp = \mathrm{Row } A$. Therefore, we are looking for a vector that is a linear combination of the rows of $A$. The rows are $\mathbf{r}_1 = (1, -1, 1, 0)$ and $\mathbf{r}_2 = (0, 1, 0, -1)$. Let's check the options: - Choice A: $(-1, 0, 1, 0)$ is in $W$ because $(-1) - (0) + (1) = 0$ and $(0) - (0) = 0$. It cannot be in $W^\perp$ unless it is the zero vector. - Choice B: $(1, 0, 1, -1)$ is the sum of the two rows: $\mathbf{r}_1 + \mathbf{r}_2$. Since it is a linear combination of the rows, it is in the row space, and thus in $W^\perp$. - Choices C and D are not linear combinations of the rows of $A$.
10. Let $T: \mathbb{R}^4 \to \mathbb{R}^3$ be a linear transformation with matrix representation $A = \begin{pmatrix} 1 & 2 & 0 & 1 \\ 0 & 1 & -1 & 2 \\ 1 & 0 & 2 & -1 \end{pmatrix}$. How are $\ker(T)$ and $(\text{Col}(A))^{\perp}$ related?
    Answer: $\ker(T) \perp \text{Row}(A)$ and $\ker(T) = (\text{Row}(A))^{\perp}$ where Row$(A) \subset \mathbb{R}^4$ 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 $T: \mathbb{R}^4 \to \mathbb{R}^3$ with matrix $A$, the kernel $\ker(T)$ consists of all vectors $\mathbf{x} \in \mathbb{R}^4$ such that $A\mathbf{x} = \mathbf{0}$. The Fundamental Theorem tells us that $\ker(T)$ is the orthogonal complement of the row space of $A$. Since the rows of $A$ are vectors in $\mathbb{R}^4$, we have $\text{Row}(A) \subset \mathbb{R}^4$ and $\ker(T) = (\text{Row}(A))^{\perp}$ in $\mathbb{R}^4$. 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: $\ker(T)$ exists in $\mathbb{R}^4$ (the domain), not $\mathbb{R}^3$, 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: $\ker(A) \perp \text{Row}(A)$ in $\mathbb{R}^n$, and $\ker(A^T) \perp \text{Col}(A)$ in $\mathbb{R}^m$. Always check which space each subspace lives in before claiming relationships.
11. Let $W$ be the subspace of $\mathbb{R}^3$ spanned by the vector $\mathbf{v} = \begin{pmatrix} 1 \\ -2 \\ 3 \end{pmatrix}$. Which of the following sets is a basis for the orthogonal complement $W^\perp$?
    Answer: $\left\{ \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} -3 \\ 0 \\ 1 \end{pmatrix} \right\}$ The orthogonal complement $W^\perp$ consists of all vectors $\mathbf{x} = (x_1, x_2, x_3)$ such that $\mathbf{v} \cdot \mathbf{x} = 0$. This gives the equation $x_1 - 2x_2 + 3x_3 = 0$. This is the equation of a plane in $\mathbb{R}^3$. Since $\dim(W) = 1$ and $\dim(\mathbb{R}^3) = 3$, the dimension of $W^\perp$ must be $\dim(\mathbb{R}^3) - \dim(W) = 3 - 1 = 2$. Therefore, a basis for $W^\perp$ must contain two linearly independent vectors. Both vectors in choice B satisfy the equation: $(2) - 2(1) + 3(0) = 0$ and $(-3) - 2(0) + 3(1) = 0$. They are also linearly independent. Thus, they form a basis for $W^\perp$.
12. Let $U$ and $W$ be subspaces of $\mathbb{R}^n$. Which of the following expressions is always equivalent to $(U^\perp + W^\perp)^\perp$?
    Answer: $U \cap W$ 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 $(U^\perp + W^\perp)^\perp$, we need to apply the crucial property that for any subspaces $V_1$ and $V_2$: $(V_1 + V_2)^\perp = V_1^\perp \cap V_2^\perp$. Applying this property with $V_1 = U^\perp$ and $V_2 = W^\perp$: $(U^\perp + W^\perp)^\perp = (U^\perp)^\perp \cap (W^\perp)^\perp$ Since taking the orthogonal complement twice returns you to the original subspace (i.e., $(V^\perp)^\perp = V$ for any subspace $V$), we have: $(U^\perp)^\perp = U$ and $(W^\perp)^\perp = W$ Therefore: $(U^\perp + W^\perp)^\perp = U \cap W$ Now for the incorrect options: A confuses the relationship—$U + W$ would actually equal $(U^\perp \cap W^\perp)^\perp$, 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 $U^\perp \cap W^\perp$, which equals $(U + W)^\perp$—essentially the "opposite" of what we want. Strategy tip: Memorize the key duality: $(V_1 + V_2)^\perp = V_1^\perp \cap V_2^\perp$ and $(V_1 \cap V_2)^\perp = V_1^\perp + V_2^\perp$. These relationships flip sums to intersections and vice versa when taking orthogonal complements.
13. In the inner product space $C[0, 1]$ of continuous functions on the interval $[0, 1]$ with the inner product $\langle f, g \rangle = \int_0^1 f(x)g(x) \,dx$, let $f(x) = 1$ and $g(x) = ax - 1$. For what value of $a$ are $f(x)$ and $g(x)$ orthogonal?
    Answer: $a = 2$ For $f(x)$ and $g(x)$ to be orthogonal, their inner product must be zero. We set up the integral: $\langle f, g \rangle = \int_0^1 (1)(ax - 1) \,dx = 0$ We evaluate the integral: $\int_0^1 (ax - 1) \,dx = \left[ \frac{ax^2}{2} - x \right]_0^1 = \left( \frac{a(1)^2}{2} - 1 \right) - \left( \frac{a(0)^2}{2} - 0 \right) = \frac{a}{2} - 1$ Setting the result to zero gives $\frac{a}{2} - 1 = 0$, which solves to $\frac{a}{2} = 1$, so $a = 2$.
14. Let $W$ be a subspace of $\mathbb{R}^n$. Which of the following statements is NOT always true?
    Answer: $W \cup W^\perp = \mathbb{R}^n$. The union of two subspaces is not generally a subspace, and it does not typically equal the entire vector space. For example, in $\mathbb{R}^3$, if $W$ is a line through the origin (a 1D subspace), then $W^\perp$ is a plane through the origin (a 2D subspace). The union $W \cup W^\perp$ consists of only the points on that line and on that plane, which does not include all points in $\mathbb{R}^3$. The other three statements are fundamental properties of orthogonal complements in $\mathbb{R}^n$.
15. Let $W$ be the subspace of $\mathbb{R}^4$ spanned by the vectors $\mathbf{v}_1 = (1, 2, -1, 3)$ and $\mathbf{v}_2 = (2, 1, 0, -1)$. If $\mathbf{u} = (a, b, c, d) \in W^{\perp}$, which of the following systems of equations must $\mathbf{u}$ satisfy?
    Answer: $a + 2b - c + 3d = 0$ and $2a + b - d = 0$ For $\mathbf{u}$ to be in $W^{\perp}$, it must be orthogonal to every vector in $W$. Since $W = \text{span}\{\mathbf{v}_1, \mathbf{v}_2\}$, we need $\mathbf{u} \cdot \mathbf{v}_1 = 0$ and $\mathbf{u} \cdot \mathbf{v}_2 = 0$. Computing: $\mathbf{u} \cdot \mathbf{v}_1 = a + 2b - c + 3d = 0$ and $\mathbf{u} \cdot \mathbf{v}_2 = 2a + b + 0c + (-1)d = 2a + b - d = 0$. Choice A has the wrong sign for $d$ in the second equation. Choice C omits the $d$ term entirely. Choice D incorrectly includes $c$ in the second equation.
16. Consider the subspace $U = \{(x, y, z, w) \in \mathbb{R}^4 : x + y - z = 0 \text{ and } 2x - y + w = 0\}$ of $\mathbb{R}^4$ with the standard inner product. Which of the following vectors spans $U^{\perp}$?
    Answer: $\{(1, 1, -1, 0), (2, -1, 0, 1)\}$ The subspace $U$ is defined by the system of linear equations. The orthogonal complement $U^{\perp}$ is spanned by the normal vectors to these hyperplanes. From $x + y - z = 0$, we get normal vector $(1, 1, -1, 0)$. From $2x - y + w = 0$, we get normal vector $(2, -1, 0, 1)$. 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.
17. In $\mathbb{R}^n$ with the standard inner product, let $S = \{\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_k\}$ be a set of vectors. If $(\text{span}(S))^{\perp} = \{\mathbf{0}\}$, which statement must be true?
    Answer: The vectors in $S$ span $\mathbb{R}^n$ but may be linearly dependent If $(\text{span}(S))^{\perp} = \{\mathbf{0}\}$, then span$(S)$ has no non-zero orthogonal vectors, which means span$(S) = \mathbb{R}^n$ (since $\dim(\text{span}(S)) + \dim((\text{span}(S))^{\perp}) = n$ and $\dim(\{\mathbf{0}\}) = 0$). However, the vectors in $S$ 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$(S) \neq \mathbb{R}^n$, then $(\text{span}(S))^{\perp} \neq \{\mathbf{0}\}$.
18. Let $U$ and $V$ be subspaces of a finite-dimensional inner product space $W$. If $U \perp V$ (meaning every vector in $U$ is orthogonal to every vector in $V$), which relationship between $(U + V)^{\perp}$ and $U^{\perp} \cap V^{\perp}$ is correct?
    Answer: $(U + V)^{\perp} = U^{\perp} \cap V^{\perp}$ regardless of the orthogonality condition The identity $(U + V)^{\perp} = U^{\perp} \cap V^{\perp}$ is a fundamental property of orthogonal complements that holds for any subspaces $U$ and $V$, regardless of whether $U \perp V$. A vector is orthogonal to $U + V$ if and only if it's orthogonal to both $U$ and $V$. 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.
19. In $\mathbb{R}^3$, let $W$ be the plane defined by the equation $2x - y + 3z = 0$. Which of the following provides a geometric description of the orthogonal complement $W^\perp$?
    Answer: The line passing through the origin with direction vector $\begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}$. The equation of the plane $W$ can be written as a dot product: $\begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} x \\ y \\ z \end{pmatrix} = 0$. This shows that every vector in the plane $W$ is orthogonal to the normal vector $\mathbf{n} = (2, -1, 3)$. The orthogonal complement $W^\perp$ is the set of all vectors that are orthogonal to every vector in $W$. This set is precisely the line spanned by the normal vector $\mathbf{n}$. Therefore, $W^\perp$ is the line through the origin with direction vector $(2, -1, 3)$.
20. Let $W$ be the column space of the matrix $A = \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 2 & -1 \end{pmatrix}$. Which of the following vectors forms a basis for the orthogonal complement $W^\perp$?
    Answer: $\begin{pmatrix} -2 \\ 1 \\ 1 \end{pmatrix}$ The orthogonal complement of the column space of $A$, denoted $(\mathrm{Col } A)^\perp$, is the null space of $A^T$, i.e., $\mathrm{Nul}(A^T)$. First, we find the transpose of $A$: $A^T = \begin{pmatrix} 1 & 0 & 2 \\ 0 & 1 & -1 \end{pmatrix}$. To find the null space, we solve $A^T\mathbf{x} = \mathbf{0}$. This corresponds to the system of equations $x_1 + 2x_3 = 0$ and $x_2 - x_3 = 0$. From these equations, we get $x_1 = -2x_3$ and $x_2 = x_3$. A vector in the null space has the form $\begin{pmatrix} -2x_3 \\ x_3 \\ x_3 \end{pmatrix} = x_3 \begin{pmatrix} -2 \\ 1 \\ 1 \end{pmatrix}$. Thus, a basis for $W^\perp$ is $\left\{ \begin{pmatrix} -2 \\ 1 \\ 1 \end{pmatrix} \right\}$.

Start this quiz