Orthonormal Bases - Linear Algebra

Let $S = \{\mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3\}$ be the standard orthonormal basis for $\mathbb{R}^3$ , and let $T = \{\mathbf{f}_1, \mathbf{f}_2, \mathbf{f}_3\}$ be another orthonormal basis where $\mathbf{f}_1 = \frac{1}{\sqrt{3}}(1, 1, 1)$ . If the transition matrix from $S$ to $T$ has the property that its second column sums to zero, which of the following could be $\mathbf{f}_2$ ?
Answer: $\frac{1}{\sqrt{6}}(2, -1, -1)$ The transition matrix from $S$ to $T$ has columns that are the coordinates of $\mathbf{f}_1, \mathbf{f}_2, \mathbf{f}_3$ in the standard basis. For the second column to sum to zero, the components of $\mathbf{f}_2$ must sum to zero. Also, $\mathbf{f}_2$ must be orthogonal to $\mathbf{f}_1$ and have unit length. Choice B: $\frac{2 + (-1) + (-1)}{\sqrt{6}} = \frac{0}{\sqrt{6}} = 0$ ✓, $\mathbf{f}_1 \cdot \mathbf{f}_2 = \frac{1}{\sqrt{18}}(1 \cdot 2 + 1 \cdot (-1) + 1 \cdot (-1)) = \frac{0}{\sqrt{18}} = 0$ ✓, and $\|\mathbf{f}_2\|^2 = \frac{4 + 1 + 1}{6} = 1$ ✓. Choice A has components summing to 0 but isn't orthogonal to $\mathbf{f}_1$ . Choices C and D don't have components summing to zero.
Let $\mathcal{B} = \{\mathbf{u}_1, \mathbf{u}_2\}$ be an orthonormal basis for a vector space $V$ . Two vectors $\mathbf{v}$ and $\mathbf{w}$ in $V$ have coordinate vectors with respect to $\mathcal{B}$ of $[\mathbf{v}]_\mathcal{B} = \begin{pmatrix} 2 \\ -5 \end{pmatrix}$ and $[\mathbf{w}]_\mathcal{B} = \begin{pmatrix} -1 \\ 3 \end{pmatrix}$ . What is the value of the inner product $\langle \mathbf{v}, \mathbf{w} \rangle$ ?
Answer: -17 A key property of orthonormal bases is that the inner product of two vectors is equal to the dot product of their coordinate vectors. Let $\mathbf{v} = c_1\mathbf{u}_1 + c_2\mathbf{u}_2$ and $\mathbf{w} = d_1\mathbf{u}_1 + d_2\mathbf{u}_2$ . Then $\langle \mathbf{v}, \mathbf{w} \rangle = \langle c_1\mathbf{u}_1 + c_2\mathbf{u}_2, d_1\mathbf{u}_1 + d_2\mathbf{u}_2 \rangle$ . Because the basis is orthonormal, $\langle \mathbf{u}_i, \mathbf{u}_j \rangle$ is 1 if $i=j$ and 0 if $i \neq j$ . The expression simplifies to $c_1 d_1 + c_2 d_2$ . Here, the coordinate vectors are $(2, -5)$ and $(-1, 3)$ . So, $\langle \mathbf{v}, \mathbf{w} \rangle = (2)(-1) + (-5)(3) = -2 - 15 = -17$ . Distractor B comes from a sign error: $-2 + 15 = 13$ , nope. $2(-1)-(-5)(3) = -2+15 = 13$ . That is C. Distractor B could be $2(-1)+5(3)=-2-15$ , oh wait, $2-15 = -13$ . Distractor D is a very common misconception for students who have not learned this property.
In $\mathbb{R}^3$ , let $\mathbf{u} = (a, b, c)$ be a unit vector, and define $\mathbf{v}_1 = \mathbf{u}$ , $\mathbf{v}_2 = \frac{1}{\sqrt{2}}(1, -1, 0)$ , $\mathbf{v}_3 = \frac{1}{\sqrt{6}}(1, 1, -2)$ . For what value of $a$ does $\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\}$ form an orthonormal basis for $\mathbb{R}^3$ with $a > 0$ ?
Answer: $\frac{1}{\sqrt{3}}$ For orthonormality, we need $\mathbf{v}_1 \perp \mathbf{v}_2$ and $\mathbf{v}_1 \perp \mathbf{v}_3$ , plus $\|\mathbf{v}_1\| = 1$ . From $\mathbf{v}_1 \cdot \mathbf{v}_2 = 0$ : $(a, b, c) \cdot \frac{1}{\sqrt{2}}(1, -1, 0) = \frac{1}{\sqrt{2}}(a - b) = 0$ , so $a = b$ . From $\mathbf{v}_1 \cdot \mathbf{v}_3 = 0$ : $(a, b, c) \cdot \frac{1}{\sqrt{6}}(1, 1, -2) = \frac{1}{\sqrt{6}}(a + b - 2c) = 0$ , so $a + b - 2c = 0$ . Since $a = b$ , we get $2a - 2c = 0$ , thus $c = a$ . From $\|\mathbf{v}_1\| = 1$ : $a^2 + b^2 + c^2 = 1$ , and since $a = b = c$ , we have $3a^2 = 1$ , so $a = \pm\frac{1}{\sqrt{3}}$ . Since $a > 0$ , we get $a = \frac{1}{\sqrt{3}}$ . Choice B would give $a = b = c = \frac{\sqrt{2}}{2}$ , but then $3 \cdot \frac{1}{2} = \frac{3}{2} \neq 1$ . Similar errors occur in choices C and D.
In the vector space of continuous functions on $[-\pi, \pi]$ , $C[-\pi, \pi]$ , with the inner product $\langle f, g \rangle = \int_{-\pi}^{\pi} f(x)g(x)dx$ , the set $\{\frac{1}{\sqrt{2\pi}}, \frac{\cos(x)}{\sqrt{\pi}}, \frac{\sin(x)}{\sqrt{\pi}}\}$ forms an orthonormal set. What is the coordinate of the function $f(x) = 3\cos(x) - 4\sin(x)$ corresponding to the basis vector $u(x) = \frac{\sin(x)}{\sqrt{\pi}}$ ?
Answer: $-4\sqrt{\pi}$ The coordinate $c$ corresponding to the basis vector $u(x)$ is found by computing the inner product $c = \langle f(x), u(x) \rangle$ . $c = \int_{-\pi}^{\pi} (3\cos(x) - 4\sin(x)) \left( \frac{\sin(x)}{\sqrt{\pi}} \right) dx$ . $c = \frac{1}{\sqrt{\pi}} \int_{-\pi}^{\pi} (3\cos(x)\sin(x) - 4\sin^2(x)) dx$ . Due to the orthogonality of $\cos(x)$ and $\sin(x)$ over $[-\pi, \pi]$ , the integral of their product is zero: $\int_{-\pi}^{\pi} 3\cos(x)\sin(x) dx = 0$ . So we are left with: $c = -\frac{4}{\sqrt{\pi}} \int_{-\pi}^{\pi} \sin^2(x) dx$ . The integral $\int_{-\pi}^{\pi} \sin^2(x) dx = \pi$ . Therefore, $c = -\frac{4}{\sqrt{\pi}} (\pi) = -4\sqrt{\pi}$ . Distractor A (-4) is a common mistake where the student simply reads the coefficient of $\sin(x)$ from the function definition, ignoring the inner product and the normalization of the basis vector. Distractor D is a result of forgetting to multiply by the result of the integral. Distractor B is a result of miscalculating the integral or the scalar multiple.
The vectors $\mathbf{u}_1 = \frac{1}{3}\begin{pmatrix} 2 \\ 1 \\ 2 \end{pmatrix}$ and $\mathbf{u}_2 = \frac{1}{3}\begin{pmatrix} -2 \\ 2 \\ 1 \end{pmatrix}$ form an orthonormal set. Which of the following vectors $\mathbf{u}_3$ completes the set to form an orthonormal basis for $\mathbb{R}^3$ ?
Answer: $\frac{1}{3}\begin{pmatrix} 1 \\ 2 \\ -2 \end{pmatrix}$ To complete the orthonormal basis, $\mathbf{u}_3$ must be a unit vector and must be orthogonal to both $\mathbf{u}_1$ and $\mathbf{u}_2$ . We check each option. A) Let $\mathbf{w} = \begin{pmatrix} 1 \\ 2 \\ -2 \end{pmatrix}$ . $\langle \mathbf{w}, \mathbf{u}_1 \rangle = \frac{1}{3}(2+2-4)=0$ . $\langle \mathbf{w}, \mathbf{u}_2 \rangle = \frac{1}{3}(-2+4-2)=0$ . So it is orthogonal. However, its norm is $\|\mathbf{w}\| = \sqrt{1^2+2^2+(-2)^2} = \sqrt{1+4+4} = \sqrt{9}=3$ , so it is not a unit vector. B) This is the vector from (A) normalized by its length, $3$ . Its norm is $1$ . It is orthogonal to $\mathbf{u}_1$ and $\mathbf{u}_2$ . This is the correct choice. C) This vector is $\mathbf{u}_1$ itself, so it is not orthogonal to $\mathbf{u}_1$ and cannot be part of the basis. D) Let $\mathbf{z} = \frac{1}{\sqrt{5}}\begin{pmatrix} 1 \\ -2 \\ 0 \end{pmatrix}$ . This is a unit vector, but $\langle \mathbf{z}, \mathbf{u}_1 \rangle = \frac{1}{3\sqrt{5}}(2-2+0)=0$ . However, $\langle \mathbf{z}, \mathbf{u}_2 \rangle = \frac{1}{3\sqrt{5}}(-2-4+0) = -\frac{6}{3\sqrt{5}} \neq 0$ . It is not orthogonal to $\mathbf{u}_2$ .
In an inner product space, let $\mathcal{B} = \{\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3\}$ be an orthonormal basis. A vector $\mathbf{v}$ is expressed as $\mathbf{v} = 4\mathbf{u}_1 - c_2\mathbf{u}_2 + 2\mathbf{u}_3$ . If the norm of $\mathbf{v}$ is $\|\mathbf{v}\| = \sqrt{29}$ , what is a possible value for the coordinate $c_2$ ?
Answer: 3 By Parseval's identity, the squared norm of a vector is the sum of the squares of its coordinates with respect to an orthonormal basis. So, $\|\mathbf{v}\|^2 = (4)^2 + (-c_2)^2 + (2)^2$ . We are given that $\|\mathbf{v}\| = \sqrt{29}$ , so $\|\mathbf{v}\|^2 = 29$ . Therefore, $29 = 16 + c_2^2 + 4$ . $29 = 20 + c_2^2$ . $c_2^2 = 9$ . $c_2 = \pm 3$ . One possible value for $c_2$ is 3. Distractor B (9) is the value of $c_2^2$ . Distractor C ( $\sqrt{5}$ ) results from an arithmetic error $29-16-4=9$ , maybe $29-16-4=5$ . $29-20=9$ . Maybe $29-4=25, 25-16=9$ . Maybe they miscalculate $29-(16+4)=29-20=9$ . Maybe they do $\sqrt{29} - 4 - 2 = \sqrt{29}-6$ . Or maybe $29 = 16+c_2+4$ , so $c_2=9$ . Let's make the sign different on the stem. $c_2$ vs $-c_2$ doesn't matter. Okay, the distractors are fine. D (5) is another plausible arithmetic error, maybe related to $3-4-5$ triangles.
Let $\mathcal{B} = \{\mathbf{u}_1, \mathbf{u}_2\}$ be an orthonormal basis for a plane $W$ in $\mathbb{R}^3$ . For a vector $\mathbf{v} \in \mathbb{R}^3$ , we find that its projection onto $W$ has a norm of $\|\text{proj}_W(\mathbf{v})\| = 13$ and its coordinate with respect to $\mathbf{u}_1$ is $c_1 = \langle \mathbf{v}, \mathbf{u}_1 \rangle = -5$ . If the coordinate with respect to $\mathbf{u}_2$ is $c_2$ , what is a possible value of $c_2$ ?
Answer: 12 The projection of $\mathbf{v}$ onto $W$ is given by $\text{proj}_W(\mathbf{v}) = c_1\mathbf{u}_1 + c_2\mathbf{u}_2$ , where $c_1 = \langle \mathbf{v}, \mathbf{u}_1 \rangle$ and $c_2 = \langle \mathbf{v}, \mathbf{u}_2 \rangle$ . Since $\{\mathbf{u}_1, \mathbf{u}_2\}$ is an orthonormal basis for $W$ , the squared norm of the projection vector is the sum of the squares of its coordinates: $\|\text{proj}_W(\mathbf{v})\|^2 = c_1^2 + c_2^2$ . We are given $\|\text{proj}_W(\mathbf{v})\| = 13$ and $c_1 = -5$ . Plugging these values in: $13^2 = (-5)^2 + c_2^2$ . $169 = 25 + c_2^2$ . $c_2^2 = 169 - 25 = 144$ . $c_2 = \pm \sqrt{144} = \pm 12$ . A possible value for $c_2$ is 12. Distractor A (8) comes from the incorrect assumption of a linear relationship, $13 - 5 = 8$ . Distractor D (144) is the value of $c_2^2$ , a common error. Distractor C (18) might arise from $13+5=18$ .
In an inner product space $V$ , let $\{\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3\}$ be an orthonormal basis. If $\mathbf{v} = 2\mathbf{u}_1 - 3\mathbf{u}_2 + \mathbf{u}_3$ and $\mathbf{w} = a\mathbf{u}_1 + b\mathbf{u}_2 + c\mathbf{u}_3$ , what values of $a$ , $b$ , and $c$ make $\langle \mathbf{v}, \mathbf{w} \rangle = 5$ and $\|\mathbf{w}\| = \sqrt{6}$ ?
Answer: Infinitely many solutions forming a circle in $\mathbb{R}^3$ Using orthonormality, $\langle \mathbf{v}, \mathbf{w} \rangle = 2a - 3b + c = 5$ and $\|\mathbf{w}\|^2 = a^2 + b^2 + c^2 = 6$ . This gives us two equations in three unknowns: a plane and a sphere intersecting in a circle. Since the plane $2a - 3b + c = 5$ doesn't pass through the origin, and the sphere $a^2 + b^2 + c^2 = 6$ has radius $\sqrt{6} > 0$ , their intersection is generically a circle (assuming the plane intersects the sphere, which can be verified by checking that the distance from origin to plane is less than the sphere radius). The distance is $\frac{|5|}{\sqrt{4 + 9 + 1}} = \frac{5}{\sqrt{14}} < \sqrt{6}$ , so intersection occurs. Choice A gives only one point, which is insufficient. Choice B gives only two points. Choice D is incorrect since solutions exist.
Let $\mathcal{B} = \{\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3\}$ be an orthonormal basis for $\mathbb{R}^3$ , where $\mathbf{u}_1 = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}$ , $\mathbf{u}_2 = \frac{1}{\sqrt{3}}\begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix}$ , and $\mathbf{u}_3 = \frac{1}{\sqrt{6}}\begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}$ . Let $\mathbf{v} = \begin{pmatrix} 2 \\ 0 \\ -1 \end{pmatrix}$ . If the coordinates of $\mathbf{v}$ with respect to $\mathcal{B}$ are $(c_1, c_2, c_3)$ , what is the value of $c_1 - c_2 + 2c_3$ ?
Answer: $0$ The coordinates are found by taking the inner (dot) product of $\mathbf{v}$ with each basis vector: $c_1 = \langle \mathbf{v}, \mathbf{u}_1 \rangle = (2)(\frac{1}{\sqrt{2}}) + (0)(\frac{1}{\sqrt{2}}) + (-1)(0) = \frac{2}{\sqrt{2}} = \sqrt{2}$ . $c_2 = \langle \mathbf{v}, \mathbf{u}_2 \rangle = (2)(-\frac{1}{\sqrt{3}}) + (0)(\frac{1}{\sqrt{3}}) + (-1)(\frac{1}{\sqrt{3}}) = -\frac{2}{\sqrt{3}} - \frac{1}{\sqrt{3}} = -\frac{3}{\sqrt{3}} = -\sqrt{3}$ . $c_3 = \langle \mathbf{v}, \mathbf{u}_3 \rangle = (2)(\frac{1}{\sqrt{6}}) + (0)(-\frac{1}{\sqrt{6}}) + (-1)(\frac{2}{\sqrt{6}}) = \frac{2}{\sqrt{6}} - \frac{2}{\sqrt{6}} = 0$ . The desired linear combination is $c_1 - c_2 + 2c_3 = \sqrt{2} - (-\sqrt{3}) + 2(0) = \sqrt{2} + \sqrt{3}$ . There seems to be a calculation error in my setup. Let's recompute. $c_2 = -\frac{2}{\sqrt{3}} - \frac{1}{\sqrt{3}} = -\frac{3}{\sqrt{3}} = -\sqrt{3}$ . $c_3=0$ . This gives $\sqrt{2}+\sqrt{3}$ . My options are incorrect. Let's adjust the question. Let $\mathbf{u}_2 = \frac{1}{\sqrt{3}}\begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}$ and $\mathbf{u}_3 = \frac{1}{\sqrt{6}}\begin{pmatrix} -1 \\ -1 \\ 2 \end{pmatrix}$ which is an ONB with the given $\mathbf{u}_1$ . Let's try again with a better basis. Let $\mathbf{u}_1 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$ , $\mathbf{u}_2 = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}$ , $\mathbf{u}_3 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$ . This is too easy. Let's use $\mathbf{u}_1 = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}$ , $\mathbf{u}_2 = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix}$ , $\mathbf{u}_3 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$ . Let $\mathbf{v} = \begin{pmatrix} 3 \\ 1 \\ 4 \end{pmatrix}$ . Then $c_1 = \frac{4}{\sqrt{2}}=2\sqrt{2}$ , $c_2 = \frac{2}{\sqrt{2}}=\sqrt{2}$ , $c_3 = 4$ . Ask for $c_1 - c_2 - c_3$ . $2\sqrt{2} - \sqrt{2} - 4 = \sqrt{2} - 4$ . This is better. I will rewrite the question with these new values. Let $\mathcal{B} = \{\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3\}$ be an orthonormal basis for $\mathbb{R}^3$ , where $\mathbf{u}_1 = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}$ , $\mathbf{u}_2 = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix}$ , and $\mathbf{u}_3 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$ . Let $\mathbf{v} = \begin{pmatrix} 3 \\ 5 \\ -2 \end{pmatrix}$ . If the coordinates of $\mathbf{v}$ with respect to $\mathcal{B}$ are $(c_1, c_2, c_3)$ , what is the value of $c_1 + 2c_2 + c_3$ ? $c_1 = \langle \mathbf{v}, \mathbf{u}_1 \rangle = \frac{3+5}{\sqrt{2}} = \frac{8}{\sqrt{2}} = 4\sqrt{2}$ . $c_2 = \langle \mathbf{v}, \mathbf{u}_2 \rangle = \frac{3-5}{\sqrt{2}} = -\frac{2}{\sqrt{2}} = -\sqrt{2}$ . $c_3 = \langle \mathbf{v}, \mathbf{u}_3 \rangle = -2$ . The desired expression is $c_1 + 2c_2 + c_3 = 4\sqrt{2} + 2(-\sqrt{2}) + (-2) = 4\sqrt{2} - 2\sqrt{2} - 2 = 2\sqrt{2} - 2$ . Let's make the numbers cleaner. Let $\mathbf{v}=\begin{pmatrix} \sqrt{2} \\ 3\sqrt{2} \\ -1 \end{pmatrix}$ . $c_1 = \frac{\sqrt{2}+3\sqrt{2}}{\sqrt{2}} = 4$ . $c_2 = \frac{\sqrt{2}-3\sqrt{2}}{\sqrt{2}} = -2$ . $c_3 = -1$ . What is $c_1+c_2+c_3$ ? $4-2-1=1$ . This is a good setup. Final explanation: The coordinates are found by taking the inner (dot) product of $\mathbf{v}$ with each basis vector: $c_1 = \langle \mathbf{v}, \mathbf{u}_1 \rangle = (\sqrt{2})(\frac{1}{\sqrt{2}}) + (3\sqrt{2})(\frac{1}{\sqrt{2}}) + (-1)(0) = 1 + 3 = 4$ . $c_2 = \langle \mathbf{v}, \mathbf{u}_2 \rangle = (\sqrt{2})(\frac{1}{\sqrt{2}}) + (3\sqrt{2})(-\frac{1}{\sqrt{2}}) + (-1)(0) = 1 - 3 = -2$ . $c_3 = \langle \mathbf{v}, \mathbf{u}_3 \rangle = (\sqrt{2})(0) + (3\sqrt{2})(0) + (-1)(1) = -1$ . The desired sum is $c_1 + c_2 + c_3 = 4 + (-2) + (-1) = 1$ . Distractor A: $4\sqrt{2}-1$ is the sum of norms of components. No. Distractor B: A miscalculation. Distractor D: sum of vector components. This is good.
Consider the orthonormal basis $\{\mathbf{q}_1, \mathbf{q}_2, \mathbf{q}_3\}$ for $\mathbb{R}^3$ where $\mathbf{q}_1 = \frac{1}{3}(2, 1, 2)$ , $\mathbf{q}_2 = \frac{1}{3}(1, 2, -2)$ , and $\mathbf{q}_3 = \frac{1}{3}(2, -2, 1)$ . If vector $\mathbf{v}$ satisfies $\|\mathbf{v}\|^2 = 14$ and has coordinates $(\alpha, \beta, \gamma)$ in this basis, what is $\alpha^2 + \beta^2 + \gamma^2$ ?
Answer: $14$ For any orthonormal basis, Parseval's identity states that $\|\mathbf{v}\|^2 = \alpha^2 + \beta^2 + \gamma^2$ where $\alpha, \beta, \gamma$ are the coordinates of $\mathbf{v}$ in that basis. Since $\{\mathbf{q}_1, \mathbf{q}_2, \mathbf{q}_3\}$ is orthonormal and $\|\mathbf{v}\|^2 = 14$ , we immediately get $\alpha^2 + \beta^2 + \gamma^2 = 14$ . Choice B incorrectly applies $\frac{\|\mathbf{v}\|^2}{9}$ thinking the basis vectors have length $\frac{1}{3}$ rather than 1. Choice C computes $14 \times 9 = 126$ , incorrectly scaling up. Choice D uses $14 \times 3 = 42$ , another scaling error.
Let $\mathcal{B} = \{\mathbf{u}_1, \mathbf{u}_2\}$ be an orthonormal basis for $\mathbb{R}^2$ . For a vector $\mathbf{v} \in \mathbb{R}^2$ , the scalar value $c_1 = \langle \mathbf{v}, \mathbf{u}_1 \rangle$ represents which geometric quantity?
Answer: The signed magnitude (scalar component) of the orthogonal projection of $\mathbf{v}$ onto the line spanned by $\mathbf{u}_1$ . The orthogonal projection of $\mathbf{v}$ onto the line spanned by the unit vector $\mathbf{u}_1$ is the vector $\text{proj}_{\mathbf{u}_1}(\mathbf{v}) = \langle \mathbf{v}, \mathbf{u}_1 \rangle \mathbf{u}_1$ . The scalar coefficient $c_1 = \langle \mathbf{v}, \mathbf{u}_1 \rangle$ is the scalar component of this projection. It represents a signed magnitude: its absolute value, $|c_1|$ , is the length of the projection vector, and its sign indicates whether the projection points in the same direction as $\mathbf{u}_1$ (positive) or the opposite direction (negative). (A) is incorrect; the length of $\mathbf{v}$ is $\sqrt{c_1^2 + c_2^2}$ . (B) is incorrect; the angle $\theta$ is related by $c_1 = \|\mathbf{v}\| \cos(\theta)$ , but $c_1$ is not the angle itself. (C) is subtly incorrect. The length of the projection is $\|c_1\mathbf{u}_1\| = |c_1|\|\mathbf{u}_1\| = |c_1|$ , which is the absolute value. The coordinate $c_1$ can be negative, while length cannot.
Which of the following sets of vectors forms an orthonormal basis for $\mathbb{R}^2$ ?
Answer: $\{\frac{1}{5}\begin{pmatrix} 3 \\ 4 \end{pmatrix}, \frac{1}{5}\begin{pmatrix} 4 \\ -3 \end{pmatrix}\}$ An orthonormal basis consists of vectors that are mutually orthogonal and have a norm of 1. A) The vectors are orthogonal since $(1)(1) + (1)(-1) = 0$ . However, their norms are $\sqrt{1^2+1^2} = \sqrt{2}$ , not 1. So this is an orthogonal basis, but not orthonormal. B) The vectors are both unit vectors. However, they are not orthogonal since $(1)(1/\sqrt{2}) + (0)(1/\sqrt{2}) = 1/\sqrt{2} \neq 0$ . C) Let $\mathbf{u}_1 = \frac{1}{5}\begin{pmatrix} 3 \\ 4 \end{pmatrix}$ and $\mathbf{u}_2 = \frac{1}{5}\begin{pmatrix} 4 \\ -3 \end{pmatrix}$ . The norms are $\|\mathbf{u}_1\| = \frac{1}{5}\sqrt{3^2+4^2} = \frac{1}{5}\sqrt{25} = 1$ and $\|\mathbf{u}_2\| = \frac{1}{5}\sqrt{4^2+(-3)^2} = \frac{1}{5}\sqrt{25} = 1$ . The vectors are orthogonal since their dot product is $\frac{1}{25}((3)(4) + (4)(-3)) = \frac{1}{25}(12-12)=0$ . This set is an orthonormal basis. D) The vectors are orthogonal. The first vector is a unit vector. However, the second vector has a norm of $\sqrt{0^2+2^2}=2$ . So the set is not orthonormal.
Let $\mathcal{B} = \{\mathbf{u}_1, \mathbf{u}_2\}$ be an orthonormal basis for a subspace $W$ of $\mathbb{R}^4$ . A vector $\mathbf{v} \in W$ has coordinates with respect to this basis given by $[\mathbf{v}]_\mathcal{B} = \begin{pmatrix} -3 \\ 5 \end{pmatrix}$ . What is the squared norm, $\|\mathbf{v}\|^2$ , of the vector $\mathbf{v}$ ?
Answer: 34 For an orthonormal basis $\mathcal{B} = \{\mathbf{u}_1, \mathbf{u}_2, ..., \mathbf{u}_k\}$ , the norm of a vector $\mathbf{v}$ can be calculated from its coordinates $(c_1, c_2, ..., c_k)$ using Parseval's identity: $\|\mathbf{v}\|^2 = c_1^2 + c_2^2 + ... + c_k^2$ . In this case, $\mathbf{v} = -3\mathbf{u}_1 + 5\mathbf{u}_2$ . Therefore, $\|\mathbf{v}\|^2 = (-3)^2 + 5^2 = 9 + 25 = 34$ . Distractor A is the sum of the coordinates, $5-3=2$ . Distractor B is the sum of the absolute values of the coordinates, $|-3|+|5|=8$ . Distractor C is the squared sum of coordinates, $(5-3)^2=4$ , this is a weak distractor. Let's make it $(5-|-3|)^2 = (2)^2=4$ . Or maybe just a calculation error. A better distractor is $5^2-3^2 = 25-9=16$ .

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