Linear Algebra Quiz: Coordinate Vectors
15 questions · exam conditions
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Coordinate VectorsQuestion 1 of 15

Let B={[12],[34]}\mathcal{B} = \left\{ \begin{bmatrix} 1 \\ 2 \end{bmatrix}, \begin{bmatrix} 3 \\ 4 \end{bmatrix} \right\} be a basis for R2\mathbb{R}^2. What is the coordinate vector [v]B[\mathbf{v}]_\mathcal{B} of the vector v=[15]\mathbf{v} = \begin{bmatrix} 1 \\ 5 \end{bmatrix} relative to B\mathcal{B}?

[11/2−3/2]\begin{bmatrix} 11/2 \\ -3/2 \end{bmatrix}
[−113]\begin{bmatrix} -11 \\ 3 \end{bmatrix}
[9/2−17/2]\begin{bmatrix} 9/2 \\ -17/2 \end{bmatrix}
[13−4]\begin{bmatrix} 13 \\ -4 \end{bmatrix}
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Linear Algebra Quiz

Linear Algebra Quiz: Coordinate Vectors

Practice Coordinate Vectors in Linear Algebra with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Coordinate Vectors, giving you a quick way to practice the rules, question types, and explanations that matter most for Linear Algebra.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

Let B={[12],[34]}\mathcal{B} = \left\{ \begin{bmatrix} 1 \\ 2 \end{bmatrix}, \begin{bmatrix} 3 \\ 4 \end{bmatrix} \right\} be a basis for R2\mathbb{R}^2. What is the coordinate vector [v]B[\mathbf{v}]_\mathcal{B} of the vector v=[15]\mathbf{v} = \begin{bmatrix} 1 \\ 5 \end{bmatrix} relative to B\mathcal{B}?

  1. [11/2−3/2]\begin{bmatrix} 11/2 \\ -3/2 \end{bmatrix} (correct answer)
  2. [−113]\begin{bmatrix} -11 \\ 3 \end{bmatrix}
  3. [9/2−17/2]\begin{bmatrix} 9/2 \\ -17/2 \end{bmatrix}
  4. [13−4]\begin{bmatrix} 13 \\ -4 \end{bmatrix}
Explanation: The coordinate vector [v]B=[c1c2][\mathbf{v}]_\mathcal{B} = \begin{bmatrix} c_1 \\ c_2 \end{bmatrix} must satisfy the equation c1[12]+c2[34]=[15]c_1 \begin{bmatrix} 1 \\ 2 \end{bmatrix} + c_2 \begin{bmatrix} 3 \\ 4 \end{bmatrix} = \begin{bmatrix} 1 \\ 5 \end{bmatrix}. This corresponds to the linear system c1+3c2=1c_1 + 3c_2 = 1 and 2c1+4c2=52c_1 + 4c_2 = 5. Solving this system, we can multiply the first equation by -2 to get −2c1−6c2=−2-2c_1 - 6c_2 = -2. Adding this to the second equation yields −2c2=3-2c_2 = 3, so c2=−3/2c_2 = -3/2. Substituting this back into the first equation gives c1+3(−3/2)=1c_1 + 3(-3/2) = 1, which simplifies to c1−9/2=1c_1 - 9/2 = 1, so c1=11/2c_1 = 11/2. Thus, the coordinate vector is [11/2−3/2]\begin{bmatrix} 11/2 \\ -3/2 \end{bmatrix}.

Question 2

Let B={b1,b2,b3,b4}\mathcal{B} = \{\mathbf{b}_1, \mathbf{b}_2, \mathbf{b}_3, \mathbf{b}_4\} be an ordered basis for a vector space VV. What is the coordinate vector of the vector v=b2\mathbf{v} = \mathbf{b}_2 relative to the basis B\mathcal{B}?

  1. [0100]\begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \end{bmatrix} (correct answer)
  2. [1111]\begin{bmatrix} 1 \\ 1 \\ 1 \\ 1 \end{bmatrix}
  3. The vector b2\mathbf{b}_2 itself.
  4. The zero vector, [0000]\begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}.
Explanation: The coordinate vector [v]B=[b2]B[\mathbf{v}]_\mathcal{B} = [\mathbf{b}_2]_\mathcal{B} is the set of coefficients (c1,c2,c3,c4)(c_1, c_2, c_3, c_4) such that b2=c1b1+c2b2+c3b3+c4b4\mathbf{b}_2 = c_1\mathbf{b}_1 + c_2\mathbf{b}_2 + c_3\mathbf{b}_3 + c_4\mathbf{b}_4. By the definition of a basis, the vectors b1,…,b4\mathbf{b}_1, \dots, \mathbf{b}_4 are linearly independent. Therefore, the only unique solution to this equation is c1=0c_1=0, c2=1c_2=1, c3=0c_3=0, and c4=0c_4=0. The resulting coordinate vector is the standard basis vector e2\mathbf{e}_2.

Question 3

Let B={[11],[1−1]}\mathcal{B} = \left\{ \begin{bmatrix} 1 \\ 1 \end{bmatrix}, \begin{bmatrix} 1 \\ -1 \end{bmatrix} \right\} be a basis for R2\mathbb{R}^2. Let u=[42]\mathbf{u} = \begin{bmatrix} 4 \\ 2 \end{bmatrix} and v=[02]\mathbf{v} = \begin{bmatrix} 0 \\ 2 \end{bmatrix}. What is the coordinate vector [u−v]B[\mathbf{u}-\mathbf{v}]_\mathcal{B}?

  1. [44]\begin{bmatrix} 4 \\ 4 \end{bmatrix}
  2. [40]\begin{bmatrix} 4 \\ 0 \end{bmatrix}
  3. [20]\begin{bmatrix} 2 \\ 0 \end{bmatrix}
  4. [22]\begin{bmatrix} 2 \\ 2 \end{bmatrix} (correct answer)
Explanation: There are two methods. Method 1: First compute w=u−v=[42]−[02]=[40]\mathbf{w} = \mathbf{u}-\mathbf{v} = \begin{bmatrix} 4 \\ 2 \end{bmatrix} - \begin{bmatrix} 0 \\ 2 \end{bmatrix} = \begin{bmatrix} 4 \\ 0 \end{bmatrix}. Then find its coordinates [w]B[\mathbf{w}]_\mathcal{B} by solving c1[11]+c2[1−1]=[40]c_1\begin{bmatrix} 1 \\ 1 \end{bmatrix} + c_2\begin{bmatrix} 1 \\ -1 \end{bmatrix} = \begin{bmatrix} 4 \\ 0 \end{bmatrix}. This gives the system c1+c2=4c_1+c_2=4 and c1−c2=0c_1-c_2=0. From the second equation, c1=c2c_1=c_2. Substituting into the first gives 2c1=42c_1=4, so c1=2c_1=2 and c2=2c_2=2. Thus, [u−v]B=[22][\mathbf{u}-\mathbf{v}]_\mathcal{B} = \begin{bmatrix} 2 \\ 2 \end{bmatrix}. Method 2: Use the linearity of coordinate mappings. [u−v]B=[u]B−[v]B[\mathbf{u}-\mathbf{v}]_\mathcal{B} = [\mathbf{u}]_\mathcal{B} - [\mathbf{v}]_\mathcal{B}. For u\mathbf{u}, we solve c1+c2=4,c1−c2=2c_1+c_2=4, c_1-c_2=2, which gives c1=3,c2=1c_1=3, c_2=1. So [u]B=[31][\mathbf{u}]_\mathcal{B} = \begin{bmatrix} 3 \\ 1 \end{bmatrix}. For v\mathbf{v}, we solve d1+d2=0,d1−d2=2d_1+d_2=0, d_1-d_2=2, which gives d1=1,d2=−1d_1=1, d_2=-1. So [v]B=[1−1][\mathbf{v}]_\mathcal{B} = \begin{bmatrix} 1 \\ -1 \end{bmatrix}. Then [u]B−[v]B=[31]−[1−1]=[22][\mathbf{u}]_\mathcal{B} - [\mathbf{v}]_\mathcal{B} = \begin{bmatrix} 3 \\ 1 \end{bmatrix} - \begin{bmatrix} 1 \\ -1 \end{bmatrix} = \begin{bmatrix} 2 \\ 2 \end{bmatrix}.

Question 4

Let B={b1,b2}\mathcal{B} = \{\mathbf{b}_1, \mathbf{b}_2\} be a basis for R2\mathbb{R}^2. The coordinate mapping T:R2→R2T: \mathbb{R}^2 \to \mathbb{R}^2 is defined by T(x)=[x]BT(\mathbf{x}) = [\mathbf{x}]_\mathcal{B}. Let PBP_\mathcal{B} be the matrix whose columns are b1\mathbf{b}_1 and b2\mathbf{b}_2. Which of the following statements about this mapping is FALSE?

  1. The mapping TT is a one-to-one linear transformation.
  2. T(cx+dy)=cT(x)+dT(y)T(c\mathbf{x} + d\mathbf{y}) = cT(\mathbf{x}) + dT(\mathbf{y}) for all scalars c,dc,d and vectors x,y\mathbf{x}, \mathbf{y} in R2\mathbb{R}^2.
  3. For any vector c\mathbf{c} in R2\mathbb{R}^2, there exists a unique vector x\mathbf{x} in R2\mathbb{R}^2 such that T(x)=cT(\mathbf{x}) = \mathbf{c}.
  4. For every vector x\mathbf{x} in R2\mathbb{R}^2, the equation T(x)=PBxT(\mathbf{x}) = P_\mathcal{B} \mathbf{x} holds. (correct answer)
Explanation: The coordinate mapping T(x)=[x]BT(\mathbf{x}) = [\mathbf{x}]_\mathcal{B} is a linear transformation that is both one-to-one and onto (an isomorphism). Therefore, statements A, B, and C are true. Statement A claims it's one-to-one and linear. Statement B is the definition of linearity. Statement C claims the mapping is onto and unique, which is also true. The relationship between a vector x\mathbf{x}, its coordinate vector [x]B[\mathbf{x}]_\mathcal{B}, and the change-of-coordinates matrix PBP_\mathcal{B} is given by the equation x=PB[x]B\mathbf{x} = P_\mathcal{B} [\mathbf{x}]_\mathcal{B}. To find the coordinate vector, we must solve for it: [x]B=PB−1x[\mathbf{x}]_\mathcal{B} = P_\mathcal{B}^{-1} \mathbf{x}. Therefore, T(x)=PB−1xT(\mathbf{x}) = P_\mathcal{B}^{-1} \mathbf{x}. Statement D incorrectly claims T(x)=PBxT(\mathbf{x}) = P_\mathcal{B} \mathbf{x}, so it is false.

Question 5

Let B={v1,v2}\mathcal{B} = \{\mathbf{v}_1, \mathbf{v}_2\} be a basis for R2\mathbb{R}^2, where v1=[13]\mathbf{v}_1 = \begin{bmatrix} 1 \\ 3 \end{bmatrix}. The vector v2\mathbf{v}_2 is unknown. If the vector x=[21]\mathbf{x} = \begin{bmatrix} 2 \\ 1 \end{bmatrix} has the coordinate vector [x]B=[4−1][\mathbf{x}]_\mathcal{B} = \begin{bmatrix} 4 \\ -1 \end{bmatrix}, what is the vector v2\mathbf{v}_2?

  1. [613]\begin{bmatrix} 6 \\ 13 \end{bmatrix}
  2. [211]\begin{bmatrix} 2 \\ 11 \end{bmatrix} (correct answer)
  3. [−2−11]\begin{bmatrix} -2 \\ -11 \end{bmatrix}
  4. [2−7]\begin{bmatrix} 2 \\ -7 \end{bmatrix}
Explanation: The definition of the coordinate vector [x]B[\mathbf{x}]_\mathcal{B} states that x=4v1−1v2\mathbf{x} = 4\mathbf{v}_1 - 1\mathbf{v}_2. We are given x\mathbf{x} and v1\mathbf{v}_1, and we need to solve for v2\mathbf{v}_2. Substitute the known vectors into the equation: [21]=4[13]−v2\begin{bmatrix} 2 \\ 1 \end{bmatrix} = 4\begin{bmatrix} 1 \\ 3 \end{bmatrix} - \mathbf{v}_2 [21]=[412]−v2\begin{bmatrix} 2 \\ 1 \end{bmatrix} = \begin{bmatrix} 4 \\ 12 \end{bmatrix} - \mathbf{v}_2 Now, rearrange the equation to isolate v2\mathbf{v}_2: v2=[412]−[21]\mathbf{v}_2 = \begin{bmatrix} 4 \\ 12 \end{bmatrix} - \begin{bmatrix} 2 \\ 1 \end{bmatrix} v2=[4−212−1]=[211]\mathbf{v}_2 = \begin{bmatrix} 4-2 \\ 12-1 \end{bmatrix} = \begin{bmatrix} 2 \\ 11 \end{bmatrix}.

Question 6

Let B={[110],[011],[101]}\mathcal{B} = \left\{ \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix}, \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} \right\} be a basis for R3\mathbb{R}^3. The coordinate vector of v=[354]\mathbf{v} = \begin{bmatrix} 3 \\ 5 \\ 4 \end{bmatrix} is [v]B=[c1c2c3][\mathbf{v}]_\mathcal{B} = \begin{bmatrix} c_1 \\ c_2 \\ c_3 \end{bmatrix}. What is the value of the component c2c_2?

  1. 1
  2. 2
  3. 3 (correct answer)
  4. 5
Explanation: We must solve the vector equation c1[110]+c2[011]+c3[101]=[354]c_1\begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} + c_2\begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix} + c_3\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 3 \\ 5 \\ 4 \end{bmatrix}. This yields the system of linear equations:
  1. c1+c3=3c_1 + c_3 = 3
  2. c1+c2=5c_1 + c_2 = 5
  3. c2+c3=4c_2 + c_3 = 4
From equation (1), c1=3−c3c_1 = 3 - c_3. From equation (3), c2=4−c3c_2 = 4 - c_3. Substitute both into equation (2): (3−c3)+(4−c3)=5(3 - c_3) + (4 - c_3) = 5. This simplifies to 7−2c3=57 - 2c_3 = 5, which gives 2c3=22c_3 = 2, so c3=1c_3 = 1. Now we can find c2c_2 using equation (3): c2+1=4c_2 + 1 = 4, which means c2=3c_2 = 3. (For completeness, c1=3−1=2c_1 = 3 - 1 = 2). The question asks for the value of c2c_2, which is 3.

Question 7

In the vector space P2P_2 of polynomials of degree at most 2, consider the basis B={1,1+t,1+t+t2}\mathcal{B} = \{1, 1+t, 1+t+t^2\}. What is the coordinate vector of the polynomial p(t)=3−t+2t2p(t) = 3 - t + 2t^2 relative to B\mathcal{B}?

  1. [3−12]\begin{bmatrix} 3 \\ -1 \\ 2 \end{bmatrix}
  2. [4−32]\begin{bmatrix} 4 \\ -3 \\ 2 \end{bmatrix} (correct answer)
  3. [2−34]\begin{bmatrix} 2 \\ -3 \\ 4 \end{bmatrix}
  4. [012]\begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix}
Explanation: We want to find scalars c1,c2,c3c_1, c_2, c_3 such that c1(1)+c2(1+t)+c3(1+t+t2)=3−t+2t2c_1(1) + c_2(1+t) + c_3(1+t+t^2) = 3 - t + 2t^2. Expanding the left side gives (c1+c2+c3)+(c2+c3)t+c3t2(c_1+c_2+c_3) + (c_2+c_3)t + c_3t^2. By equating coefficients of like powers of tt, we get a system of equations:
  1. t2:c3=2t^2: c_3 = 2
  2. t1:c2+c3=−1t^1: c_2 + c_3 = -1
  3. t0:c1+c2+c3=3t^0: c_1 + c_2 + c_3 = 3
From (1), c3=2c_3=2. Substituting into (2), c2+2=−1c_2 + 2 = -1, so c2=−3c_2 = -3. Substituting c2c_2 and c3c_3 into (3), c1+(−3)+2=3c_1 + (-3) + 2 = 3, so c1−1=3c_1 - 1 = 3, which gives c1=4c_1 = 4. The coordinate vector is [4−32]\begin{bmatrix} 4 \\ -3 \\ 2 \end{bmatrix}.

Question 8

In R2\mathbb{R}^2, consider the basis B={b1,b2}\mathcal{B} = \left\{ \mathbf{b}_1, \mathbf{b}_2 \right\} where b1=[21]\mathbf{b}_1 = \begin{bmatrix} 2 \\ 1 \end{bmatrix} and b2=[−11]\mathbf{b}_2 = \begin{bmatrix} -1 \\ 1 \end{bmatrix}. A vector v\mathbf{v} has coordinate vector [v]B=[3−2][\mathbf{v}]_\mathcal{B} = \begin{bmatrix} 3 \\ -2 \end{bmatrix}. Which of the following vectors represents v\mathbf{v} in the standard basis?

  1. [3−2]\begin{bmatrix} 3 \\ -2 \end{bmatrix}
  2. [−71]\begin{bmatrix} -7 \\ 1 \end{bmatrix}
  3. [81]\begin{bmatrix} 8 \\ 1 \end{bmatrix} (correct answer)
  4. [55]\begin{bmatrix} 5 \\ 5 \end{bmatrix}
Explanation: The vector v\mathbf{v} is found by taking the linear combination of the basis vectors with the coefficients from its coordinate vector. The relationship is v=3b1+(−2)b2\mathbf{v} = 3\mathbf{b}_1 + (-2)\mathbf{b}_2. Substituting the basis vectors: v=3[21]−2[−11]\mathbf{v} = 3\begin{bmatrix} 2 \\ 1 \end{bmatrix} - 2\begin{bmatrix} -1 \\ 1 \end{bmatrix} Performing the scalar multiplication and vector subtraction: v=[63]−[−22]=[6−(−2)3−2]=[81]\mathbf{v} = \begin{bmatrix} 6 \\ 3 \end{bmatrix} - \begin{bmatrix} -2 \\ 2 \end{bmatrix} = \begin{bmatrix} 6 - (-2) \\ 3 - 2 \end{bmatrix} = \begin{bmatrix} 8 \\ 1 \end{bmatrix} Thus, the vector v\mathbf{v} in the standard basis is [81]\begin{bmatrix} 8 \\ 1 \end{bmatrix}.

Question 9

Let the set S={[12],[−2−4]}\mathcal{S} = \left\{ \begin{bmatrix} 1 \\ 2 \end{bmatrix}, \begin{bmatrix} -2 \\ -4 \end{bmatrix} \right\} be used to represent vectors in R2\mathbb{R}^2. For the vector v=[36]\mathbf{v} = \begin{bmatrix} 3 \\ 6 \end{bmatrix}, what is its coordinate vector [v]S[\mathbf{v}]_\mathcal{S}?

  1. [30]\begin{bmatrix} 3 \\ 0 \end{bmatrix}
  2. The coordinate vector is not uniquely defined because S\mathcal{S} is not a basis. (correct answer)
  3. The coordinate vector does not exist because v\mathbf{v} is not in the span of S\mathcal{S}.
  4. [1−1]\begin{bmatrix} 1 \\ -1 \end{bmatrix}
Explanation: For S\mathcal{S} to be a basis, its vectors must be linearly independent. However, [−2−4]=−2[12]\begin{bmatrix} -2 \\ -4 \end{bmatrix} = -2 \begin{bmatrix} 1 \\ 2 \end{bmatrix}, so the vectors are linearly dependent. Thus, S\mathcal{S} is not a basis for R2\mathbb{R}^2. To find the coordinate vector, we solve c1[12]+c2[−2−4]=[36]c_1 \begin{bmatrix} 1 \\ 2 \end{bmatrix} + c_2 \begin{bmatrix} -2 \\ -4 \end{bmatrix} = \begin{bmatrix} 3 \\ 6 \end{bmatrix}. This simplifies to (c1−2c2)[12]=3[12](c_1 - 2c_2)\begin{bmatrix} 1 \\ 2 \end{bmatrix} = 3\begin{bmatrix} 1 \\ 2 \end{bmatrix}, which means c1−2c2=3c_1 - 2c_2 = 3. This is one equation with two variables, so there are infinitely many solutions for (c1,c2)(c_1, c_2). For example, (3,0)(3, 0) and (1,−1)(1, -1) are both valid solutions. Because the representation is not unique, the coordinate vector is not uniquely defined.

Question 10

In the vector space M2×2M_{2\times2} of 2×22 \times 2 matrices, consider the ordered basis C={[1000],[0001],[0100],[0010]}\mathcal{C} = \left\{ \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}, \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix} \right\}. What is the coordinate vector of the matrix A=[3−105]A = \begin{bmatrix} 3 & -1 \\ 0 & 5 \end{bmatrix} relative to C\mathcal{C}?

  1. [3−105]\begin{bmatrix} 3 \\ -1 \\ 0 \\ 5 \end{bmatrix}
  2. [30−15]\begin{bmatrix} 3 \\ 0 \\ -1 \\ 5 \end{bmatrix}
  3. [35−10]\begin{bmatrix} 3 \\ 5 \\ -1 \\ 0 \end{bmatrix} (correct answer)
  4. [350−1]\begin{bmatrix} 3 \\ 5 \\ 0 \\ -1 \end{bmatrix}
Explanation: Let the basis vectors be c1,c2,c3,c4\mathbf{c}_1, \mathbf{c}_2, \mathbf{c}_3, \mathbf{c}_4. We need to find scalars k1,k2,k3,k4k_1, k_2, k_3, k_4 such that A=k1c1+k2c2+k3c3+k4c4A = k_1\mathbf{c}_1 + k_2\mathbf{c}_2 + k_3\mathbf{c}_3 + k_4\mathbf{c}_4. [3−105]=k1[1000]+k2[0001]+k3[0100]+k4[0010]\begin{bmatrix} 3 & -1 \\ 0 & 5 \end{bmatrix} = k_1\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} + k_2\begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} + k_3\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} + k_4\begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix} Combining the matrices on the right side gives [k1k3k4k2]\begin{bmatrix} k_1 & k_3 \\ k_4 & k_2 \end{bmatrix}. By equating the corresponding entries with matrix A, we find:
  • Top-left (1,1) entry: k1=3k_1 = 3
  • Bottom-right (2,2) entry: k2=5k_2 = 5
  • Top-right (1,2) entry: k3=−1k_3 = -1
  • Bottom-left (2,1) entry: k4=0k_4 = 0
The coordinate vector is formed by these scalars in the order they correspond to the basis vectors in C\mathcal{C}. Thus, [A]C=[k1k2k3k4]=[35−10][A]_\mathcal{C} = \begin{bmatrix} k_1 \\ k_2 \\ k_3 \\ k_4 \end{bmatrix} = \begin{bmatrix} 3 \\ 5 \\ -1 \\ 0 \end{bmatrix}.

Question 11

Let B={[101],[011],[110]}\mathcal{B} = \left\{ \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix}, \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} \right\} be a basis for R3\mathbb{R}^3. If a vector v\mathbf{v} has the coordinate vector [v]B=[2−13][\mathbf{v}]_\mathcal{B} = \begin{bmatrix} 2 \\ -1 \\ 3 \end{bmatrix}, what is the vector v\mathbf{v} in the standard basis?

  1. [521]\begin{bmatrix} 5 \\ 2 \\ 1 \end{bmatrix} (correct answer)
  2. [2−13]\begin{bmatrix} 2 \\ -1 \\ 3 \end{bmatrix}
  3. [512]\begin{bmatrix} 5 \\ 1 \\ 2 \end{bmatrix}
  4. [543]\begin{bmatrix} 5 \\ 4 \\ 3 \end{bmatrix}
Explanation: The vector v\mathbf{v} is the linear combination of the basis vectors using the components of [v]B[\mathbf{v}]_\mathcal{B} as coefficients. Let B={b1,b2,b3}\mathcal{B} = \{ \mathbf{b}_1, \mathbf{b}_2, \mathbf{b}_3 \}. Then v=2b1−1b2+3b3\mathbf{v} = 2\mathbf{b}_1 - 1\mathbf{b}_2 + 3\mathbf{b}_3. v=2[101]−1[011]+3[110]=[202]−[011]+[330]=[2−0+30−1+32−1+0]=[521]\mathbf{v} = 2\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} - 1\begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix} + 3\begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 2 \\ 0 \\ 2 \end{bmatrix} - \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix} + \begin{bmatrix} 3 \\ 3 \\ 0 \end{bmatrix} = \begin{bmatrix} 2 - 0 + 3 \\ 0 - 1 + 3 \\ 2 - 1 + 0 \end{bmatrix} = \begin{bmatrix} 5 \\ 2 \\ 1 \end{bmatrix}.

Question 12

Let B={b1,b2}\mathcal{B} = \{\mathbf{b}_1, \mathbf{b}_2\} be a basis for a vector space VV. Consider a second basis C={b1+b2,b1−b2}\mathcal{C} = \{\mathbf{b}_1 + \mathbf{b}_2, \mathbf{b}_1 - \mathbf{b}_2\}. If a vector v\mathbf{v} has the coordinate vector [v]C=[24][\mathbf{v}]_\mathcal{C} = \begin{bmatrix} 2 \\ 4 \end{bmatrix}, what is its coordinate vector [v]B[\mathbf{v}]_\mathcal{B}?

  1. [6−2]\begin{bmatrix} 6 \\ -2 \end{bmatrix} (correct answer)
  2. [3−1]\begin{bmatrix} 3 \\ -1 \end{bmatrix}
  3. [66]\begin{bmatrix} 6 \\ 6 \end{bmatrix}
  4. [24]\begin{bmatrix} 2 \\ 4 \end{bmatrix}
Explanation: We are given [v]C=[24][\mathbf{v}]_\mathcal{C} = \begin{bmatrix} 2 \\ 4 \end{bmatrix}. By the definition of a coordinate vector, this means v=2(b1+b2)+4(b1−b2)\mathbf{v} = 2(\mathbf{b}_1 + \mathbf{b}_2) + 4(\mathbf{b}_1 - \mathbf{b}_2). We need to express v\mathbf{v} as a linear combination of the vectors in B\mathcal{B}. Distributing the scalars, we get v=(2b1+2b2)+(4b1−4b2)\mathbf{v} = (2\mathbf{b}_1 + 2\mathbf{b}_2) + (4\mathbf{b}_1 - 4\mathbf{b}_2). Combining like terms, we have v=(2+4)b1+(2−4)b2=6b1−2b2\mathbf{v} = (2+4)\mathbf{b}_1 + (2-4)\mathbf{b}_2 = 6\mathbf{b}_1 - 2\mathbf{b}_2. Therefore, the coordinate vector of v\mathbf{v} relative to the basis B\mathcal{B} is [v]B=[6−2][\mathbf{v}]_\mathcal{B} = \begin{bmatrix} 6 \\ -2 \end{bmatrix}.

Question 13

Let S={p(x)∈P3:p(1)=0}S = \{p(x) \in P_3 : p(1) = 0\} be a subspace of the vector space P3P_3 of polynomials of degree at most 3. Consider the basis B={x−1,x2−1,x3−1}B = \{x-1, x^2-1, x^3-1\} for SS. If q(x)=2x3−x2−x+1q(x) = 2x^3 - x^2 - x + 1, determine whether q(x)∈Sq(x) \in S and, if so, find [q(x)]B[q(x)]_B.

  1. q(x)∉Sq(x) \notin S (correct answer)
  2. [q(x)]B=(1,−1,2)[q(x)]_B = (1, -1, 2)
  3. [q(x)]B=(−1,1,2)[q(x)]_B = (-1, 1, 2)
  4. [q(x)]B=(2,−1,1)[q(x)]_B = (2, -1, 1)
Explanation: First, check if q(x)∈Sq(x) \in S by evaluating q(1)q(1): q(1)=2(1)3−(1)2−(1)+1=2−1−1+1=1≠0q(1) = 2(1)^3 - (1)^2 - (1) + 1 = 2 - 1 - 1 + 1 = 1 \neq 0. Since q(1)≠0q(1) \neq 0, we have q(x)∉Sq(x) \notin S. Therefore, it's impossible to express q(x)q(x) as a linear combination of the basis vectors in BB, since every vector in SS must satisfy p(1)=0p(1) = 0. Choice B would give 1(x−1)−1(x2−1)+2(x3−1)=x−1−x2+1+2x3−2=2x3−x2+x−21(x-1) - 1(x^2-1) + 2(x^3-1) = x - 1 - x^2 + 1 + 2x^3 - 2 = 2x^3 - x^2 + x - 2, which is not equal to q(x)q(x). Choices C and D can be similarly eliminated by direct computation, and more importantly, they're conceptually wrong since q(x)q(x) doesn't belong to the subspace SS.

Question 14

Let VV be the vector space of 2×22 \times 2 symmetric matrices over R\mathbb{R}. Consider the basis B={E1=(1000),E2=(0110),E3=(0001)}B = \left\{E_1 = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, E_2 = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, E_3 = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}\right\} for VV. If A=(233−1)A = \begin{pmatrix} 2 & 3 \\ 3 & -1 \end{pmatrix} and B=(1−1−12)B = \begin{pmatrix} 1 & -1 \\ -1 & 2 \end{pmatrix} , what is [A+2B]B[A + 2B]_B?

  1. (4,1,3)(4, 1, 3) (correct answer)
  2. (4,3,1)(4, 3, 1)
  3. (2,1,3)(2, 1, 3)
  4. (2,3,1)(2, 3, 1)
Explanation: First, compute A+2B=(233−1)+2(1−1−12)=(233−1)+(2−2−24)=(4113)A + 2B = \begin{pmatrix} 2 & 3 \\ 3 & -1 \end{pmatrix} + 2\begin{pmatrix} 1 & -1 \\ -1 & 2 \end{pmatrix} = \begin{pmatrix} 2 & 3 \\ 3 & -1 \end{pmatrix} + \begin{pmatrix} 2 & -2 \\ -2 & 4 \end{pmatrix} = \begin{pmatrix} 4 & 1 \\ 1 & 3 \end{pmatrix} . To find the coordinate vector, we express this matrix as aE1+bE2+cE3aE_1 + bE_2 + cE_3 where (4113)=a(1000)+b(0110)+c(0001)=(abbc)\begin{pmatrix} 4 & 1 \\ 1 & 3 \end{pmatrix} = a\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} + b\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} + c\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} a & b \\ b & c \end{pmatrix}. Comparing entries: a=4a = 4, b=1b = 1, c=3c = 3. Therefore, [A+2B]B=(4,1,3)[A + 2B]_B = (4, 1, 3). Choice B swaps the second and third coordinates. Choices C and D have the wrong first coordinate.

Question 15

In the vector space R2×2\mathbb{R}^{2 \times 2} of 2×22 \times 2 matrices, consider the basis B={(1000),(0100),(0010),(0001)}B = \left\{\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}\right\}. If A=(2−134)A = \begin{pmatrix} 2 & -1 \\ 3 & 4 \end{pmatrix} and [A]B=(a,b,c,d)[A]_B = (a, b, c, d), what is [2AT−A]B[2A^T - A]_B?

  1. (2,5,−7,4)(2, 5, -7, 4)
  2. (2,−5,7,4)(2, -5, 7, 4)
  3. (2,7,−5,4)(2, 7, -5, 4) (correct answer)
  4. (−2,5,7,−4)(-2, 5, 7, -4)
Explanation: First, [A]B=(2,−1,3,4)[A]_B = (2, -1, 3, 4) since AA has entries 2, -1, 3, 4 in positions (1,1), (1,2), (2,1), (2,2) respectively. Next, AT=(23−14)A^T = \begin{pmatrix} 2 & 3 \\ -1 & 4 \end{pmatrix} , so [AT]B=(2,3,−1,4)[A^T]_B = (2, 3, -1, 4). Then 2AT−A=2(23−14)−(2−134)=(46−28)−(2−134)=(27−54)2A^T - A = 2\begin{pmatrix} 2 & 3 \\ -1 & 4 \end{pmatrix} - \begin{pmatrix} 2 & -1 \\ 3 & 4 \end{pmatrix} = \begin{pmatrix} 4 & 6 \\ -2 & 8 \end{pmatrix} - \begin{pmatrix} 2 & -1 \\ 3 & 4 \end{pmatrix} = \begin{pmatrix} 2 & 7 \\ -5 & 4 \end{pmatrix} . Therefore, [2AT−A]B=(2,7,−5,4)[2A^T - A]_B = (2, 7, -5, 4).