Linear Algebra Quiz: Gram Schmidt Process
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Gram Schmidt ProcessQuestion 1 of 18

The Gram-Schmidt process is applied to the vectors v⃗1=(1,2,2)\vec{v}_1 = (1, 2, 2) and v⃗2=(1,0,1)\vec{v}_2 = (1, 0, 1) to produce an orthonormal set {u⃗1,u⃗2}\{\vec{u}_1, \vec{u}_2\}. Which of the following is the vector u⃗2\vec{u}_2?

(23,−23,13)(\frac{2}{3}, -\frac{2}{3}, \frac{1}{3})
(−23,23,−13)(-\frac{2}{3}, \frac{2}{3}, -\frac{1}{3})
(23,13,−23)(\frac{2}{3}, \frac{1}{3}, -\frac{2}{3})
(12,0,12)(\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}})
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Linear Algebra Quiz

Linear Algebra Quiz: Gram Schmidt Process

Practice Gram Schmidt Process 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 Gram Schmidt Process, giving you a quick way to practice the rules, question types, and explanations that matter most for Linear Algebra.

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Question 1

The Gram-Schmidt process is applied to the vectors v⃗1=(1,2,2)\vec{v}_1 = (1, 2, 2) and v⃗2=(1,0,1)\vec{v}_2 = (1, 0, 1) to produce an orthonormal set {u⃗1,u⃗2}\{\vec{u}_1, \vec{u}_2\}. Which of the following is the vector u⃗2\vec{u}_2?

  1. (23,−23,13)(\frac{2}{3}, -\frac{2}{3}, \frac{1}{3}) (correct answer)
  2. (−23,23,−13)(-\frac{2}{3}, \frac{2}{3}, -\frac{1}{3})
  3. (23,13,−23)(\frac{2}{3}, \frac{1}{3}, -\frac{2}{3})
  4. (12,0,12)(\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}})
Explanation: First, find u⃗1\vec{u}_1: w⃗1=v⃗1=(1,2,2)\vec{w}_1 = \vec{v}_1 = (1, 2, 2). ∥w⃗1∥=12+22+22=3\|\vec{w}_1\| = \sqrt{1^2+2^2+2^2} = 3. So, u⃗1=(13,23,23)\vec{u}_1 = (\frac{1}{3}, \frac{2}{3}, \frac{2}{3}). Next, find w⃗2=v⃗2−proju⃗1(v⃗2)=v⃗2−⟨v⃗2,u⃗1⟩u⃗1\vec{w}_2 = \vec{v}_2 - \text{proj}_{\vec{u}_1}(\vec{v}_2) = \vec{v}_2 - \langle \vec{v}_2, \vec{u}_1 \rangle \vec{u}_1. The dot product is ⟨v⃗2,u⃗1⟩=1(13)+0(23)+1(23)=1\langle \vec{v}_2, \vec{u}_1 \rangle = 1(\frac{1}{3}) + 0(\frac{2}{3}) + 1(\frac{2}{3}) = 1. So, w⃗2=(1,0,1)−1⋅(13,23,23)=(23,−23,13)\vec{w}_2 = (1, 0, 1) - 1 \cdot (\frac{1}{3}, \frac{2}{3}, \frac{2}{3}) = (\frac{2}{3}, -\frac{2}{3}, \frac{1}{3}). Finally, normalize w⃗2\vec{w}_2. ∥w⃗2∥=(23)2+(−23)2+(13)2=4+4+19=1\|\vec{w}_2\| = \sqrt{(\frac{2}{3})^2 + (-\frac{2}{3})^2 + (\frac{1}{3})^2} = \sqrt{\frac{4+4+1}{9}} = 1. Since its norm is 1, u⃗2=w⃗2=(23,−23,13)\vec{u}_2 = \vec{w}_2 = (\frac{2}{3}, -\frac{2}{3}, \frac{1}{3}).

Question 2

Let S={v⃗1,v⃗2}S = \{\vec{v}_1, \vec{v}_2\} be a linearly independent set of non-orthogonal vectors in R3\mathbb{R}^3. Let B1={u⃗1,u⃗2}B_1 = \{\vec{u}_1, \vec{u}_2\} be the orthonormal basis obtained by applying the Gram-Schmidt process to SS in the order v⃗1,v⃗2\vec{v}_1, \vec{v}_2. Let B2={u′⃗1,u′⃗2}B_2 = \{\vec{u'}_1, \vec{u'}_2\} be the basis obtained by applying the process to SS in the order v⃗2,v⃗1\vec{v}_2, \vec{v}_1. Which of the following statements is always true?

  1. B1B_1 and B2B_2 are the same basis; that is, u⃗1=u′⃗1\vec{u}_1 = \vec{u'}_1 and u⃗2=u′⃗2\vec{u}_2 = \vec{u'}_2.
  2. The basis vectors are swapped; that is, u⃗1=u′⃗2\vec{u}_1 = \vec{u'}_2 and u⃗2=u′⃗1\vec{u}_2 = \vec{u'}_1.
  3. The first vectors will differ (i.e., u⃗1≠u′⃗1\vec{u}_1 \neq \vec{u'}_1), but the subspaces they span will be identical. (correct answer)
  4. The subspaces spanned by B1B_1 and B2B_2 will be different.
Explanation: The Gram-Schmidt process is order-dependent. The first vector in the orthonormal set is always the normalized version of the first vector from the original set. Since v⃗1≠v⃗2\vec{v}_1 \neq \vec{v}_2 (and they are not scalar multiples), u⃗1=v⃗1∥v⃗1∥\vec{u}_1 = \frac{\vec{v}_1}{\|\vec{v}_1\|} will be different from u′⃗1=v⃗2∥v⃗2∥\vec{u'}_1 = \frac{\vec{v}_2}{\|\vec{v}_2\|}. However, both resulting orthonormal bases, B1B_1 and B2B_2, must span the exact same subspace as the original set of vectors, span{v⃗1,v⃗2}\text{span}\{\vec{v}_1, \vec{v}_2\}.

Question 3

A student is applying the Gram-Schmidt process to the basis {v⃗1,v⃗2,v⃗3}\{\vec{v}_1, \vec{v}_2, \vec{v}_3\} of R3\mathbb{R}^3. They have correctly computed the first two orthonormal vectors, u⃗1\vec{u}_1 and u⃗2\vec{u}_2, such that span{v⃗1,v⃗2}=span{u⃗1,u⃗2}\text{span}\{\vec{v}_1, \vec{v}_2\} = \text{span}\{\vec{u}_1, \vec{u}_2\}. What is the correct expression for the third orthogonal (but not yet normalized) vector, w⃗3\vec{w}_3?

  1. w⃗3=v⃗3−⟨v⃗3,u⃗1⟩u⃗1\vec{w}_3 = \vec{v}_3 - \langle \vec{v}_3, \vec{u}_1 \rangle \vec{u}_1
  2. w⃗3=v⃗3−⟨v⃗3,v⃗1⟩v⃗1−⟨v⃗3,v⃗2⟩v⃗2\vec{w}_3 = \vec{v}_3 - \langle \vec{v}_3, \vec{v}_1 \rangle \vec{v}_1 - \langle \vec{v}_3, \vec{v}_2 \rangle \vec{v}_2
  3. w⃗3=v⃗3∥v⃗3∥−u⃗1−u⃗2\vec{w}_3 = \frac{\vec{v}_3}{\|\vec{v}_3\|} - \vec{u}_1 - \vec{u}_2
  4. w⃗3=v⃗3−⟨v⃗3,u⃗1⟩u⃗1−⟨v⃗3,u⃗2⟩u⃗2\vec{w}_3 = \vec{v}_3 - \langle \vec{v}_3, \vec{u}_1 \rangle \vec{u}_1 - \langle \vec{v}_3, \vec{u}_2 \rangle \vec{u}_2 (correct answer)
Explanation: To find the next orthogonal vector in the sequence, one must take the next original vector (v⃗3\vec{v}_3) and subtract its projections onto all previously found orthonormal vectors (both u⃗1\vec{u}_1 and u⃗2\vec{u}_2). Choice A only subtracts the projection onto u⃗1\vec{u}_1. Choice B incorrectly uses the original, non-orthogonal vectors v⃗1\vec{v}_1 and v⃗2\vec{v}_2 for the projection. Choice C is not part of the Gram-Schmidt algorithm.

Question 4

Consider the set of already-orthogonal vectors in R3\mathbb{R}^3 given by S={v⃗1,v⃗2}S = \{\vec{v}_1, \vec{v}_2\}, where v⃗1=(4,0,0)\vec{v}_1 = (4, 0, 0) and v⃗2=(0,−2,0)\vec{v}_2 = (0, -2, 0). If the Gram-Schmidt orthonormalization process is applied to SS, what is the resulting vector u⃗2\vec{u}_2?

  1. (0,−2,0)(0, -2, 0)
  2. (0,1,0)(0, 1, 0)
  3. (0,−1,0)(0, -1, 0) (correct answer)
  4. The zero vector, since the projection is zero.
Explanation: The Gram-Schmidt process first creates an orthogonal set, then normalizes each vector. Since the initial vectors are already orthogonal, the projection of v⃗2\vec{v}_2 onto v⃗1\vec{v}_1 is zero. So, the orthogonal vector w⃗2\vec{w}_2 is just v⃗2\vec{v}_2 itself. The final step is to normalize w⃗2=(0,−2,0)\vec{w}_2 = (0, -2, 0). The norm is ∥w⃗2∥=02+(−2)2+02=2\|\vec{w}_2\| = \sqrt{0^2 + (-2)^2 + 0^2} = 2. Therefore, u⃗2=w⃗2∥w⃗2∥=(0,−2,0)2=(0,−1,0)\vec{u}_2 = \frac{\vec{w}_2}{\|\vec{w}_2\|} = \frac{(0, -2, 0)}{2} = (0, -1, 0).

Question 5

The Gram-Schmidt process can be used to find the QR factorization of a matrix AA, where A=QRA=QR, QQ is orthogonal, and RR is upper-triangular. If A=[v⃗1∣v⃗2∣v⃗3]A = [\vec{v}_1 | \vec{v}_2 | \vec{v}_3] and the columns of QQ are the orthonormal vectors u⃗1,u⃗2,u⃗3\vec{u}_1, \vec{u}_2, \vec{u}_3, then the columns of AA can be written as a linear combination of the columns of QQ. Specifically, v⃗2=r12u⃗1+r22u⃗2\vec{v}_2 = r_{12}\vec{u}_1 + r_{22}\vec{u}_2. What is the value of the coefficient r12r_{12}?

  1. ⟨v⃗2,u⃗1⟩\langle \vec{v}_2, \vec{u}_1 \rangle (correct answer)
  2. ⟨v⃗1,u⃗2⟩\langle \vec{v}_1, \vec{u}_2 \rangle
  3. ⟨u⃗2,u⃗1⟩\langle \vec{u}_2, \vec{u}_1 \rangle
  4. ∥v⃗2∥\|\vec{v}_2\|
Explanation: From the equation v⃗2=r12u⃗1+r22u⃗2\vec{v}_2 = r_{12}\vec{u}_1 + r_{22}\vec{u}_2, we can find r12r_{12} by taking the inner product of both sides with u⃗1\vec{u}_1. This gives ⟨v⃗2,u⃗1⟩=⟨r12u⃗1+r22u⃗2,u⃗1⟩\langle \vec{v}_2, \vec{u}_1 \rangle = \langle r_{12}\vec{u}_1 + r_{22}\vec{u}_2, \vec{u}_1 \rangle. By linearity, this is r12⟨u⃗1,u⃗1⟩+r22⟨u⃗2,u⃗1⟩r_{12}\langle \vec{u}_1, \vec{u}_1 \rangle + r_{22}\langle \vec{u}_2, \vec{u}_1 \rangle. Since the u⃗\vec{u} vectors are orthonormal, ⟨u⃗1,u⃗1⟩=1\langle \vec{u}_1, \vec{u}_1 \rangle = 1 and ⟨u⃗2,u⃗1⟩=0\langle \vec{u}_2, \vec{u}_1 \rangle = 0. The equation simplifies to ⟨v⃗2,u⃗1⟩=r12(1)+r22(0)\langle \vec{v}_2, \vec{u}_1 \rangle = r_{12}(1) + r_{22}(0), so r12=⟨v⃗2,u⃗1⟩r_{12} = \langle \vec{v}_2, \vec{u}_1 \rangle.

Question 6

When applying the Gram-Schmidt process to a set of vectors {v⃗1,v⃗2,…,v⃗k}\{\vec{v}_1, \vec{v}_2, \dots, \vec{v}_k\}, which condition on the original set of vectors is necessary for the process to yield a set of kk non-zero orthogonal vectors?

  1. The set of vectors must be orthonormal.
  2. The set of vectors must be linearly independent. (correct answer)
  3. The set of vectors must be orthogonal.
  4. The set of vectors must span the entire vector space.
Explanation: If the set of vectors is linearly dependent, then at some step jj, the vector v⃗j\vec{v}_j will be a linear combination of the preceding vectors {v⃗1,…,v⃗j−1}\{\vec{v}_1, \dots, \vec{v}_{j-1}\}. When this happens, the process will produce a zero vector at the jj-th step, and we will not get kk non-zero vectors. Therefore, linear independence is the necessary condition. The vectors do not need to be orthogonal or orthonormal to begin with (that is the purpose of the process). They also do not need to span the entire space, only a kk-dimensional subspace.

Question 7

Consider the polynomial space P2\mathcal{P}_2 with inner product ⟨p,q⟩=∫01p(x)q(x)dx\langle p, q \rangle = \int_0^1 p(x)q(x) dx. If the Gram-Schmidt process is applied to the basis {1,x,x2}\{1, x, x^2\} in that order, which of the following represents the second orthogonal polynomial (before normalization)?

  1. x−12x - \frac{1}{2} (correct answer)
  2. x−13x - \frac{1}{3}
  3. x−23x - \frac{2}{3}
  4. x−1x - 1
Explanation: The Gram-Schmidt process starts with u1=1∣∣1∣∣=1∫0112dx=11=1\mathbf{u}_1 = \frac{1}{||1||} = \frac{1}{\sqrt{\int_0^1 1^2 dx}} = \frac{1}{\sqrt{1}} = 1. For the second step, we compute w2=x−⟨x,1⟩⋅1\mathbf{w}_2 = x - \langle x, 1 \rangle \cdot 1. We have ⟨x,1⟩=∫01x⋅1dx=∫01xdx=x22∣01=12\langle x, 1 \rangle = \int_0^1 x \cdot 1 dx = \int_0^1 x dx = \frac{x^2}{2}\Big|_0^1 = \frac{1}{2}. Therefore, w2=x−12⋅1=x−12\mathbf{w}_2 = x - \frac{1}{2} \cdot 1 = x - \frac{1}{2}. Choice B gives the projection coefficient for x2x^2 onto 11, choice C is 1−13=231 - \frac{1}{3} = \frac{2}{3} which is incorrect, and choice D would result from incorrectly computing the integral.

Question 8

The Gram-Schmidt process is used to convert a linearly independent set of vectors {v⃗1,v⃗2,...,v⃗k}\{\vec{v}_1, \vec{v}_2, ..., \vec{v}_k\} into an orthonormal set {u⃗1,u⃗2,...,u⃗k}\{\vec{u}_1, \vec{u}_2, ..., \vec{u}_k\}. Which of the following properties is guaranteed to be true for any integer jj such that 1≤j≤k1 \le j \le k?

  1. v⃗j\vec{v}_j is a scalar multiple of u⃗j\vec{u}_j.
  2. The span of {v⃗1,...,v⃗j}\{\vec{v}_1, ..., \vec{v}_j\} is equal to the span of {u⃗1,...,u⃗j}\{\vec{u}_1, ..., \vec{u}_j\}. (correct answer)
  3. The inner product ⟨v⃗j,u⃗j⟩=1\langle \vec{v}_j, \vec{u}_j \rangle = 1.
  4. The span of the single vector {v⃗j}\{\vec{v}_j\} is equal to the span of the single vector {u⃗j}\{\vec{u}_j\}.
Explanation: A fundamental property of the Gram-Schmidt process is that for any jj, the subspace spanned by the first jj original vectors is the same as the subspace spanned by the first jj orthonormal vectors. The vector u⃗j\vec{u}_j is a linear combination of v⃗1,...,v⃗j\vec{v}_1, ..., \vec{v}_j, so their individual spans are not generally equal (except for j=1j=1), making D incorrect. A and C are also not generally true.

Question 9

Let R3\mathbb{R}^3 be an inner product space with the standard dot product. Given two linearly independent vectors v⃗1\vec{v}_1 and v⃗2\vec{v}_2, the Gram-Schmidt process begins by setting u⃗1=v⃗1∥v⃗1∥\vec{u}_1 = \frac{\vec{v}_1}{\|\vec{v}_1\|}. The second orthogonal vector, w⃗2\vec{w}_2, is then computed as w⃗2=v⃗2−⟨v⃗2,u⃗1⟩u⃗1\vec{w}_2 = \vec{v}_2 - \langle \vec{v}_2, \vec{u}_1 \rangle \vec{u}_1. Which statement provides the best geometric interpretation of w⃗2\vec{w}_2?

  1. w⃗2\vec{w}_2 is the projection of v⃗2\vec{v}_2 onto the subspace spanned by v⃗1\vec{v}_1.
  2. w⃗2\vec{w}_2 is the component of v⃗2\vec{v}_2 that is orthogonal to the subspace spanned by v⃗1\vec{v}_1. (correct answer)
  3. w⃗2\vec{w}_2 is a unit vector in the same direction as v⃗2\vec{v}_2.
  4. w⃗2\vec{w}_2 is a vector that is linearly dependent on u⃗1\vec{u}_1.
Explanation: The term ⟨v⃗2,u⃗1⟩u⃗1\langle \vec{v}_2, \vec{u}_1 \rangle \vec{u}_1 represents the projection of v⃗2\vec{v}_2 onto the direction of u⃗1\vec{u}_1 (which is the same as the direction of v⃗1\vec{v}_1). By subtracting this projection from the original vector v⃗2\vec{v}_2, we are left with the component of v⃗2\vec{v}_2 that is orthogonal (perpendicular) to u⃗1\vec{u}_1. This is the core principle of the orthogonalization step.

Question 10

A student makes a mistake while applying the Gram-Schmidt process to vectors v⃗1=(1,1)\vec{v}_1 = (1, 1) and v⃗2=(2,0)\vec{v}_2 = (2, 0). Their work is shown below: Step 1: w⃗1=v⃗1=(1,1)\vec{w}_1 = \vec{v}_1 = (1, 1) Step 2: w⃗2=v⃗2−⟨v⃗2,w⃗1⟩∥w⃗1∥w⃗1=(2,0)−22(1,1)=(2−2,−2)\vec{w}_2 = \vec{v}_2 - \frac{\langle \vec{v}_2, \vec{w}_1 \rangle}{\|\vec{w}_1\|} \vec{w}_1 = (2, 0) - \frac{2}{\sqrt{2}}(1, 1) = (2-\sqrt{2}, -\sqrt{2}) Step 3: Normalize w⃗1\vec{w}_1 and w⃗2\vec{w}_2 to get u⃗1\vec{u}_1 and u⃗2\vec{u}_2. What is the conceptual error in the student's work?

  1. In Step 1, the student should have normalized v⃗1\vec{v}_1 immediately.
  2. In Step 2, the dot product ⟨v⃗2,w⃗1⟩\langle \vec{v}_2, \vec{w}_1 \rangle was calculated incorrectly.
  3. In Step 2, the denominator of the projection coefficient should be ∥w⃗1∥2\|\vec{w}_1\|^2, not ∥w⃗1∥\|\vec{w}_1\|. (correct answer)
  4. In Step 2, the vector being subtracted from should be v⃗1\vec{v}_1, not v⃗2\vec{v}_2.
Explanation: The formula for the projection of a vector a⃗\vec{a} onto a vector b⃗\vec{b} is ⟨a⃗,b⃗⟩∥b⃗∥2b⃗\frac{\langle \vec{a}, \vec{b} \rangle}{\|\vec{b}\|^2}\vec{b}. The student incorrectly used ∥w⃗1∥\|\vec{w}_1\| in the denominator instead of its square, ∥w⃗1∥2\|\vec{w}_1\|^2. The dot product ⟨(2,0),(1,1)⟩=2\langle(2,0), (1,1)\rangle = 2 is correct, and the norm ∥(1,1)∥=2\|(1,1)\| = \sqrt{2} is correct, but the formula itself was misremembered.

Question 11

Consider the vector space P1P_1 of polynomials of degree at most 1, with the inner product defined as ⟨p,q⟩=∫01p(x)q(x)dx\langle p, q \rangle = \int_0^1 p(x)q(x) dx. If the Gram-Schmidt process is applied to the standard basis {1,x}\{1, x\}, what is the resulting second vector, u2(x)u_2(x), of the orthonormal basis?

  1. x−12x - \frac{1}{2}
  2. xx
  3. 3(2x−1)\sqrt{3}(2x - 1) (correct answer)
  4. 2x−12x - 1
Explanation: Let v1=1,v2=xv_1=1, v_2=x. First, normalize v1v_1: ∥v1∥2=∫0112dx=1\|v_1\|^2 = \int_0^1 1^2 dx = 1, so u1=1u_1 = 1. Next, find w2=v2−⟨v2,u1⟩u1w_2 = v_2 - \langle v_2, u_1 \rangle u_1. The inner product is ⟨x,1⟩=∫01xdx=12\langle x, 1 \rangle = \int_0^1 x dx = \frac{1}{2}. So, w2(x)=x−12w_2(x) = x - \frac{1}{2}. Finally, normalize w2w_2: ∥w2∥2=∫01(x−12)2dx=[13(x−12)3]01=13((12)3−(−12)3)=13(18+18)=112\|w_2\|^2 = \int_0^1 (x - \frac{1}{2})^2 dx = [\frac{1}{3}(x-\frac{1}{2})^3]_0^1 = \frac{1}{3}((\frac{1}{2})^3 - (-\frac{1}{2})^3) = \frac{1}{3}(\frac{1}{8} + \frac{1}{8}) = \frac{1}{12}. Thus, ∥w2∥=112=123\|w_2\| = \frac{1}{\sqrt{12}} = \frac{1}{2\sqrt{3}}. Then u2=w2∥w2∥=x−1/21/(23)=23(x−1/2)=3(2x−1)u_2 = \frac{w_2}{\|w_2\|} = \frac{x - 1/2}{1/(2\sqrt{3})} = 2\sqrt{3}(x - 1/2) = \sqrt{3}(2x-1).

Question 12

Let S={v⃗1,v⃗2,v⃗3}S = \{\vec{v}_1, \vec{v}_2, \vec{v}_3\} be a set of vectors in R4\mathbb{R}^4 where v⃗3=2v⃗1−5v⃗2\vec{v}_3 = 2\vec{v}_1 - 5\vec{v}_2. The vectors v⃗1\vec{v}_1 and v⃗2\vec{v}_2 are linearly independent. If the Gram-Schmidt process is applied to SS in the order v⃗1,v⃗2,v⃗3\vec{v}_1, \vec{v}_2, \vec{v}_3 to produce an orthogonal set {w⃗1,w⃗2,w⃗3}\{\vec{w}_1, \vec{w}_2, \vec{w}_3\}, what will be the result for w⃗3\vec{w}_3?

  1. w⃗3\vec{w}_3 will be a non-zero vector orthogonal to w⃗1\vec{w}_1 and w⃗2\vec{w}_2.
  2. The process will fail due to a division by zero when calculating the projection.
  3. w⃗3\vec{w}_3 will be the zero vector. (correct answer)
  4. w⃗3\vec{w}_3 will be equal to 2w⃗1−5w⃗22\vec{w}_1 - 5\vec{w}_2.
Explanation: The Gram-Schmidt process produces a third orthogonal vector w⃗3\vec{w}_3 by taking v⃗3\vec{v}_3 and subtracting its projections onto the subspace spanned by the previous vectors, span{v⃗1,v⃗2}\text{span}\{\vec{v}_1, \vec{v}_2\}. Since v⃗3\vec{v}_3 is already a linear combination of v⃗1\vec{v}_1 and v⃗2\vec{v}_2, it lies entirely within that subspace. Therefore, subtracting its projection will leave the zero vector. Formally, w⃗3=v⃗3−projspan{w⃗1,w⃗2}(v⃗3)=v⃗3−v⃗3=0⃗\vec{w}_3 = \vec{v}_3 - \text{proj}_{\text{span}\{\vec{w}_1, \vec{w}_2\}}(\vec{v}_3) = \vec{v}_3 - \vec{v}_3 = \vec{0}.

Question 13

In applying the Gram-Schmidt process to vectors v1,v2,v3\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3 in that order, suppose that after the first two steps we have orthonormal vectors u1,u2\mathbf{u}_1, \mathbf{u}_2. If ⟨v3,u1⟩=−2\langle \mathbf{v}_3, \mathbf{u}_1 \rangle = -2 and ⟨v3,u2⟩=3\langle \mathbf{v}_3, \mathbf{u}_2 \rangle = 3, and the resulting u3\mathbf{u}_3 has norm 1, what is ∣∣v3∣∣2||\mathbf{v}_3||^2?

  1. 1313
  2. 1414 (correct answer)
  3. 1212
  4. 14\sqrt{14}
Explanation: In the Gram-Schmidt process, w3=v3−⟨v3,u1⟩u1−⟨v3,u2⟩u2=v3−(−2)u1−3u2=v3+2u1−3u2\mathbf{w}_3 = \mathbf{v}_3 - \langle \mathbf{v}_3, \mathbf{u}_1 \rangle \mathbf{u}_1 - \langle \mathbf{v}_3, \mathbf{u}_2 \rangle \mathbf{u}_2 = \mathbf{v}_3 - (-2)\mathbf{u}_1 - 3\mathbf{u}_2 = \mathbf{v}_3 + 2\mathbf{u}_1 - 3\mathbf{u}_2. Since u3=w3∣∣w3∣∣\mathbf{u}_3 = \frac{\mathbf{w}_3}{||\mathbf{w}_3||} and ∣∣u3∣∣=1||\mathbf{u}_3|| = 1, we have ∣∣w3∣∣=1||\mathbf{w}_3|| = 1. By the Pythagorean theorem (since the projection and orthogonal component are perpendicular), ∣∣v3∣∣2=∣∣projspan{u1,u2}v3∣∣2+∣∣w3∣∣2||\mathbf{v}_3||^2 = ||\text{proj}_{\text{span}\{\mathbf{u}_1,\mathbf{u}_2\}}\mathbf{v}_3||^2 + ||\mathbf{w}_3||^2. The projection is −2u1+3u2-2\mathbf{u}_1 + 3\mathbf{u}_2, which has squared norm (−2)2+32=4+9=13(-2)^2 + 3^2 = 4 + 9 = 13 (since u1,u2\mathbf{u}_1, \mathbf{u}_2 are orthonormal). Therefore, ∣∣v3∣∣2=13+12=14||\mathbf{v}_3||^2 = 13 + 1^2 = 14.

Question 14

Suppose vectors v1,v2,v3\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3 in an inner product space undergo Gram-Schmidt orthonormalization to produce u1,u2,u3\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3. If the intermediate unnormalized orthogonal vectors are w1=v1\mathbf{w}_1 = \mathbf{v}_1, w2\mathbf{w}_2, and w3\mathbf{w}_3, and we know that ∣∣w1∣∣=2||\mathbf{w}_1|| = 2, ∣∣w2∣∣=3||\mathbf{w}_2|| = 3, and ∣∣w3∣∣=1||\mathbf{w}_3|| = 1, what is the volume of the parallelepiped formed by the original vectors v1,v2,v3\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3?

  1. 22
  2. 1212
  3. 33
  4. 66 (correct answer)
Explanation: When you encounter Gram-Schmidt problems involving volumes, recognize that there's a fundamental relationship between orthogonalization and geometric properties. The key insight is that the Gram-Schmidt process preserves the volume of the parallelepiped formed by the original vectors. The volume of a parallelepiped formed by vectors equals the product of the norms of the orthogonal vectors produced by Gram-Schmidt. This happens because orthogonalization essentially "untangles" the vectors without changing the overall volume they enclose. Since ∣∣w1∣∣=2||\mathbf{w}_1|| = 2, ∣∣w2∣∣=3||\mathbf{w}_2|| = 3, and ∣∣w3∣∣=1||\mathbf{w}_3|| = 1, the volume is 2×3×1=62 \times 3 \times 1 = 6. Looking at the wrong answers: Choice (A) gives 22, which would be correct if you mistakenly thought only the first vector's norm mattered. Choice (B) gives 1212, which you might get if you incorrectly calculated 2×3×22 \times 3 \times 2 instead of 2×3×12 \times 3 \times 1. Choice (C) gives 33, which could result from taking just the largest norm or averaging incorrectly. The correct answer is (D) 66. Study tip: Remember that volumes and determinants are preserved under orthogonalization processes. When you see Gram-Schmidt with volume questions, immediately think "multiply the norms of the orthogonal vectors." This connection between orthogonalization and volume preservation appears frequently in linear algebra applications.

Question 15

In the vector space of 2×22 \times 2 matrices with the inner product ⟨A,B⟩=tr(ATB)\langle A, B \rangle = \text{tr}(A^T B), consider the matrices M1=(1001)M_1 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} , M2=(1100)M_2 = \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix} , and M3=(0011)M_3 = \begin{pmatrix} 0 & 0 \\ 1 & 1 \end{pmatrix} . If the Gram-Schmidt process is applied to orthonormalize these matrices in the given order, what is tr(U2)\text{tr}(U_2) where U2U_2 is the second orthonormal matrix?

  1. 12\frac{1}{2}
  2. 12\frac{1}{\sqrt{2}}
  3. 00 (correct answer)
  4. 11
Explanation: When you encounter Gram-Schmidt orthonormalization in matrix spaces, remember that you're applying the same process as with vectors, but using the given inner product for matrices. The Gram-Schmidt process starts with the first matrix. Since M1=(1001)M_1 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} is already normalized under the inner product ⟨A,B⟩=tr(ATB)\langle A, B \rangle = \text{tr}(A^T B) (you can verify ⟨M1,M1⟩=tr(M1TM1)=tr(I)=2\langle M_1, M_1 \rangle = \text{tr}(M_1^T M_1) = \text{tr}(I) = 2), we have U1=12M1U_1 = \frac{1}{\sqrt{2}}M_1. For the second orthonormal matrix U2U_2, we need to orthogonalize M2M_2 against U1U_1. First, calculate the projection: projU1(M2)=⟨M2,U1⟩⟨U1,U1⟩U1\text{proj}_{U_1}(M_2) = \frac{\langle M_2, U_1 \rangle}{\langle U_1, U_1 \rangle} U_1. Computing ⟨M2,U1⟩=tr(M2TU1)=tr(1010)⋅12(1001)=12\langle M_2, U_1 \rangle = \text{tr}(M_2^T U_1) = \text{tr}\begin{pmatrix} 1 & 0 \\ 1 & 0 \end{pmatrix} \cdot \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \frac{1}{\sqrt{2}}. The orthogonal component is M2−projU1(M2)=(1100)−12(1001)=(1/210−1/2)M_2 - \text{proj}_{U_1}(M_2) = \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix} - \frac{1}{2}\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1/2 & 1 \\ 0 & -1/2 \end{pmatrix} . After normalizing, U2U_2 has trace 1/2−1/2norm=0\frac{1/2 - 1/2}{\text{norm}} = 0. Choice A (12\frac{1}{2}) ignores the normalization step. Choice B (12\frac{1}{\sqrt{2}}) incorrectly assumes the trace is preserved from M1M_1. Choice D (11) forgets the orthogonalization against U1U_1. Study tip: In Gram-Schmidt problems, always check your orthogonal component before normalizing—the trace often reveals structural properties of your result.

Question 16

Let v1=(1100)\mathbf{v}_1 = \begin{pmatrix} 1 \\ 1 \\ 0 \\ 0 \end{pmatrix} , v2=(1010)\mathbf{v}_2 = \begin{pmatrix} 1 \\ 0 \\ 1 \\ 0 \end{pmatrix} , and v3=(1111)\mathbf{v}_3 = \begin{pmatrix} 1 \\ 1 \\ 1 \\ 1 \end{pmatrix} . After applying Gram-Schmidt orthonormalization to these vectors in the given order, what is the coefficient of u1\mathbf{u}_1 in the expression v3=au1+bu2+cu3\mathbf{v}_3 = a\mathbf{u}_1 + b\mathbf{u}_2 + c\mathbf{u}_3 where {u1,u2,u3}\{\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3\} is the resulting orthonormal set?

  1. 22\frac{2}{\sqrt{2}}
  2. 22\frac{\sqrt{2}}{2}
  3. 22
  4. 2\sqrt{2} (correct answer)
Explanation: When you encounter Gram-Schmidt problems, remember that this process creates an orthonormal basis where each new vector is built by removing projections onto previously constructed vectors. The key insight is that the coefficients in the final expansion directly relate to the dot products computed during the process. Let's work through the Gram-Schmidt procedure step by step. First, u1=v1∣∣v1∣∣=12(1100)\mathbf{u}_1 = \frac{\mathbf{v}_1}{||\mathbf{v}_1||} = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ 1 \\ 0 \\ 0 \end{pmatrix} since ∣∣v1∣∣=12+12=2||\mathbf{v}_1|| = \sqrt{1^2 + 1^2} = \sqrt{2}. For u2\mathbf{u}_2, we remove the projection of v2\mathbf{v}_2 onto u1\mathbf{u}_1: v2−(v2⋅u1)u1\mathbf{v}_2 - (\mathbf{v}_2 \cdot \mathbf{u}_1)\mathbf{u}_1. Since v2⋅u1=12\mathbf{v}_2 \cdot \mathbf{u}_1 = \frac{1}{\sqrt{2}}, this gives us $$\mathbf{u}_2 = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \ -1 \ 2 \ 0 \end{pmatrix} The crucial step: when we express $$\mathbf{v}_3$$ in terms of the orthonormal basis, the coefficient of $$\mathbf{u}_1$$ is simply $$\mathbf{v}_3 \cdot \mathbf{u}_1$$. Computing this: $$\mathbf{v}_3 \cdot \mathbf{u}_1 = \begin{pmatrix} 1 \\ 1 \\ 1 \\ 1 \end{pmatrix} \cdot \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ 1 \\ 0 \\ 0 \end{pmatrix} = \frac{2}{\sqrt{2}} = \sqrt{2}$$. Choice A gives $$\frac{2}{\sqrt{2}}$$ without simplifying the radical. Choice B represents $$\frac{\sqrt{2}}{2}$$, which would result from incorrectly taking the reciprocal. Choice C gives 2, likely from forgetting the normalization factor entirely. The correct answer is D: $$\sqrt{2}$$. **Study tip**: In orthonormal expansions, coefficients are always dot products with the corresponding basis vector. Master this connection to streamline Gram-Schmidt calculations.

Question 17

Consider applying the Gram-Schmidt process to vectors a,b,c\mathbf{a}, \mathbf{b}, \mathbf{c} to obtain orthonormal vectors u1,u2,u3\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3. If it's known that b−projab\mathbf{b} - \text{proj}_{\mathbf{a}} \mathbf{b} has the same magnitude as a\mathbf{a}, and a⋅b=34∣∣a∣∣2\mathbf{a} \cdot \mathbf{b} = \frac{3}{4}||\mathbf{a}||^2, what is the ratio ∣∣b∣∣∣∣a∣∣\frac{||\mathbf{b}||}{||\mathbf{a}||}?

  1. 53\frac{5}{3}
  2. 74\frac{\sqrt{7}}{4}
  3. 54\frac{5}{4} (correct answer)
  4. 254\frac{\sqrt{25}}{4}
Explanation: When you encounter Gram-Schmidt problems involving projections and magnitudes, focus on the geometric relationships between vectors. The key insight is that b−projab\mathbf{b} - \text{proj}_{\mathbf{a}} \mathbf{b} represents the component of b\mathbf{b} orthogonal to a\mathbf{a}. Let's work with the given information systematically. We know that projab=a⋅b∣∣a∣∣2a=3∣∣a∣∣2/4∣∣a∣∣2a=34a\text{proj}_{\mathbf{a}} \mathbf{b} = \frac{\mathbf{a} \cdot \mathbf{b}}{||\mathbf{a}||^2}\mathbf{a} = \frac{3||\mathbf{a}||^2/4}{||\mathbf{a}||^2}\mathbf{a} = \frac{3}{4}\mathbf{a}. The orthogonal component is b−projab=b−34a\mathbf{b} - \text{proj}_{\mathbf{a}} \mathbf{b} = \mathbf{b} - \frac{3}{4}\mathbf{a}, and we're told this has magnitude ∣∣a∣∣||\mathbf{a}||. Since the projection and orthogonal components are perpendicular, we can use the Pythagorean theorem: ∣∣b∣∣2=∣∣projab∣∣2+∣∣b−projab∣∣2||\mathbf{b}||^2 = ||\text{proj}_{\mathbf{a}} \mathbf{b}||^2 + ||\mathbf{b} - \text{proj}_{\mathbf{a}} \mathbf{b}||^2. Substituting: ∣∣b∣∣2=∣∣34a∣∣2+∣∣a∣∣2=916∣∣a∣∣2+∣∣a∣∣2=2516∣∣a∣∣2||\mathbf{b}||^2 = ||\frac{3}{4}\mathbf{a}||^2 + ||\mathbf{a}||^2 = \frac{9}{16}||\mathbf{a}||^2 + ||\mathbf{a}||^2 = \frac{25}{16}||\mathbf{a}||^2. Therefore, ∣∣b∣∣∣∣a∣∣=54\frac{||\mathbf{b}||}{||\mathbf{a}||} = \frac{5}{4}, which is answer C. Looking at the wrong answers: A) 53\frac{5}{3} results from incorrectly adding the magnitudes instead of their squares. B) 74\frac{\sqrt{7}}{4} comes from computational errors in the Pythagorean theorem. D) 254\frac{\sqrt{25}}{4} fails to simplify 25=5\sqrt{25} = 5. Remember: when vectors are orthogonal, always use the Pythagorean theorem with squared magnitudes, not the magnitudes themselves.

Question 18

Let v⃗1=(1,0,1)\vec{v}_1 = (1, 0, 1) and v⃗2=(0,1,1)\vec{v}_2 = (0, 1, 1). The Gram-Schmidt process is applied to obtain an orthogonal set {w⃗1,w⃗2}\{\vec{w}_1, \vec{w}_2\}. What is the vector w⃗2\vec{w}_2?

  1. (0,1,1)(0, 1, 1)
  2. (12,1,12)(\frac{1}{2}, 1, \frac{1}{2})
  3. (−1,2,1)(-1, 2, 1)
  4. (−12,1,12)(-\frac{1}{2}, 1, \frac{1}{2}) (correct answer)
Explanation: First, set w⃗1=v⃗1=(1,0,1)\vec{w}_1 = \vec{v}_1 = (1, 0, 1). The second orthogonal vector is found using the formula w⃗2=v⃗2−projw⃗1(v⃗2)=v⃗2−⟨v⃗2,w⃗1⟩∥w⃗1∥2w⃗1\vec{w}_2 = \vec{v}_2 - \text{proj}_{\vec{w}_1}(\vec{v}_2) = \vec{v}_2 - \frac{\langle \vec{v}_2, \vec{w}_1 \rangle}{\|\vec{w}_1\|^2}\vec{w}_1. We calculate the components: ⟨v⃗2,w⃗1⟩=(0)(1)+(1)(0)+(1)(1)=1\langle \vec{v}_2, \vec{w}_1 \rangle = (0)(1) + (1)(0) + (1)(1) = 1. ∥w⃗1∥2=12+02+12=2\|\vec{w}_1\|^2 = 1^2 + 0^2 + 1^2 = 2. Plugging these in: w⃗2=(0,1,1)−12(1,0,1)=(0,1,1)−(12,0,12)=(−12,1,12)\vec{w}_2 = (0, 1, 1) - \frac{1}{2}(1, 0, 1) = (0, 1, 1) - (\frac{1}{2}, 0, \frac{1}{2}) = (-\frac{1}{2}, 1, \frac{1}{2}). The question asks for the orthogonal vector w⃗2\vec{w}_2, not the normalized vector u⃗2\vec{u}_2.