Projections & Decompositions - Linear Algebra

1. Let $V$ be a finite-dimensional inner product space and $W$ a subspace of $V$. If $v \in V$ satisfies $\text{proj}_W(v) = \frac{2}{3}v$, which of the following must be true?
   Answer: $v \notin W$ and $\|\text{proj}_{W^\perp}(v)\| = \frac{1}{3}\|v\|$ Since $\text{proj}_W(v) = \frac{2}{3}v \neq v$ (unless $v = 0$), we have $v \notin W$. By the orthogonal decomposition, $v = \text{proj}_W(v) + \text{proj}_{W^\perp}(v) = \frac{2}{3}v + \text{proj}_{W^\perp}(v)$, so $\text{proj}_{W^\perp}(v) = v - \frac{2}{3}v = \frac{1}{3}v$. Therefore, $\|\text{proj}_{W^\perp}(v)\| = \frac{1}{3}\|v\|$. Choice A is wrong because $v \notin W$. Choice C has the wrong coefficient. Choice D is wrong because if $v \in W$, then $\text{proj}_W(v) = v$, not $\frac{2}{3}v$.
2. Let $V$ be an inner product space and $W_1, W_2$ be subspaces of $V$ such that $V = W_1 \oplus W_2$ (orthogonal direct sum). For any $v \in V$, if $\|\text{proj}_{W_1}(v)\| = \frac{3}{5}\|v\|$ and $\|v\| = 10$, what is $\|\text{proj}_{W_2}(v)\|$?
   Answer: $8$ Since $V = W_1 \oplus W_2$ (orthogonal direct sum), we have $v = \text{proj}_{W_1}(v) + \text{proj}_{W_2}(v)$ where the projections onto $W_1$ and $W_2$ are orthogonal to each other. Given that $\|\text{proj}_{W_1}(v)\| = \frac{3}{5}\|v\| = \frac{3}{5} \cdot 10 = 6$. By the Pythagorean theorem for orthogonal components: $\|v\|^2 = \|\text{proj}_{W_1}(v)\|^2 + \|\text{proj}_{W_2}(v)\|^2$. Substituting: $10^2 = 6^2 + \|\text{proj}_{W_2}(v)\|^2$, so $100 = 36 + \|\text{proj}_{W_2}(v)\|^2$. Therefore $\|\text{proj}_{W_2}(v)\|^2 = 64$, which gives $\|\text{proj}_{W_2}(v)\| = 8$.
3. Let $W$ be the plane in $\mathbb{R}^3$ spanned by the vectors $\vec{u}_1 = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}$ and $\vec{u}_2 = \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}$. Let $\vec{y} = \begin{pmatrix} 0 \\ 0 \\ 6 \end{pmatrix}$. Find the orthogonal projection of $\vec{y}$ onto $W$.
   Answer: $\begin{pmatrix} -2 \\ 2 \\ 4 \end{pmatrix}$ The basis vectors $\vec{u}_1$ and $\vec{u}_2$ are not orthogonal, since $\vec{u}_1 \cdot \vec{u}_2 = 1+2+0 = 3 \neq 0$. Therefore, we cannot use the simple sum-of-projections formula directly. A correct method is to find the orthogonal complement $W^\perp$, project $\vec{y}$ onto it, and subtract this from $\vec{y}$. 1. Find a basis for $W^\perp$. The vector $\vec{n} = \vec{u}_1 \times \vec{u}_2$ is orthogonal to both $\vec{u}_1$ and $\vec{u}_2$ and thus spans $W^\perp$. $\vec{n} = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} \times \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix} = \begin{pmatrix} (1)(1) - (0)(2) \\ (0)(1) - (1)(1) \\ (1)(2) - (1)(1) \end{pmatrix} = \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}$. 2. Project $\vec{y}$ onto $W^\perp$ (which is the line spanned by $\vec{n}$): $\text{proj}_{W^\perp}(\vec{y}) = \frac{\vec{y} \cdot \vec{n}}{\vec{n} \cdot \vec{n}}\vec{n} = \frac{(0)(1) + (0)(-1) + (6)(1)}{(1)^2 + (-1)^2 + (1)^2}\vec{n} = \frac{6}{3}\vec{n} = 2\begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix} = \begin{pmatrix} 2 \\ -2 \\ 2 \end{pmatrix}$. 3. The projection onto $W$ is $\text{proj}_W(\vec{y}) = \vec{y} - \text{proj}_{W^\perp}(\vec{y})$. $\text{proj}_W(\vec{y}) = \begin{pmatrix} 0 \\ 0 \\ 6 \end{pmatrix} - \begin{pmatrix} 2 \\ -2 \\ 2 \end{pmatrix} = \begin{pmatrix} -2 \\ 2 \\ 4 \end{pmatrix}$. (A) is incorrect; this is the result of incorrectly applying the projection formula for an orthogonal basis. (B) is incorrect; this is the projection onto the orthogonal complement $W^\perp$, not onto $W$. (D) is incorrect; this results from a sign error, calculating $\text{proj}_{W^\perp}(\vec{y}) - \vec{y}$.
4. Let $\vec{v} = \begin{pmatrix} 7 \\ 1 \end{pmatrix}$ and $\vec{u} = \begin{pmatrix} 4 \\ -2 \end{pmatrix}$. The vector $\vec{v}$ can be written as the sum $\vec{v} = \vec{p} + \vec{z}$, where $\vec{p}$ is parallel to $\vec{u}$ and $\vec{z}$ is orthogonal to $\vec{u}$. What is the vector $\vec{z}$?
   Answer: $\begin{pmatrix} 1.8 \\ 3.6 \end{pmatrix}$ The vector $\vec{p}$ is the orthogonal projection of $\vec{v}$ onto $\vec{u}$. The formula is $\vec{p} = \text{proj}_{\vec{u}}(\vec{v}) = \frac{\vec{v} \cdot \vec{u}}{\vec{u} \cdot \vec{u}}\vec{u}$. First, compute the dot products: $\vec{v} \cdot \vec{u} = (7)(4) + (1)(-2) = 28 - 2 = 26$. $\vec{u} \cdot \vec{u} = (4)(4) + (-2)(-2) = 16 + 4 = 20$. Now, find the projection $\vec{p}$: $\vec{p} = \frac{26}{20} \vec{u} = \frac{13}{10} \begin{pmatrix} 4 \\ -2 \end{pmatrix} = \begin{pmatrix} 52/10 \\ -26/10 \end{pmatrix} = \begin{pmatrix} 5.2 \\ -2.6 \end{pmatrix}$. The vector $\vec{z}$ is the component of $\vec{v}$ orthogonal to $\vec{u}$, which is found by $\vec{z} = \vec{v} - \vec{p}$. $\vec{z} = \begin{pmatrix} 7 \\ 1 \end{pmatrix} - \begin{pmatrix} 5.2 \\ -2.6 \end{pmatrix} = \begin{pmatrix} 7 - 5.2 \\ 1 - (-2.6) \end{pmatrix} = \begin{pmatrix} 1.8 \\ 3.6 \end{pmatrix}$. (A) is incorrect because it is the projection vector $\vec{p}$, not the orthogonal component $\vec{z}$. (C) is incorrect because it is the negative of the correct answer, which would result from calculating $\vec{p} - \vec{v}$. (D) is incorrect because it results from a sign error in the dot product calculation: $\vec{v} \cdot \vec{u} = 28 + 2 = 30$, leading to $\vec{p} = \frac{30}{20}\vec{u} = \begin{pmatrix} 6 \\ -3 \end{pmatrix}$ and $\vec{z} = \begin{pmatrix} 7 \\ 1 \end{pmatrix} - \begin{pmatrix} 6 \\ -3 \end{pmatrix} = \begin{pmatrix} 1 \\ 4 \end{pmatrix}$.
5. In $\mathbb{R}^4$ with the standard inner product, let $U = \text{span}\{(1,1,0,1), (0,1,1,0), (1,0,1,1)\}$. If $v = (2,3,1,4)$, which vector represents $v - \text{proj}_U(v)$?
   Answer: $(\frac{1}{3}, \frac{2}{3}, -\frac{1}{3}, \frac{1}{3})$ We need to find an orthogonal basis for U using Gram-Schmidt. Let $u_1 = (1,1,0,1)$, $u_2 = (0,1,1,0)$, $u_3 = (1,0,1,1)$. First, $w_1 = u_1 = (1,1,0,1)$ with $\|w_1\|^2 = 3$. Next, $w_2 = u_2 - \frac{\langle u_2, w_1 \rangle}{\|w_1\|^2} w_1 = (0,1,1,0) - \frac{1}{3}(1,1,0,1) = (-\frac{1}{3}, \frac{2}{3}, 1, -\frac{1}{3})$ with $\|w_2\|^2 = \frac{5}{3}$. For $w_3$, we get $w_3 = (1,0,1,1) - \frac{2}{3}(1,1,0,1) - \frac{\frac{4}{3}}{\frac{5}{3}}(-\frac{1}{3}, \frac{2}{3}, 1, -\frac{1}{3}) = (\frac{1}{5}, -\frac{3}{5}, \frac{1}{5}, \frac{1}{5})$. Computing $\text{proj}_U(v)$ and subtracting from $v$ gives $(\frac{1}{3}, \frac{2}{3}, -\frac{1}{3}, \frac{1}{3})$.
6. Let $W$ be the subspace of $\mathbb{R}^3$ with orthogonal basis $\mathcal{B} = \left\{\vec{u}_1 = \begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix}, \vec{u}_2 = \begin{pmatrix} 1 \\ 1 \\ -2 \end{pmatrix}\right\}$. Let $\vec{y} = \begin{pmatrix} 3 \\ 1 \\ 4 \end{pmatrix}$. If $\hat{\vec{y}} = \text{proj}_W(\vec{y})$, what is the coordinate vector $[\hat{\vec{y}}]_\mathcal{B}$?
   Answer: $\begin{pmatrix} 1 \\ -2/3 \end{pmatrix}$ The coordinate vector $[\hat{\vec{y}}]_\mathcal{B}$ consists of the scalar coefficients used to build $\hat{\vec{y}}$ from the orthogonal basis vectors of $W$. These scalars are found using the projection formula. Let $[\hat{\vec{y}}]_\mathcal{B} = \begin{pmatrix} c_1 \\ c_2 \end{pmatrix}$. The coefficients are: $c_1 = \frac{\vec{y} \cdot \vec{u}_1}{\vec{u}_1 \cdot \vec{u}_1}$ and $c_2 = \frac{\vec{y} \cdot \vec{u}_2}{\vec{u}_2 \cdot \vec{u}_2}$. Calculate $c_1$: $\vec{y} \cdot \vec{u}_1 = (3)(1) + (1)(-1) + (4)(0) = 3 - 1 = 2$. $\vec{u}_1 \cdot \vec{u}_1 = (1)^2 + (-1)^2 + (0)^2 = 1 + 1 = 2$. $c_1 = \frac{2}{2} = 1$. Calculate $c_2$: $\vec{y} \cdot \vec{u}_2 = (3)(1) + (1)(1) + (4)(-2) = 3 + 1 - 8 = -4$. $\vec{u}_2 \cdot \vec{u}_2 = (1)^2 + (1)^2 + (-2)^2 = 1 + 1 + 4 = 6$. $c_2 = \frac{-4}{6} = -\frac{2}{3}$. So, the coordinate vector is $[\hat{\vec{y}}]_\mathcal{B} = \begin{pmatrix} 1 \\ -2/3 \end{pmatrix}$. (A) is incorrect; these are the numerators of the coefficient fractions ($\{\vec{y} \cdot \vec{u}_i\}$) before division by the squared norms. (C) is incorrect; this vector is $\hat{\vec{y}}$ itself, not its coordinate vector with respect to $\mathcal{B}$. It is calculated as $1\vec{u}_1 - \frac{2}{3}\vec{u}_2 = \begin{pmatrix} 1/3 \\ -5/3 \\ 4/3 \end{pmatrix}$ (note: choice in question was slightly different from this calculation, but represents the same type of error). (D) is incorrect; this results from a sign error in the dot product for $c_2$, i.e., $3+1+8=12$, leading to $c_2=12/6=2$.
7. In $\mathbb{R}^3$, the orthogonal projection of the vector $\vec{y} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$ onto a line $W$ is $\text{proj}_W(\vec{y}) = \begin{pmatrix} 1.5 \\ 1.5 \\ 0 \end{pmatrix}$. What is the orthogonal projection of $\vec{y}$ onto the plane $W^\perp$?
   Answer: $\begin{pmatrix} -0.5 \\ 0.5 \\ 3 \end{pmatrix}$ The Orthogonal Decomposition Theorem states that any vector $\vec{y}$ can be written as the sum of its projection onto a subspace $W$ and its projection onto the orthogonal complement $W^\perp$. That is: $\vec{y} = \text{proj}_W(\vec{y}) + \text{proj}_{W^\perp}(\vec{y})$ We are given $\vec{y}$ and $\text{proj}_W(\vec{y})$, and we want to find $\text{proj}_{W^\perp}(\vec{y})$. We can rearrange the formula: $\text{proj}_{W^\perp}(\vec{y}) = \vec{y} - \text{proj}_W(\vec{y})$ Substituting the given values: $\text{proj}_{W^\perp}(\vec{y}) = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} - \begin{pmatrix} 1.5 \\ 1.5 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 - 1.5 \\ 2 - 1.5 \\ 3 - 0 \end{pmatrix} = \begin{pmatrix} -0.5 \\ 0.5 \\ 3 \end{pmatrix}$. (B) is incorrect because it is the result of the incorrect subtraction $\text{proj}_W(\vec{y}) - \vec{y}$. (C) is incorrect; it is the result of adding the two vectors instead of subtracting. (D) is incorrect; it is the projection onto $W$, not $W^\perp$.
8. Let $W$ be the subspace of $\mathbb{R}^3$ spanned by the orthogonal basis ${\vec{u}_1, \vec{u}_2}$, where $\vec{u}_1 = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}$ and $\vec{u}_2 = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}$. Let $\vec{y} = \begin{pmatrix} 3 \\ 4 \\ 5 \end{pmatrix}$. Find the orthogonal projection of $\vec{y}$ onto $W$.
   Answer: $\begin{pmatrix} 5 \\ 2 \\ 3 \end{pmatrix}$ Since ${\vec{u}_1, \vec{u}_2}$ is an orthogonal basis for $W$, the projection of $\vec{y}$ onto $W$ is the sum of the projections of $\vec{y}$ onto each basis vector: $\text{proj}_W(\vec{y}) = \frac{\vec{y} \cdot \vec{u}_1}{\vec{u}_1 \cdot \vec{u}_1}\vec{u}_1 + \frac{\vec{y} \cdot \vec{u}_2}{\vec{u}_2 \cdot \vec{u}_2}\vec{u}_2$. Calculate the dot products: $\vec{y} \cdot \vec{u}_1 = (3)(1) + (4)(1) + (5)(0) = 7$. $\vec{u}_1 \cdot \vec{u}_1 = (1)^2 + (1)^2 + (0)^2 = 2$. $\vec{y} \cdot \vec{u}_2 = (3)(1) + (4)(-1) + (5)(2) = 3 - 4 + 10 = 9$. $\vec{u}_2 \cdot \vec{u}_2 = (1)^2 + (-1)^2 + (2)^2 = 1 + 1 + 4 = 6$. Now substitute these values into the formula: $\text{proj}_W(\vec{y}) = \frac{7}{2}\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} + \frac{9}{6}\begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix} = \frac{7}{2}\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} + \frac{3}{2}\begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}$. Sum the vectors: $\text{proj}_W(\vec{y}) = \begin{pmatrix} 7/2 + 3/2 \\ 7/2 - 3/2 \\ 0 + 6/2 \end{pmatrix} = \begin{pmatrix} 10/2 \\ 4/2 \\ 6/2 \end{pmatrix} = \begin{pmatrix} 5 \\ 2 \\ 3 \end{pmatrix}$. (B) is incorrect because it is the orthogonal component $\vec{z} = \vec{y} - \text{proj}_W(\vec{y}) = \begin{pmatrix} 3 \\ 4 \\ 5 \end{pmatrix} - \begin{pmatrix} 5 \\ 2 \\ 3 \end{pmatrix} = \begin{pmatrix} -2 \\ 2 \\ 2 \end{pmatrix}$. (C) is incorrect because it results from forgetting to divide by the norms squared ($\|\vec{u}_1\|^2$ and $\|\vec{u}_2\|^2$). (D) is incorrect because it results from an arithmetic error, such as calculating the second scalar as $\frac{9}{2}$ instead of $\frac{9}{6}$.
9. Let $V$ be an inner product space with orthonormal basis $\{u_1, u_2, u_3, u_4\}$. If $v = 2u_1 - u_2 + 3u_3 + u_4$ and $W = \text{span}\{u_1 + u_2, u_2 + u_3\}$, what is $\|\text{proj}_W(v)\|^2$?
   Answer: $\frac{5}{2}$ First, we need an orthogonal basis for W. Let $w_1 = u_1 + u_2$ and apply Gram-Schmidt to $w_2 = u_2 + u_3$. We have $\langle w_2, w_1 \rangle = \langle u_2 + u_3, u_1 + u_2 \rangle = 1$, and $\|w_1\|^2 = 2$. So $w_2' = w_2 - \frac{\langle w_2, w_1 \rangle}{\|w_1\|^2} w_1 = (u_2 + u_3) - \frac{1}{2}(u_1 + u_2) = -\frac{1}{2}u_1 + \frac{1}{2}u_2 + u_3$. Now $\text{proj}_W(v) = \frac{\langle v, w_1 \rangle}{\|w_1\|^2} w_1 + \frac{\langle v, w_2' \rangle}{\|w_2'\|^2} w_2'$. We have $\langle v, w_1 \rangle = 1$, $\langle v, w_2' \rangle = -2$, and $\|w_2'\|^2 = \frac{3}{2}$. Thus $\text{proj}_W(v) = \frac{1}{2}(u_1 + u_2) - \frac{4}{3}(-\frac{1}{2}u_1 + \frac{1}{2}u_2 + u_3) = \frac{5}{6}u_1 - \frac{1}{6}u_2 - \frac{4}{3}u_3$. Therefore $\|\text{proj}_W(v)\|^2 = \frac{25}{36} + \frac{1}{36} + \frac{16}{9} = \frac{5}{2}$.
10. Let $P$ be the standard matrix for the orthogonal projection onto a proper subspace $W$ of $\mathbb{R}^n$ (where $W \neq \{\vec{0}\}$ and $W \neq \mathbb{R}^n$). Which of the following matrix equations must be true?
    Answer: $P^2 = P$ A projection matrix $P$ has the property that applying it a second time does not change the result. Geometrically, once a vector $\vec{x}$ is projected into the subspace $W$, its image $P\vec{x}$ is already in $W$. Projecting it again will leave it unchanged. Therefore, $P(P\vec{x}) = P\vec{x}$ for all $\vec{x}$ in $\mathbb{R}^n$. This implies that the matrix equation $P^2 = P$ must hold. This property is called idempotence. (A) is incorrect. Since $W$ is a proper subspace, $P$ maps some non-zero vectors to zero (any vector in $W^\perp$), so its null space is non-trivial. Thus, $P$ is singular and not invertible. (B) is incorrect. The property $P^2=I$ defines an involution, such as a reflection matrix. A projection is not generally a reflection. (D) is incorrect. This would imply $P=-I$, which is not a projection matrix.
11. Let $W$ be the subspace of $\mathbb{R}^3$ spanned by $\vec{u}_1 = \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}$ and $\vec{u}_2 = \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}$. For which vector $\vec{y}$ is the orthogonal projection onto $W$ equal to $\vec{0}$?
    Answer: $\begin{pmatrix} 3 \\ 0 \\ -3 \end{pmatrix}$ When you encounter orthogonal projections in linear algebra, remember that the projection of a vector onto a subspace is zero if and only if that vector is orthogonal (perpendicular) to the entire subspace. This means the vector must be orthogonal to every vector that spans the subspace. To find which vector has zero projection onto $W$, you need to check which option is orthogonal to both spanning vectors $\vec{u}_1 = \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}$ and $\vec{u}_2 = \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}$. Two vectors are orthogonal when their dot product equals zero. For option D, $\begin{pmatrix} 3 \\ 0 \\ -3 \end{pmatrix}$: - $\begin{pmatrix} 3 \\ 0 \\ -3 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix} = 3(1) + 0(2) + (-3)(1) = 0$ - $\begin{pmatrix} 3 \\ 0 \\ -3 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix} = 3(1) + 0(-1) + (-3)(1) = 0$ Since both dot products are zero, option D is orthogonal to the subspace $W$. Options A, B, and C all fail this test. For example, option A gives dot products of 1 and 1 with the spanning vectors, option B gives 6 and 0, and option C gives 4 and 0. Since these vectors aren't orthogonal to both spanning vectors, their projections onto $W$ are non-zero. Key strategy: When asked about zero projections, immediately think "orthogonal to the subspace" and test dot products with all spanning vectors.
12. Let $W$ be a subspace of $\mathbb{R}^4$. Suppose a vector $\vec{y}$ is decomposed as $\vec{y} = \vec{w} + \vec{z}$, where $\vec{w}$ is in $W$ and $\vec{z}$ is in $W^\perp$. If $\vec{w} = \begin{pmatrix} 1 \\ 0 \\ 2 \\ -1 \end{pmatrix}$ and $\vec{z} = \begin{pmatrix} 2 \\ 3 \\ -1 \\ -1 \end{pmatrix}$, what is $\text{proj}_W(\vec{y})$?
    Answer: $\begin{pmatrix} 1 \\ 0 \\ 2 \\ -1 \end{pmatrix}$ The Orthogonal Decomposition Theorem states that any vector $\vec{y}$ can be uniquely decomposed into the sum of a vector in a subspace $W$ and a vector in its orthogonal complement $W^\perp$. This decomposition is given by $\vec{y} = \text{proj}_W(\vec{y}) + \vec{z}_{\perp}$, where $\text{proj}_W(\vec{y})$ is in $W$ and $\vec{z}_{\perp}$ is in $W^\perp$. The problem provides such a decomposition: $\vec{y} = \vec{w} + \vec{z}$, with $\vec{w} \in W$ and $\vec{z} \in W^\perp$. By the uniqueness of this decomposition, we can directly identify the components: $\text{proj}_W(\vec{y}) = \vec{w}$ and the orthogonal component is $\vec{z}$. Therefore, $\text{proj}_W(\vec{y}) = \vec{w} = \begin{pmatrix} 1 \\ 0 \\ 2 \\ -1 \end{pmatrix}$. No calculation is needed. (B) is incorrect because it is the component in $W^\perp$, not the projection onto $W$. (C) is incorrect because it is the vector $\vec{y}$ itself, which is the sum of the two components. (D) is incorrect; it is the result of overthinking the problem and incorrectly calculating the projection of $\vec{y}$ onto the line spanned by the single vector $\vec{w}$.
13. Consider the vector space $\mathbb{R}^4$ with the standard inner product. Let $S = \{(1,0,1,0), (0,1,0,1), (1,1,0,0)\}$ and $T = \{(1,0,0,1), (0,1,1,0)\}$. If $v = (2,1,3,2)$, what is $\|\text{proj}_{\text{span}(S)}(v) - \text{proj}_{\text{span}(T)}(v)\|^2$?
    Answer: $\frac{7}{2}$ First, we find orthogonal bases for both subspaces using Gram-Schmidt. For span(S): Starting with $(1,0,1,0)$, then orthogonalizing $(0,1,0,1)$ and $(1,1,0,0)$, we get an orthogonal basis. For span(T): The vectors $(1,0,0,1)$ and $(0,1,1,0)$ are already orthogonal since their inner product is $0$. Computing the projections: $\text{proj}_{\text{span}(S)}(v) = (\frac{5}{2}, \frac{3}{2}, \frac{5}{2}, \frac{3}{2})$ and $\text{proj}_{\text{span}(T)}(v) = (2, \frac{3}{2}, \frac{1}{2}, 2)$. The difference is $(-\frac{1}{2}, 0, 2, -\frac{1}{2})$, so $\|\text{proj}_{\text{span}(S)}(v) - \text{proj}_{\text{span}(T)}(v)\|^2 = \frac{1}{4} + 0 + 4 + \frac{1}{4} = \frac{7}{2}$.
14. In the inner product space $C[0,1]$ with inner product $\langle f,g \rangle = \int_0^1 f(x)g(x) dx$, consider the subspace $W = \text{span}\{1, x\}$. What is the projection of $f(x) = x^2$ onto $W$?
    Answer: $-\frac{1}{6} + x$ We use the Gram-Schmidt process to orthogonalize the basis $\{1, x\}$. Let $u_1 = 1$ with $\|u_1\|^2 = 1$. Then $u_2 = x - \frac{\langle x, 1 \rangle}{\|1\|^2} \cdot 1 = x - \frac{1/2}{1} = x - \frac{1}{2}$. We have $\|u_2\|^2 = \int_0^1 (x - \frac{1}{2})^2 dx = \frac{1}{12}$. The projection is $\text{proj}_W(x^2) = \frac{\langle x^2, 1 \rangle}{\|1\|^2} \cdot 1 + \frac{\langle x^2, x - \frac{1}{2} \rangle}{\|x - \frac{1}{2}\|^2} \cdot (x - \frac{1}{2})$. Computing: $\langle x^2, 1 \rangle = \frac{1}{3}$ and $\langle x^2, x - \frac{1}{2} \rangle = \frac{1}{4} - \frac{1}{6} = \frac{1}{12}$. Thus $\text{proj}_W(x^2) = \frac{1}{3} + \frac{1/12}{1/12}(x - \frac{1}{2}) = \frac{1}{3} + x - \frac{1}{2} = x - \frac{1}{6}$.
15. Let $W$ be the column space of the matrix $A = \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 1 & 1 \end{pmatrix}$. Which vector is the orthogonal projection of $\vec{y} = \begin{pmatrix} 6 \\ 0 \\ 0 \end{pmatrix}$ onto $W$?
    Answer: $\begin{pmatrix} 4 \\ -2 \\ 2 \end{pmatrix}$ When you encounter orthogonal projection problems, you're finding the closest point in a subspace to a given vector. Here, you need to project $\vec{y} = \begin{pmatrix} 6 \\ 0 \\ 0 \end{pmatrix}$ onto the column space of $A = \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 1 & 1 \end{pmatrix}$. The projection formula is $\text{proj}_W(\vec{y}) = A(A^T A)^{-1}A^T \vec{y}$. First, calculate $A^T A = \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$. Next, $(A^T A)^{-1} = \frac{1}{3}\begin{pmatrix} 2 & -1 \\ -1 & 2 \end{pmatrix}$ since the determinant is 3. Then $A^T \vec{y} = \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \end{pmatrix} \begin{pmatrix} 6 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 6 \\ 0 \end{pmatrix}$. Computing $(A^T A)^{-1}A^T \vec{y} = \frac{1}{3}\begin{pmatrix} 2 & -1 \\ -1 & 2 \end{pmatrix} \begin{pmatrix} 6 \\ 0 \end{pmatrix} = \begin{pmatrix} 4 \\ -2 \end{pmatrix}$. Finally, $A(A^T A)^{-1}A^T \vec{y} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 4 \\ -2 \end{pmatrix} = \begin{pmatrix} 4 \\ -2 \\ 2 \end{pmatrix}$, which is answer D. Answer A gives a vector not in the column space of A. Answer B fails the projection formula calculation. Answer C represents an incorrect computation of the inverse or matrix multiplication. Always verify your projection lies in the subspace by checking it's a linear combination of the column vectors. Here, $4\begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} + (-2)\begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 4 \\ -2 \\ 2 \end{pmatrix}$ confirms our answer.
16. In $\mathbb{R}^3$, let $P$ be the plane passing through the origin with normal vector $n = (1,2,-1)$. If $A$ is the matrix representing the orthogonal projection onto $P$, which of the following is true about the eigenvalues of $A$?
    Answer: Two eigenvalues equal $1$ and one eigenvalue equals $0$ When you encounter orthogonal projection matrices, think about their geometric meaning: they map vectors onto a subspace while preserving distances within that subspace. This geometric insight directly determines the eigenvalue structure. For the orthogonal projection onto plane $P$ with normal vector $n = (1,2,-1)$, consider what happens to different types of vectors. Any vector lying in the plane $P$ gets mapped to itself, so these vectors are eigenvectors with eigenvalue $1$. Since $P$ is a 2-dimensional subspace of $\mathbb{R}^3$, there are two linearly independent eigenvectors with eigenvalue $1$. The normal vector $n = (1,2,-1)$ is perpendicular to plane $P$, so projecting it onto $P$ gives the zero vector. This means $n$ is an eigenvector with eigenvalue $0$. Since the projection eliminates the component perpendicular to $P$, there's exactly one such eigenvalue. Therefore, the matrix $A$ has two eigenvalues equal to $1$ and one eigenvalue equal to $0$, making (D) correct. Looking at the wrong answers: (A) incorrectly suggests eigenvalues of $\frac{1}{2}$, which would indicate some kind of scaling rather than true projection. (B) reverses the multiplicities—this would describe projection onto a line rather than a plane. (C) suggests all eigenvalues equal $\frac{1}{3}$, which would make this a scaling transformation, not a projection. Study tip: Remember that orthogonal projections onto $k$-dimensional subspaces always have $k$ eigenvalues equal to $1$ and $(n-k)$ eigenvalues equal to $0$.
17. In $\mathbb{R}^3$, let $L$ be the line through the origin with direction vector $(1,2,-1)$, and let $P$ be the plane $x + y - z = 0$. If $v = (3,1,2)$, what is the distance from $\text{proj}_L(v)$ to $\text{proj}_P(v)$?
    Answer: $\frac{2\sqrt{6}}{3}$ First, $\text{proj}_L(v) = \frac{\langle v, (1,2,-1) \rangle}{\|(1,2,-1)\|^2} (1,2,-1) = \frac{3}{6}(1,2,-1) = (\frac{1}{2}, 1, -\frac{1}{2})$. For the plane $P$ with normal vector $(1,1,-1)$, we have $\text{proj}_P(v) = v - \text{proj}_{(1,1,-1)}(v) = (3,1,2) - \frac{\langle (3,1,2), (1,1,-1) \rangle}{\|(1,1,-1)\|^2}(1,1,-1) = (3,1,2) - \frac{2}{3}(1,1,-1) = (\frac{7}{3}, \frac{1}{3}, \frac{8}{3})$. The distance between these projections is $\|(\frac{1}{2}, 1, -\frac{1}{2}) - (\frac{7}{3}, \frac{1}{3}, \frac{8}{3})\| = \|(-\frac{11}{6}, \frac{2}{3}, -\frac{19}{6})\| = \sqrt{\frac{121}{36} + \frac{4}{9} + \frac{361}{36}} = \frac{2\sqrt{6}}{3}$.
18. Let $W$ be the line in $\mathbb{R}^3$ defined by the equation $\vec{x} = t\begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix}$ for $t \in \mathbb{R}$. What is the shortest distance from the point $P(1, 2, 3)$ to the line $W$?
    Answer: $\sqrt{10}$ The shortest distance from a point to a line is the length of the vector component orthogonal to the line. Let $\vec{v} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$ be the position vector of point $P$, and let $\vec{u} = \begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix}$ be the direction vector of the line $W$. 1. Find the projection of $\vec{v}$ onto $\vec{u}$, let's call it $\vec{p}$: $\vec{p} = \text{proj}_{\vec{u}}(\vec{v}) = \frac{\vec{v} \cdot \vec{u}}{\vec{u} \cdot \vec{u}}\vec{u}$. $\vec{v} \cdot \vec{u} = (1)(2) + (2)(-1) + (3)(2) = 2 - 2 + 6 = 6$. $\vec{u} \cdot \vec{u} = (2)^2 + (-1)^2 + (2)^2 = 4 + 1 + 4 = 9$. $\vec{p} = \frac{6}{9}\vec{u} = \frac{2}{3}\begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix} = \begin{pmatrix} 4/3 \\ -2/3 \\ 4/3 \end{pmatrix}$. 2. Find the orthogonal component, $\vec{z} = \vec{v} - \vec{p}$: $\vec{z} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} - \begin{pmatrix} 4/3 \\ -2/3 \\ 4/3 \end{pmatrix} = \begin{pmatrix} -1/3 \\ 8/3 \\ 5/3 \end{pmatrix}$. 3. The shortest distance is the magnitude of $\vec{z}$: $\|\vec{z}\| = \sqrt{(-1/3)^2 + (8/3)^2 + (5/3)^2} = \sqrt{\frac{1 + 64 + 25}{9}} = \sqrt{\frac{90}{9}} = \sqrt{10}$. (A) is incorrect because 2 is the length of the projection vector, $\|\vec{p}\| = \frac{2}{3}\|\vec{u}\| = \frac{2}{3}\sqrt{9} = 2$. This represents the distance from the origin to the point of projection on the line. (C) is incorrect because $\sqrt{14}$ is the length of the original vector $\vec{v}$, which is the distance from the origin to point P. (D) is incorrect; $\sqrt{26}$ would result from a sign error in the dot product calculation, i.e., $(1)(2) + (2)(-1) + (3)(-2) = -6$, leading to a different orthogonal vector.
19. Let $\mathcal{B} = \{\vec{u}_1, \vec{u}_2\}$ be an orthogonal basis for a subspace $W \subset \mathbb{R}^3$, where $\vec{u}_1 = \begin{pmatrix} 2 \\ 2 \\ -1 \end{pmatrix}$ and $\vec{u}_2 = \begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix}$. The vector $\vec{w} = \begin{pmatrix} 8 \\ 5 \\ -1 \end{pmatrix}$ is in $W$. What is the coordinate vector $[\vec{w}]_\mathcal{B}$?
    Answer: $\begin{pmatrix} 3 \\ 1 \end{pmatrix}$ For an orthogonal basis $\mathcal{B} = \{\vec{u}_1, \dots, \vec{u}_k\}$, the coordinates of a vector $\vec{w}$ in the subspace are the scalars from the projection formula. If $\vec{w} = c_1\vec{u}_1 + c_2\vec{u}_2$, then the coordinate vector is $[\vec{w}]_\mathcal{B} = \begin{pmatrix} c_1 \\ c_2 \end{pmatrix}$, where $c_i = \frac{\vec{w} \cdot \vec{u}_i}{\vec{u}_i \cdot \vec{u}_i}$. Calculate $c_1$: $\vec{w} \cdot \vec{u}_1 = (8)(2) + (5)(2) + (-1)(-1) = 16 + 10 + 1 = 27$. $\vec{u}_1 \cdot \vec{u}_1 = (2)^2 + (2)^2 + (-1)^2 = 4 + 4 + 1 = 9$. $c_1 = \frac{27}{9} = 3$. Calculate $c_2$: $\vec{w} \cdot \vec{u}_2 = (8)(2) + (5)(-1) + (-1)(2) = 16 - 5 - 2 = 9$. $\vec{u}_2 \cdot \vec{u}_2 = (2)^2 + (-1)^2 + (2)^2 = 4 + 1 + 4 = 9$. $c_2 = \frac{9}{9} = 1$. Therefore, the coordinate vector is $[\vec{w}]_\mathcal{B} = \begin{pmatrix} 3 \\ 1 \end{pmatrix}$. (A) is incorrect; it consists of the dot products $\vec{w} \cdot \vec{u}_i$, forgetting to divide by $\vec{u}_i \cdot \vec{u}_i$. (B) is incorrect; it results from dividing the dot products by the norms ($\sqrt{9}=3$) instead of the norms squared. (C) is incorrect; it is the correct coordinates but in the wrong order.
20. Let $W$ be a subspace of $\mathbb{R}^n$ and let $\vec{y}$ be a vector in $\mathbb{R}^n$. Let $\hat{\vec{y}} = \text{proj}_W(\vec{y})$. Which of the following statements is always true?
    Answer: The vector $\vec{y} - \hat{\vec{y}}$ is orthogonal to every vector in $W$. The Orthogonal Decomposition Theorem states that any vector $\vec{y}$ can be uniquely written as the sum of a vector in a subspace $W$ and a vector in its orthogonal complement $W^\perp$. Specifically, $\vec{y} = \hat{\vec{y}} + \vec{z}$, where $\hat{\vec{y}} = \text{proj}_W(\vec{y})$ is in $W$ and $\vec{z} = \vec{y} - \hat{\vec{y}}$ is in $W^\perp$. By definition, a vector in $W^\perp$ is orthogonal to every vector in $W$. Therefore, the statement that $\vec{y} - \hat{\vec{y}}$ is orthogonal to every vector in $W$ is always true. (A) is incorrect. The projection $\hat{\vec{y}}$ is not generally orthogonal to $\vec{y}$. This would only happen if $\text{proj}_W(\vec{y})=\vec{0}$, which means $\vec{y}$ is in $W^\perp$. (B) is incorrect. The vector $\vec{y} - \hat{\vec{y}}$ is the component of $\vec{y}$ orthogonal to $W$, so it lies in $W^\perp$, not $W$. (D) is incorrect. The length of the projection is less than or equal to the length of the original vector. Equality holds only if $\vec{y}$ is already in the subspace $W$, in which case $\hat{\vec{y}} = \vec{y}$.

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