Geometric Applications - Linear Algebra

1. The vector $\vec{y} = [7, -4, -1]^T$ can be written as the sum $\vec{y} = \vec{w} + \vec{z}$, where $\vec{w}$ is in the subspace $W$ spanned by $\vec{u} = [1, 2, 2]^T$ and $\vec{z}$ is in the orthogonal complement $W^\perp$. What is the vector $\vec{z}$?
   Answer: $\begin{pmatrix} 22/3 \\ -10/3 \\ -1/3 \end{pmatrix}$ First, find the projection of $\vec{y}$ onto the subspace $W$, which is the vector $\vec{w}$. $\vec{w} = \text{proj}_W \vec{y} = \frac{\vec{y} \cdot \vec{u}}{\|\vec{u}\|^2}\vec{u}$. $\vec{y} \cdot \vec{u} = (7)(1) + (-4)(2) + (-1)(2) = 7 - 8 - 2 = -3$. $\|\vec{u}\|^2 = 1^2 + 2^2 + 2^2 = 1+4+4=9$. $\vec{w} = \frac{-3}{9}\vec{u} = -\frac{1}{3}\begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix} = \begin{pmatrix} -1/3 \\ -2/3 \\ -2/3 \end{pmatrix}$. The vector $\vec{z}$ is the component of $\vec{y}$ orthogonal to $W$, which is calculated as $\vec{z} = \vec{y} - \vec{w}$. $\vec{z} = \begin{pmatrix} 7 \\ -4 \\ -1 \end{pmatrix} - \begin{pmatrix} -1/3 \\ -2/3 \\ -2/3 \end{pmatrix} = \begin{pmatrix} 21/3 + 1/3 \\ -12/3 + 2/3 \\ -3/3 + 2/3 \end{pmatrix} = \begin{pmatrix} 22/3 \\ -10/3 \\ -1/3 \end{pmatrix}$. (A) is the vector $\vec{w}$ in the subspace $W$, not the vector $\vec{z}$ in $W^\perp$. (C) results from forgetting to divide by the squared norm in the projection formula. (D) is the vector $\vec{w} - \vec{y}$, a sign error.
2. Which matrix $P$ performs an orthogonal projection of vectors in $\mathbb{R}^3$ onto the line spanned by the vector $\vec{a} = [1, -2, 3]^T$?
   Answer: $\frac{1}{14} \begin{pmatrix} 1 & -2 & 3 \\ -2 & 4 & -6 \\ 3 & -6 & 9 \end{pmatrix}$ The projection matrix $P$ that projects onto the column space of a matrix $A$ is $P=A(A^TA)^{-1}A^T$. For a line spanned by a single vector $\vec{a}$, we let $A=\vec{a}$. The formula simplifies to $P = \frac{1}{\vec{a}^T\vec{a}} \vec{a}\vec{a}^T = \frac{1}{\|\vec{a}\|^2} \vec{a}\vec{a}^T$. First, calculate $\|\vec{a}\|^2 = 1^2 + (-2)^2 + 3^2 = 1+4+9 = 14$. Next, calculate the outer product $\vec{a}\vec{a}^T$: $\vec{a}\vec{a}^T = \begin{pmatrix} 1 \\ -2 \\ 3 \end{pmatrix} \begin{pmatrix} 1 & -2 & 3 \end{pmatrix} = \begin{pmatrix} 1 & -2 & 3 \\ -2 & 4 & -6 \\ 3 & -6 & 9 \end{pmatrix}$. Finally, $P = \frac{1}{14} \begin{pmatrix} 1 & -2 & 3 \\ -2 & 4 & -6 \\ 3 & -6 & 9 \end{pmatrix}$. (A) is incorrect because it omits the normalization factor $1/\|\vec{a}\|^2$. (C) incorrectly divides by $\|\vec{a}\|$ instead of $\|\vec{a}\|^2$. (D) is the matrix $I-P$, which projects onto the orthogonal complement of the line.
3. Let $W$ be the subspace of $\mathbb{R}^3$ spanned by the vectors $\vec{u}_1 = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}$ and $\vec{u}_2 = \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}$. Which vector in $W$ is closest to the vector $\vec{y} = \begin{pmatrix} 2 \\ 0 \\ -1 \end{pmatrix}$?
   Answer: $\begin{pmatrix} 5/3 \\ 1/3 \\ -4/3 \end{pmatrix}$ The vector in $W$ closest to $\vec{y}$ is the orthogonal projection of $\vec{y}$ onto $W$. Since the basis vectors $\vec{u}_1$ and $\vec{u}_2$ are not orthogonal ($\vec{u}_1 \cdot \vec{u}_2 = 1 \neq 0$), we must solve the normal equations $A^TA\hat{\vec{x}} = A^T\vec{y}$ where $A = \begin{pmatrix} \vec{u}_1 & \vec{u}_2 \end{pmatrix}$. $A = \begin{pmatrix} 1 & 0 \\ 1 & 1 \\ 0 & 1 \end{pmatrix}$. $A^TA = \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 1 & 1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$. $A^T\vec{y} = \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \end{pmatrix} \begin{pmatrix} 2 \\ 0 \\ -1 \end{pmatrix} = \begin{pmatrix} 2 \\ -1 \end{pmatrix}$. We solve $\begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix} \hat{\vec{x}} = \begin{pmatrix} 2 \\ -1 \end{pmatrix}$, which yields $\hat{\vec{x}} = \begin{pmatrix} 5/3 \\ -4/3 \end{pmatrix}$. The projection is $A\hat{\vec{x}} = \frac{5}{3}\vec{u}_1 - \frac{4}{3}\vec{u}_2 = \frac{5}{3}\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} - \frac{4}{3}\begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 5/3 \\ 1/3 \\ -4/3 \end{pmatrix}$. (A) is incorrect because it is the sum of projections onto $\vec{u}_1$ and $\vec{u}_2$ as if they were an orthogonal basis. (C) is the error vector $\vec{y} - \text{proj}_W \vec{y}$. (D) is the least-squares solution vector $\hat{\vec{x}}$, not the projection itself.
4. To find the best-fit parabola of the form $y = c_0 + c_1 x + c_2 x^2$ for the data points $(-1, 0)$, $(0, 1)$, $(1, 3)$, and $(2, 5)$, one sets up a least-squares problem $A\vec{c} = \vec{y}$. What is the design matrix $A$?
   Answer: $\begin{pmatrix} 1 & -1 & 1 \\ 1 & 0 & 0 \\ 1 & 1 & 1 \\ 1 & 2 & 4 \end{pmatrix}$ The model is $y = c_0 + c_1 x + c_2 x^2$. We substitute each data point $(x, y)$ to form a row of the system $A\vec{c}=\vec{y}$, where $\vec{c}=[c_0, c_1, c_2]^T$. The columns of $A$ correspond to the coefficients of $c_0$, $c_1$, and $c_2$, which are $1$, $x$, and $x^2$ respectively. For $(-1, 0)$: $c_0(1) + c_1(-1) + c_2(-1)^2 = c_0 - c_1 + c_2$. Row is $[1, -1, 1]$. For $(0, 1)$: $c_0(1) + c_1(0) + c_2(0)^2 = c_0$. Row is $[1, 0, 0]$. For $(1, 3)$: $c_0(1) + c_1(1) + c_2(1)^2 = c_0 + c_1 + c_2$. Row is $[1, 1, 1]$. For $(2, 5)$: $c_0(1) + c_1(2) + c_2(2)^2 = c_0 + 2c_1 + 4c_2$. Row is $[1, 2, 4]$. Assembling these rows gives the matrix $A = \begin{pmatrix} 1 & -1 & 1 \\ 1 & 0 & 0 \\ 1 & 1 & 1 \\ 1 & 2 & 4 \end{pmatrix}$. (B) incorrectly calculates the third column ($x^2$) by not squaring the x-values. (C) incorrectly constructs the matrix using data vectors as columns. (D) omits the first column of ones, which corresponds to the constant term $c_0$.
5. To find the best-fit line of the form $y = c_0 + c_1x$ for the data points $(0, 1)$, $(1, 3)$, and $(2, 4)$, one must solve the normal equations $A^TA\hat{\vec{x}} = A^T\vec{b}$ for $\hat{\vec{x}} = [c_0, c_1]^T$. Which of the following is the correct system?
   Answer: $\begin{pmatrix} 3 & 3 \\ 3 & 5 \end{pmatrix} \begin{pmatrix} c_0 \\ c_1 \end{pmatrix} = \begin{pmatrix} 8 \\ 11 \end{pmatrix}$ The linear system $A\vec{x} = \vec{b}$ is constructed from the data points: $c_0 + c_1(0) = 1$ $c_0 + c_1(1) = 3$ $c_0 + c_1(2) = 4$ This gives $A = \begin{pmatrix} 1 & 0 \\ 1 & 1 \\ 1 & 2 \end{pmatrix}$ and $\vec{b} = \begin{pmatrix} 1 \\ 3 \\ 4 \end{pmatrix}$. Next, we compute $A^TA$ and $A^T\vec{b}$. $A^TA = \begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 2 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 1 & 1 \\ 1 & 2 \end{pmatrix} = \begin{pmatrix} 3 & 3 \\ 3 & 5 \end{pmatrix}$. $A^T\vec{b} = \begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 2 \end{pmatrix} \begin{pmatrix} 1 \\ 3 \\ 4 \end{pmatrix} = \begin{pmatrix} 1+3+4 \\ 0+3+8 \end{pmatrix} = \begin{pmatrix} 8 \\ 11 \end{pmatrix}$. The resulting normal equations are $\begin{pmatrix} 3 & 3 \\ 3 & 5 \end{pmatrix} \begin{pmatrix} c_0 \\ c_1 \end{pmatrix} = \begin{pmatrix} 8 \\ 11 \end{pmatrix}$. (B) results from defining the vector of unknowns as $[c_1, c_0]^T$ and swapping the columns of $A$. (C) uses an incorrect calculation for $A^T\vec{b}$. (D) incorrectly sets up the equation as $A^T\vec{x}=\vec{b}$, which is dimensionally incorrect.
6. In a 3D graphics application, a camera positioned at the origin looks along the negative $z$-axis. A 3D point $Q = (3, 4, -8)$ is projected onto the plane $z = -5$ (the image plane). Using perspective projection with the camera at the origin, what are the coordinates of the projected point on the image plane?
   Answer: The projected point has coordinates $\left(\frac{15}{8}, \frac{20}{8}, -5\right)$ because perspective projection scales coordinates proportionally to depth In perspective projection from the origin onto the plane $z = -5$, a point $(x, y, z)$ is projected to $\left(\frac{-5x}{z}, \frac{-5y}{z}, -5\right)$. This is because the line from the origin through $(x, y, z)$ has parametric form $(tx, ty, tz)$, and we need $tz = -5$, giving $t = \frac{-5}{z}$. For point $Q = (3, 4, -8)$, we have $t = \frac{-5}{-8} = \frac{5}{8}$. The projected point is $\left(3 \cdot \frac{5}{8}, 4 \cdot \frac{5}{8}, -5\right) = \left(\frac{15}{8}, \frac{20}{8}, -5\right) = \left(\frac{15}{8}, \frac{5}{2}, -5\right)$. Note that $\frac{20}{8} = \frac{5}{2}$. Choice B incorrectly uses the image plane distance in the denominator instead of the object's $z$-coordinate. Choice C makes an error in the perspective division formula. Choice D uses an incorrect scaling relationship.
7. Let $W$ be the subspace of $\mathbb{R}^4$ spanned by $\vec{u}_1 = [1, 1, 0, 0]^T$ and $\vec{u}_2 = [0, 1, 1, 0]^T$. Let $\vec{y} = [2, 0, 3, 4]^T$. If $\hat{\vec{y}}$ is the orthogonal projection of $\vec{y}$ onto $W$, which statement is true about the error vector $\vec{z} = \vec{y} - \hat{\vec{y}}$?
   Answer: $\vec{z} \cdot \vec{u}_1 = 0$ and $\vec{z} \cdot \vec{u}_2 = 0$. By the definition of an orthogonal projection, the error vector $\vec{z} = \vec{y} - \hat{\vec{y}}$ is orthogonal to the subspace $W$. This means $\vec{z}$ must be orthogonal to every vector in $W$, including the basis vectors that span $W$. Therefore, $\vec{z} \cdot \vec{u}_1 = 0$ and $\vec{z} \cdot \vec{u}_2 = 0$. (A) is not necessarily true. The orthogonal complement $W^\perp$ is two-dimensional, so $\vec{z}$ could be any linear combination of a basis for $W^\perp$, not just a multiple of a single vector. (B) is false. Any vector $\hat{\vec{y}}$ in $W$ is a linear combination of $\vec{u}_1$ and $\vec{u}_2$ and must have a fourth component of 0. Thus, the fourth component of $\vec{z}$ is $y_4 - \hat{y}_4 = 4 - 0 = 4$. (C) is false because $\vec{y}$ is not in $W$ (its fourth component is 4, while all vectors in $W$ have a fourth component of 0), so its projection is not equal to itself.
8. If $\hat{\vec{x}}$ is the least-squares solution to an inconsistent system $A\vec{x}=\vec{b}$, which of the following vectors must lie in the null space of $A^T$, which is also the orthogonal complement of the column space of $A$?
   Answer: The residual (or error) vector, $\vec{b} - A\hat{\vec{x}}$. The least-squares solution $\hat{\vec{x}}$ is defined by the property that the vector $A\hat{\vec{x}}$ is the orthogonal projection of $\vec{b}$ onto the column space of $A$, denoted $\text{Col}(A)$. The definition of an orthogonal projection implies that the difference between the original vector and its projection, which is the residual vector $\vec{b} - A\hat{\vec{x}}$, must be orthogonal to the subspace being projected onto. Therefore, $\vec{b} - A\hat{\vec{x}}$ is in $(\text{Col}(A))^\perp$. The fundamental subspaces theorem states that $(\text{Col}(A))^\perp = \text{Nul}(A^T)$. (A) $\hat{\vec{x}}$ is in the domain of the transformation, not necessarily in $\text{Nul}(A^T)$. (B) $A\hat{\vec{x}}$ is in $\text{Col}(A)$, not its orthogonal complement. (D) $\vec{b}$ is generally not in $(\text{Col}(A))^\perp$; if it were, its projection would be the zero vector.
9. Let $W$ be a plane through the origin in $\mathbb{R}^3$, and let $\vec{b}$ be a vector representing a point not in $W$. If $\vec{p}$ is the orthogonal projection of $\vec{b}$ onto $W$, what is the geometric meaning of the quantity $\|\vec{b} - \vec{p}\|$?
   Answer: The shortest distance from the point represented by the vector $\vec{b}$ to the plane $W$. The orthogonal decomposition theorem states that $\vec{b}$ can be uniquely written as $\vec{b} = \vec{p} + \vec{z}$, where $\vec{p}$ is in $W$ (the projection) and $\vec{z}$ is in $W^\perp$ (the orthogonal component). Here, $\vec{z} = \vec{b} - \vec{p}$. The vector $\vec{z}$ is a normal vector from the plane $W$ to the point represented by $\vec{b}$. Its length, $\|\vec{z}\| = \|\vec{b} - \vec{p}\|$, represents the shortest possible distance from the point to the plane. (A) describes $\|\vec{p}\|$, not $\|\vec{b} - \vec{p}\|$. (B) is incorrect; the area would be $\frac{1}{2}\|\vec{p} \times (\vec{b}-\vec{p})\|$. (D) is incorrect; the sine of the angle between $\vec{b}$ and the plane is given by $\|\vec{b} - \vec{p}\| / \|\vec{b}\|$.
10. What is the shortest distance from the point $(3, 1, 5)$ to the plane in $\mathbb{R}^3$ spanned by the orthogonal vectors $\vec{v}_1 = [1, -2, 1]^T$ and $\vec{v}_2 = [1, 1, 1]^T$?
    Answer: $\sqrt{2}$ The distance is the magnitude of the component of $\vec{y}=[3, 1, 5]^T$ orthogonal to the subspace $W$ spanned by $\vec{v}_1$ and $\vec{v}_2$. This component is $\vec{z} = \vec{y} - \text{proj}_W \vec{y}$. Since the basis is orthogonal, the projection $\vec{p}$ is: $\vec{p} = \frac{\vec{y} \cdot \vec{v}_1}{\|\vec{v}_1\|^2}\vec{v}_1 + \frac{\vec{y} \cdot \vec{v}_2}{\|\vec{v}_2\|^2}\vec{v}_2$ $\vec{y} \cdot \vec{v}_1 = 3 - 2 + 5 = 6$; $\|\vec{v}_1\|^2 = 1+4+1=6$. $\vec{y} \cdot \vec{v}_2 = 3 + 1 + 5 = 9$; $\|\vec{v}_2\|^2 = 1+1+1=3$. $\vec{p} = \frac{6}{6}\vec{v}_1 + \frac{9}{3}\vec{v}_2 = 1\begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix} + 3\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 4 \\ 1 \\ 4 \end{pmatrix}$. The orthogonal component is $\vec{z} = \vec{y} - \vec{p} = \begin{pmatrix} 3 \\ 1 \\ 5 \end{pmatrix} - \begin{pmatrix} 4 \\ 1 \\ 4 \end{pmatrix} = \begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix}$. The distance is $\|\vec{z}\| = \sqrt{(-1)^2 + 0^2 + 1^2} = \sqrt{2}$. (A) is the magnitude of the projection vector $\vec{p}$. (B) is the magnitude of the original vector $\vec{y}$. (C) results from ignoring one of the basis vectors during the projection calculation.
11. Consider the linear system $A\vec{x} = \vec{b}$ where $A = \begin{pmatrix} 1 & 2 \\ -1 & 4 \\ 1 & 2 \end{pmatrix}$ and $\vec{b} = \begin{pmatrix} 3 \\ 2 \\ 5 \end{pmatrix}$. What is the least-squares solution $\hat{\vec{x}}$?
    Answer: $\begin{pmatrix} 2 \\ 1 \end{pmatrix}$ The least-squares solution $\hat{\vec{x}}$ is found by solving the normal equations $A^TA\hat{\vec{x}} = A^T\vec{b}$. First, compute $A^TA$: $A^TA = \begin{pmatrix} 1 & -1 & 1 \\ 2 & 4 & 2 \end{pmatrix} \begin{pmatrix} 1 & 2 \\ -1 & 4 \\ 1 & 2 \end{pmatrix} = \begin{pmatrix} 3 & 0 \\ 0 & 24 \end{pmatrix}$. Next, compute $A^T\vec{b}$: $A^T\vec{b} = \begin{pmatrix} 1 & -1 & 1 \\ 2 & 4 & 2 \end{pmatrix} \begin{pmatrix} 3 \\ 2 \\ 5 \end{pmatrix} = \begin{pmatrix} 3-2+5 \\ 6+8+10 \end{pmatrix} = \begin{pmatrix} 6 \\ 24 \end{pmatrix}$. Now, solve the system $\begin{pmatrix} 3 & 0 \\ 0 & 24 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} 6 \\ 24 \end{pmatrix}$. This gives $3x_1 = 6 \implies x_1 = 2$, and $24x_2 = 24 \implies x_2 = 1$. So, $\hat{\vec{x}} = \begin{pmatrix} 2 \\ 1 \end{pmatrix}$. (A) is the projection vector $A\hat{\vec{x}}$, not the solution vector $\hat{\vec{x}}$. (B) is the vector $A^T\vec{b}$. (D) results from a sign error when computing $A^T\vec{b}$.
12. A structural engineer analyzes stress distribution in a beam by decomposing the stress vector $\mathbf{s} = \begin{pmatrix} 8 \\ -3 \\ 6 \end{pmatrix}$ into components along two orthogonal directions: $\mathbf{u}_1 = \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix}$ and $\mathbf{u}_2 = \begin{pmatrix} -1 \\ 2 \\ 0 \end{pmatrix}$. After projecting $\mathbf{s}$ onto the subspace spanned by $\{\mathbf{u}_1, \mathbf{u}_2\}$, what is the magnitude of the component of $\mathbf{s}$ that lies outside this subspace?
    Answer: The magnitude of the external component is $\sqrt{36}$ because this equals the norm of the rejection vector from the spanned subspace Since $\mathbf{u}_1 = (2, 1, 0)$ and $\mathbf{u}_2 = (-1, 2, 0)$ both lie in the $xy$-plane ($z = 0$), the subspace they span is exactly the $xy$-plane. The component of $\mathbf{s} = (8, -3, 6)$ that lies outside this subspace is simply the $z$-component, which is the vector $(0, 0, 6)$. The magnitude of this external component is $\sqrt{0^2 + 0^2 + 6^2} = \sqrt{36} = 6$. Alternatively, we can verify this by computing the projection of $\mathbf{s}$ onto the subspace. First, check if $\mathbf{u}_1$ and $\mathbf{u}_2$ are orthogonal: $\mathbf{u}_1 \cdot \mathbf{u}_2 = (2)(-1) + (1)(2) + (0)(0) = -2 + 2 + 0 = 0$. Yes, they are orthogonal. The projection of $\mathbf{s}$ onto the subspace is $\text{proj}_{\text{span}}\mathbf{s} = \frac{\mathbf{s} \cdot \mathbf{u}_1}{\mathbf{u}_1 \cdot \mathbf{u}_1}\mathbf{u}_1 + \frac{\mathbf{s} \cdot \mathbf{u}_2}{\mathbf{u}_2 \cdot \mathbf{u}_2}\mathbf{u}_2$. Computing: $\mathbf{s} \cdot \mathbf{u}_1 = (8)(2) + (-3)(1) + (6)(0) = 16 - 3 + 0 = 13$, $\mathbf{u}_1 \cdot \mathbf{u}_1 = 4 + 1 + 0 = 5$, $\mathbf{s} \cdot \mathbf{u}_2 = (8)(-1) + (-3)(2) + (6)(0) = -8 - 6 + 0 = -14$, $\mathbf{u}_2 \cdot \mathbf{u}_2 = 1 + 4 + 0 = 5$. So $\text{proj}_{\text{span}}\mathbf{s} = \frac{13}{5}(2, 1, 0) + \frac{-14}{5}(-1, 2, 0) = \left(\frac{26}{5}, \frac{13}{5}, 0\right) + \left(\frac{14}{5}, \frac{-28}{5}, 0\right) = \left(\frac{40}{5}, \frac{-15}{5}, 0\right) = (8, -3, 0)$. The component outside the subspace is $\mathbf{s} - \text{proj}_{\text{span}}\mathbf{s} = (8, -3, 6) - (8, -3, 0) = (0, 0, 6)$, which has magnitude $\sqrt{36} = 6$. Choices B, C, D represent computational errors in the projection process.
13. Let $\vec{y}$ be a vector in $\mathbb{R}^n$ and let $W$ be a subspace of $\mathbb{R}^n$. Let $\hat{\vec{y}}$ be the orthogonal projection of $\vec{y}$ onto $W$. Which of the following statements is NOT always true?
    Answer: The projection of $\hat{\vec{y}}$ onto $W$ is the zero vector. Let's analyze each statement. (A) True. This is the definition of the orthogonal projection; the error vector $\vec{y} - \hat{\vec{y}}$ is orthogonal to the subspace $W$. (B) True. This is the Pythagorean theorem applied to the orthogonal decomposition $\vec{y} = \hat{\vec{y}} + (\vec{y} - \hat{\vec{y}})$. (C) False. Since $\hat{\vec{y}}$ is already in $W$, its projection onto $W$ is itself, i.e., $\text{proj}_W(\hat{\vec{y}}) = \hat{\vec{y}}$. This is the zero vector only in the trivial case where $\hat{\vec{y}} = \vec{0}$. (D) True. This is the Best Approximation Theorem, which states that the orthogonal projection $\hat{\vec{y}}$ is the unique vector in $W$ closest to $\vec{y}$.

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