Linear Algebra Quiz: Finding A Basis
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Finding A BasisQuestion 1 of 16

Let WW be the subspace of R4\mathbb{R}^4 spanned by the vectors v⃗1=(120−1)\vec{v}_1 = \begin{pmatrix} 1 \\ 2 \\ 0 \\ -1 \end{pmatrix}, v⃗2=(−2−415)\vec{v}_2 = \begin{pmatrix} -2 \\ -4 \\ 1 \\ 5 \end{pmatrix}, and v⃗3=(36−1−6)\vec{v}_3 = \begin{pmatrix} 3 \\ 6 \\ -1 \\ -6 \end{pmatrix}. Which of the following sets is a basis for WW?

{(120−1),(−2−415)}\{\begin{pmatrix} 1 \\ 2 \\ 0 \\ -1 \end{pmatrix}, \begin{pmatrix} -2 \\ -4 \\ 1 \\ 5 \end{pmatrix}\}
{(120−1),(−2−415),(36−1−6)}\{\begin{pmatrix} 1 \\ 2 \\ 0 \\ -1 \end{pmatrix}, \begin{pmatrix} -2 \\ -4 \\ 1 \\ 5 \end{pmatrix}, \begin{pmatrix} 3 \\ 6 \\ -1 \\ -6 \end{pmatrix}\}
{(1001),(0013)}\{\begin{pmatrix} 1 \\ 0 \\ 0 \\ 1 \end{pmatrix}, \begin{pmatrix} 0 \\ 0 \\ 1 \\ 3 \end{pmatrix}\}
{(120−1)}\{\begin{pmatrix} 1 \\ 2 \\ 0 \\ -1 \end{pmatrix}\}
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Linear Algebra Quiz

Linear Algebra Quiz: Finding A Basis

Practice Finding A Basis 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 Finding A Basis, 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

Let WW be the subspace of R4\mathbb{R}^4 spanned by the vectors v⃗1=(120−1)\vec{v}_1 = \begin{pmatrix} 1 \\ 2 \\ 0 \\ -1 \end{pmatrix}, v⃗2=(−2−415)\vec{v}_2 = \begin{pmatrix} -2 \\ -4 \\ 1 \\ 5 \end{pmatrix}, and v⃗3=(36−1−6)\vec{v}_3 = \begin{pmatrix} 3 \\ 6 \\ -1 \\ -6 \end{pmatrix}. Which of the following sets is a basis for WW?

  1. {(120−1),(−2−415)}\{\begin{pmatrix} 1 \\ 2 \\ 0 \\ -1 \end{pmatrix}, \begin{pmatrix} -2 \\ -4 \\ 1 \\ 5 \end{pmatrix}\} (correct answer)
  2. {(120−1),(−2−415),(36−1−6)}\{\begin{pmatrix} 1 \\ 2 \\ 0 \\ -1 \end{pmatrix}, \begin{pmatrix} -2 \\ -4 \\ 1 \\ 5 \end{pmatrix}, \begin{pmatrix} 3 \\ 6 \\ -1 \\ -6 \end{pmatrix}\}
  3. {(1001),(0013)}\{\begin{pmatrix} 1 \\ 0 \\ 0 \\ 1 \end{pmatrix}, \begin{pmatrix} 0 \\ 0 \\ 1 \\ 3 \end{pmatrix}\}
  4. {(120−1)}\{\begin{pmatrix} 1 \\ 2 \\ 0 \\ -1 \end{pmatrix}\}
Explanation: To find a basis for the subspace spanned by a set of vectors, we form a matrix with these vectors as columns and row-reduce it to find the pivot columns. The corresponding original vectors form the basis. Let A=[v⃗1  v⃗2  v⃗3]=(1−232−4601−1−15−6)A = [\vec{v}_1 \; \vec{v}_2 \; \vec{v}_3] = \begin{pmatrix} 1 & -2 & 3 \\ 2 & -4 & 6 \\ 0 & 1 & -1 \\ -1 & 5 & -6 \end{pmatrix}. Row reduction proceeds as: (1−2300001−103−3)→(1−2301−1000000)\begin{pmatrix} 1 & -2 & 3 \\ 0 & 0 & 0 \\ 0 & 1 & -1 \\ 0 & 3 & -3 \end{pmatrix} \to \begin{pmatrix} 1 & -2 & 3 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}. The pivot columns are the first and second columns. Therefore, a basis for WW is given by the first two original vectors, {v⃗1,v⃗2}\{\vec{v}_1, \vec{v}_2\}.

Question 2

Which of the following sets forms a basis for the null space of the matrix A=(1−205−121−8)A = \begin{pmatrix} 1 & -2 & 0 & 5 \\ -1 & 2 & 1 & -8 \end{pmatrix}?

  1. {(1−1),(01)}\{\begin{pmatrix} 1 \\ -1 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \end{pmatrix}\}
  2. {(103),(010)}\{\begin{pmatrix} 1 \\ 0 \\ 3 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}\}
  3. {(2100),(−5031)}\{\begin{pmatrix} 2 \\ 1 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} -5 \\ 0 \\ 3 \\ 1 \end{pmatrix}\} (correct answer)
  4. {(−2100),(50−31)}\{\begin{pmatrix} -2 \\ 1 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 5 \\ 0 \\ -3 \\ 1 \end{pmatrix}\}
Explanation: To find a basis for the null space of AA, we solve the homogeneous system Ax⃗=0⃗A\vec{x} = \vec{0}. The augmented matrix is (1−2050−121−80)\begin{pmatrix} 1 & -2 & 0 & 5 & 0 \\ -1 & 2 & 1 & -8 & 0 \end{pmatrix}. Adding the first row to the second gives (1−2050001−30)\begin{pmatrix} 1 & -2 & 0 & 5 & 0 \\ 0 & 0 & 1 & -3 & 0 \end{pmatrix}. This is the reduced row echelon form. The corresponding equations are x1−2x2+5x4=0x_1 - 2x_2 + 5x_4 = 0 and x3−3x4=0x_3 - 3x_4 = 0. The free variables are x2x_2 and x4x_4. We can write the solution as x1=2x2−5x4x_1 = 2x_2 - 5x_4 and x3=3x4x_3 = 3x_4. In parametric vector form, this is x⃗=x2(2100)+x4(−5031)\vec{x} = x_2 \begin{pmatrix} 2 \\ 1 \\ 0 \\ 0 \end{pmatrix} + x_4 \begin{pmatrix} -5 \\ 0 \\ 3 \\ 1 \end{pmatrix}. The vectors in this form constitute a basis for Nul AA.

Question 3

Let U=Span{(110),(011)}U = \text{Span}\{\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}\} and V=Span{(10−1),(121)}V = \text{Span}\{\begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}, \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}\} be subspaces of R3\mathbb{R}^3. Which of the following is a basis for the intersection U∩VU \cap V?

  1. {(121)}\{\begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}\} (correct answer)
  2. {(110),(011)}\{\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}\}
  3. {(10−1)}\{\begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}\}
  4. {(000)}\{\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}\}
Explanation: A vector x⃗\vec{x} is in the intersection U∩VU \cap V if it can be written as a linear combination of the basis vectors for both UU and VV. So, c1(110)+c2(011)=c3(10−1)+c4(121)c_1\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} + c_2\begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} = c_3\begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix} + c_4\begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}. This gives the system c1−c3−c4=0c_1 - c_3 - c_4 = 0, c1+c2−2c4=0c_1 + c_2 - 2c_4 = 0, and c2+c3−c4=0c_2 + c_3 - c_4 = 0. Solving this system, we find solutions where c3=−c4c_3 = -c_4. For a non-trivial solution, let c4=1c_4=1, then c3=−1c_3=-1, c2=2c_2=2, c1=0c_1=0. This is not quite right. A simpler way is to note that the vector (121)\begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix} from the basis of VV can be written as 1(110)+1(011)1\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} + 1\begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}, which means it is in UU. Since it is also in VV, it must be in the intersection. As the intersection is a subspace of dimension at most 1 (since the subspaces are distinct planes), this single non-zero vector forms a basis for U∩VU \cap V.

Question 4

Let WW be the subspace of R3\mathbb{R}^3 spanned by {v⃗1,v⃗2,v⃗3}\{\vec{v}_1, \vec{v}_2, \vec{v}_3\}, where v⃗1=(10−2)\vec{v}_1 = \begin{pmatrix} 1 \\ 0 \\ -2 \end{pmatrix}, v⃗2=(011)\vec{v}_2 = \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}, and v⃗3=(2−3−7)\vec{v}_3 = \begin{pmatrix} 2 \\ -3 \\ -7 \end{pmatrix}. Which statement accurately describes a basis for WW?

  1. The set {v⃗1,v⃗2,v⃗3}\{\vec{v}_1, \vec{v}_2, \vec{v}_3\} is a basis because it contains three vectors in R3\mathbb{R}^3.
  2. A basis for WW is {v⃗1,v⃗2}\{\vec{v}_1, \vec{v}_2\} because v⃗3\vec{v}_3 is a linear combination of v⃗1\vec{v}_1 and v⃗2\vec{v}_2. (correct answer)
  3. A basis for WW is {v⃗3}\{\vec{v}_3\} because it is the vector with the largest magnitude.
  4. A basis for WW is {v⃗1,v⃗3}\{\vec{v}_1, \vec{v}_3\} because v⃗2\vec{v}_2 is a linear combination of v⃗1\vec{v}_1 and v⃗3\vec{v}_3.
Explanation: To determine the basis, we must check for linear dependence. We can see if v⃗3\vec{v}_3 is a linear combination of v⃗1\vec{v}_1 and v⃗2\vec{v}_2 by solving c1v⃗1+c2v⃗2=v⃗3c_1\vec{v}_1 + c_2\vec{v}_2 = \vec{v}_3. This gives the system of equations: c1=2c_1 = 2, c2=−3c_2 = -3, and −2c1+c2=−7-2c_1 + c_2 = -7. Substituting the first two equations into the third gives −2(2)+(−3)=−4−3=−7-2(2) + (-3) = -4 - 3 = -7. The equation holds. Since v⃗3=2v⃗1−3v⃗2\vec{v}_3 = 2\vec{v}_1 - 3\vec{v}_2, v⃗3\vec{v}_3 is redundant. The remaining vectors, v⃗1\vec{v}_1 and v⃗2\vec{v}_2, are not scalar multiples, so they are linearly independent and form a basis for WW. Thus, WW is a plane in R3\mathbb{R}^3.

Question 5

Let v⃗1=(1−12)\vec{v}_1 = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}, v⃗2=(−23−1)\vec{v}_2 = \begin{pmatrix} -2 \\ 3 \\ -1 \end{pmatrix}, and v⃗3=(10k)\vec{v}_3 = \begin{pmatrix} 1 \\ 0 \\ k \end{pmatrix}. For which value of kk does the set {v⃗1,v⃗2,v⃗3}\{\vec{v}_1, \vec{v}_2, \vec{v}_3\} NOT form a basis for R3\mathbb{R}^3?

  1. k=0k = 0
  2. k=−1k = -1
  3. k=5k = 5 (correct answer)
  4. k=3k = 3
Explanation: The set will not form a basis for R3\mathbb{R}^3 if the vectors are linearly dependent. This occurs if the matrix formed by these vectors has a determinant of zero. A=(1−21−1302−1k)A = \begin{pmatrix} 1 & -2 & 1 \\ -1 & 3 & 0 \\ 2 & -1 & k \end{pmatrix}. The determinant is 1(3k−0)−(−2)(−k−0)+1(1−6)=3k−2k−5=k−51(3k - 0) - (-2)(-k - 0) + 1(1 - 6) = 3k - 2k - 5 = k - 5. The set is linearly dependent when the determinant is zero, which means k−5=0k - 5 = 0, or k=5k=5. For any other value of kk, the determinant is non-zero, and the vectors form a basis for R3\mathbb{R}^3.

Question 6

Let WW be the subspace of R3\mathbb{R}^3 spanned by the set {(110),(011),(121),(220)}\{\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}, \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}, \begin{pmatrix} 2 \\ 2 \\ 0 \end{pmatrix}\}. Which of the following sets is a basis for WW?

  1. {(110),(220)}\{\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 2 \\ 2 \\ 0 \end{pmatrix}\}
  2. {(110),(011),(121)}\{\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}, \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}\}
  3. {(110),(011)}\{\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}\} (correct answer)
  4. {(10−1)}\{\begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}\}
Explanation: The given set of four vectors spans WW. To find a basis, we must find a linearly independent subset that still spans WW. Let the vectors be v⃗1,v⃗2,v⃗3,v⃗4\vec{v}_1, \vec{v}_2, \vec{v}_3, \vec{v}_4. By inspection, v⃗3=v⃗1+v⃗2\vec{v}_3 = \vec{v}_1 + \vec{v}_2 and v⃗4=2v⃗1\vec{v}_4 = 2\vec{v}_1. Thus, v⃗3\vec{v}_3 and v⃗4\vec{v}_4 are redundant and can be removed. The remaining set is {v⃗1,v⃗2}={(110),(011)}\{\vec{v}_1, \vec{v}_2\} = \{\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}\}. These two vectors are not scalar multiples of each other, so they are linearly independent. Therefore, they form a basis for WW. Choice A is linearly dependent. Choice B is linearly dependent. Choice D is a vector that is not in the subspace WW.

Question 7

Consider the subspace SS of M2×2(R)M_{2 \times 2}(\mathbb{R}) consisting of all 2×22 \times 2 matrices AA such that tr(A)=0\text{tr}(A) = 0 and AT=AA^T = A. What is the dimension of SS, and which set forms a basis for SS?

  1. Dimension 2; basis: {(100−1),(0110)}\left\{\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\right\} (correct answer)
  2. Dimension 3; basis: {(1000),(0110),(000−1)}\left\{\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 0 & -1 \end{pmatrix}\right\}
  3. Dimension 1; basis: {(111−1)}\left\{\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}\right\}
  4. Dimension 2; basis: {(0110),(1111)}\left\{\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}\right\}
Explanation: For a matrix A=(abcd)A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} to be in S, we need AT=AA^T = A (symmetric) and tr(A)=0\text{tr}(A) = 0 (traceless). Symmetry requires b=cb = c, and trace zero requires a+d=0a + d = 0, so d=−ad = -a. Thus $$A = \begin{pmatrix} a & b \ b & -a \end{pmatrix} = a\begin{pmatrix} 1 & 0 \ 0 & -1 \end{pmatrix} + b\begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix}

Question 8

Let WW be the subspace of M3×3(R)M_{3 \times 3}(\mathbb{R}) consisting of all matrices AA such that A+AT=2IA + A^T = 2I, where II is the 3×33 \times 3 identity matrix. What is the dimension of WW?

  1. dim⁡(W)=3\dim(W) = 3 (correct answer)
  2. dim⁡(W)=6\dim(W) = 6
  3. dim⁡(W)=4\dim(W) = 4
  4. dim⁡(W)=5\dim(W) = 5
Explanation: For a matrix A=(abcdefghi)A = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} to satisfy A+AT=2IA + A^T = 2I, we need (abcdefghi)+(adgbehcfi)=(200020002)\begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} + \begin{pmatrix} a & d & g \\ b & e & h \\ c & f & i \end{pmatrix} = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{pmatrix}. This gives us: 2a=22a = 2 (so a=1a = 1), 2e=22e = 2 (so e=1e = 1), 2i=22i = 2 (so i=1i = 1), b+d=0b + d = 0 (so d=−bd = -b), c+g=0c + g = 0 (so g=−cg = -c), and f+h=0f + h = 0 (so h=−fh = -f). Therefore, A=(1bc−b1f−c−f1)A = \begin{pmatrix} 1 & b & c \\ -b & 1 & f \\ -c & -f & 1 \end{pmatrix} . The matrix depends on 3 free parameters (bb, cc, ff), so dim⁡(W)=3\dim(W) = 3.

Question 9

Consider the subspace VV of R6\mathbb{R}^6 spanned by the vectors u1=(1,0,1,1,0,1)u_1 = (1, 0, 1, 1, 0, 1), u2=(0,1,1,0,1,1)u_2 = (0, 1, 1, 0, 1, 1), u3=(1,1,2,1,1,2)u_3 = (1, 1, 2, 1, 1, 2), u4=(2,1,3,2,1,3)u_4 = (2, 1, 3, 2, 1, 3), and u5=(1,2,3,1,2,3)u_5 = (1, 2, 3, 1, 2, 3). After determining linear dependencies, which subset forms a basis for VV?

  1. {u1,u2,u3}\{u_1, u_2, u_3\}
  2. {u1,u2,u5}\{u_1, u_2, u_5\} (correct answer)
  3. {u1,u2,u4}\{u_1, u_2, u_4\}
  4. {u2,u3,u4}\{u_2, u_3, u_4\}
Explanation: To find a basis, we need to identify which vectors are linearly independent. Notice that u3=u1+u2u_3 = u_1 + u_2 since (1,0,1,1,0,1)+(0,1,1,0,1,1)=(1,1,2,1,1,2)=u3(1, 0, 1, 1, 0, 1) + (0, 1, 1, 0, 1, 1) = (1, 1, 2, 1, 1, 2) = u_3. Also, u4=2u1+u2u_4 = 2u_1 + u_2 since 2(1,0,1,1,0,1)+(0,1,1,0,1,1)=(2,1,3,2,1,3)=u42(1, 0, 1, 1, 0, 1) + (0, 1, 1, 0, 1, 1) = (2, 1, 3, 2, 1, 3) = u_4. We can verify that u1u_1, u2u_2, and u5u_5 are linearly independent by checking that no nontrivial linear combination equals zero. Since u3u_3 and u4u_4 are linear combinations of u1u_1 and u2u_2, and u5u_5 is not in the span of u1u_1 and u2u_2, the vectors {u1,u2,u5}\{u_1, u_2, u_5\} form a basis for the 3-dimensional subspace V. The other options either include dependent vectors or cannot span the full space.

Question 10

Let WW be the subspace of R4\mathbb{R}^4 consisting of all vectors x⃗=(x1,x2,x3,x4)\vec{x} = (x_1, x_2, x_3, x_4) such that x1−2x2+x3=0x_1 - 2x_2 + x_3 = 0. Which of the following is a basis for WW?

  1. {(1−210)}\{\begin{pmatrix} 1 \\ -2 \\ 1 \\ 0 \end{pmatrix}\}
  2. {(2100),(−1010),(0001)}\{\begin{pmatrix} 2 \\ 1 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} -1 \\ 0 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix}\} (correct answer)
  3. {(2100),(0001)}\{\begin{pmatrix} 2 \\ 1 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix}\}
  4. {(10−10),(0120),(0001),(1111)}\{\begin{pmatrix} 1 \\ 0 \\ -1 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ 2 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix}, \begin{pmatrix} 1 \\ 1 \\ 1 \\ 1 \end{pmatrix}\}
Explanation: The condition x1−2x2+x3=0x_1 - 2x_2 + x_3 = 0 can be rewritten as x1=2x2−x3x_1 = 2x_2 - x_3. The variables x2x_2, x3x_3, and x4x_4 are free. We can express any vector x⃗\vec{x} in WW as: x⃗=(2x2−x3x2x3x4)=x2(2100)+x3(−1010)+x4(0001)\vec{x} = \begin{pmatrix} 2x_2 - x_3 \\ x_2 \\ x_3 \\ x_4 \end{pmatrix} = x_2 \begin{pmatrix} 2 \\ 1 \\ 0 \\ 0 \end{pmatrix} + x_3 \begin{pmatrix} -1 \\ 0 \\ 1 \\ 0 \end{pmatrix} + x_4 \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix} The three vectors in this parametric form are linearly independent and span WW, so they form a basis. Distractor A is the normal vector to the hyperplane, which is not in WW. Distractor C does not span WW. Distractor D is linearly dependent as it contains four vectors in a three-dimensional subspace.

Question 11

Let HH be the subspace of P2\mathbb{P}_2 (polynomials of degree at most 2) defined by H={p(t)∈P2∣p(1)=0}H = \{ p(t) \in \mathbb{P}_2 \mid p(1) = 0 \}. Which of the following is a basis for HH?

  1. {t−1,t2−1,t2−t}\{t-1, t^2-1, t^2-t\}
  2. {t−1,t+1}\{t-1, t+1\}
  3. {t2−1}\{t^2-1\}
  4. {t−1,t2−t}\{t-1, t^2-t\} (correct answer)
Explanation: A polynomial p(t)=a2t2+a1t+a0p(t) = a_2t^2 + a_1t + a_0 is in HH if p(1)=a2+a1+a0=0p(1) = a_2 + a_1 + a_0 = 0. This implies a0=−a1−a2a_0 = -a_1 - a_2. Substituting back, we get p(t)=a2t2+a1t−(a1+a2)=a1(t−1)+a2(t2−1)p(t) = a_2t^2 + a_1t - (a_1+a_2) = a_1(t-1) + a_2(t^2-1). So, HH is spanned by {t−1,t2−1}\{t-1, t^2-1\}. The set in choice D, {t−1,t2−t}\{t-1, t^2-t\}, is also a valid basis. Both polynomials are in HH since they are zero at t=1t=1. They are linearly independent because neither is a scalar multiple of the other. The dimension of HH is 2, so this set forms a basis. Choice A is linearly dependent since t2−t=(t2−1)−(t−1)t^2-t = (t^2-1) - (t-1). Choice B contains t+1t+1, which is not in HH because 1+1=2≠01+1=2 \neq 0. Choice C does not span HH, as its dimension is 2.

Question 12

Consider the matrix A=(135−1−1−2−540103)A = \begin{pmatrix} 1 & 3 & 5 & -1 \\ -1 & -2 & -5 & 4 \\ 0 & 1 & 0 & 3 \end{pmatrix}. Which of the following sets forms a basis for the column space of AA?

  1. {(100),(310)}\{\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 3 \\ 1 \\ 0 \end{pmatrix}\}
  2. {(135−1),(−1−2−54)}\{\begin{pmatrix} 1 \\ 3 \\ 5 \\ -1 \end{pmatrix}, \begin{pmatrix} -1 \\ -2 \\ -5 \\ 4 \end{pmatrix}\}
  3. {(1−10),(3−21)}\{\begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix}, \begin{pmatrix} 3 \\ -2 \\ 1 \end{pmatrix}\} (correct answer)
  4. {(1−10),(3−21),(5−50),(−143)}\{\begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix}, \begin{pmatrix} 3 \\ -2 \\ 1 \end{pmatrix}, \begin{pmatrix} 5 \\ -5 \\ 0 \end{pmatrix}, \begin{pmatrix} -1 \\ 4 \\ 3 \end{pmatrix}\}
Explanation: To find a basis for the column space, we identify the pivot columns of the matrix AA. We row-reduce AA: (135−1−1−2−540103)→R2+R1(135−101030103)→R3−R2(135−101030000)\begin{pmatrix} 1 & 3 & 5 & -1 \\ -1 & -2 & -5 & 4 \\ 0 & 1 & 0 & 3 \end{pmatrix} \xrightarrow{R_2+R_1} \begin{pmatrix} 1 & 3 & 5 & -1 \\ 0 & 1 & 0 & 3 \\ 0 & 1 & 0 & 3 \end{pmatrix} \xrightarrow{R_3-R_2} \begin{pmatrix} 1 & 3 & 5 & -1 \\ 0 & 1 & 0 & 3 \\ 0 & 0 & 0 & 0 \end{pmatrix} The pivot positions are in the first and second columns. A basis for the column space is formed by the corresponding columns from the original matrix AA. Thus, a basis is {(1−10),(3−21)}\{ \begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix}, \begin{pmatrix} 3 \\ -2 \\ 1 \end{pmatrix} \}. Choice A incorrectly uses columns from the echelon form. Choice B incorrectly uses rows. Choice D includes all columns, which is a spanning set but is linearly dependent.

Question 13

Let SS be the subspace of M2×2M_{2 \times 2}, the space of 2×22 \times 2 matrices, consisting of all matrices AA such that AT=AA^T = A (symmetric matrices). Which of the following is a basis for SS?

  1. {(1000),(0100),(0010),(0001)}\{\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}\}
  2. {(01−10)}\{\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}\}
  3. {(1000),(0110),(0001),(1001)}\{\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}, \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\}
  4. {(1000),(0110),(0001)}\{\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}\} (correct answer)
Explanation: A general 2×22 \times 2 matrix is A=(abcd)A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}. The condition for symmetry, AT=AA^T = A, implies that (acbd)=(abcd)\begin{pmatrix} a & c \\ b & d \end{pmatrix} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}, which requires b=cb=c. Thus, any symmetric 2×22 \times 2 matrix has the form (abbd)\begin{pmatrix} a & b \\ b & d \end{pmatrix}. This can be written as a linear combination: a(1000)+b(0110)+d(0001)a\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} + b\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} + d\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} The three matrices in this combination are linearly independent and span the space of symmetric matrices, so they form a basis. Choice A is the standard basis for all of M2×2M_{2 \times 2}. Choice B is a basis for skew-symmetric matrices. Choice C is linearly dependent because the last matrix is the sum of the first and third.

Question 14

Let S={v⃗1,v⃗2,v⃗3}S = \{\vec{v}_1, \vec{v}_2, \vec{v}_3\} be a basis for a vector space WW. Which of the following statements must be true?

  1. The set {v⃗1,v⃗2}\{\vec{v}_1, \vec{v}_2\} is also a basis for WW.
  2. The set {v⃗1,v⃗2,v⃗3,v⃗1−v⃗3}\{\vec{v}_1, \vec{v}_2, \vec{v}_3, \vec{v}_1 - \vec{v}_3\} is linearly independent.
  3. The set {v⃗1,v⃗2,2v⃗3}\{\vec{v}_1, \vec{v}_2, 2\vec{v}_3\} is another basis for WW. (correct answer)
  4. Any set of two vectors from WW must be linearly independent.
Explanation: A basis must be linearly independent and span the space. Choice C is correct because replacing a basis vector with a non-zero scalar multiple of itself results in a new set that is still linearly independent and spans the same space. Choice A is false because {v⃗1,v⃗2}\{\vec{v}_1, \vec{v}_2\} is linearly independent but cannot span the three-dimensional space WW. Choice B is false because the set is linearly dependent; the fourth vector is a linear combination of the first and third. Choice D is false; for example, the set {v⃗1,2v⃗1}\{\vec{v}_1, 2\vec{v}_1\} consists of two vectors from WW but is linearly dependent.

Question 15

A student is finding a basis for the column space of a 4×54 \times 5 matrix AA. Which of the following procedures is guaranteed to produce a correct basis?

  1. Select the columns of AA that correspond to the pivot columns in the reduced row echelon form of AA. (correct answer)
  2. Select the non-zero rows from the reduced row echelon form of AA.
  3. Compute the solution to Ax⃗=0⃗A\vec{x} = \vec{0} and write it in parametric vector form; the vectors in this form are the basis.
  4. Select the pivot columns from the reduced row echelon form of AA.
Explanation: The standard algorithm for finding a basis for the column space (Col AA) is to row-reduce AA to an echelon form, identify the columns that contain pivots, and then select the corresponding columns from the original matrix AA. Choice A correctly describes this procedure. Choice B describes a procedure for finding a basis for the row space of AA. Choice C describes the procedure for finding a basis for the null space of AA. Choice D is a common mistake; the pivot columns of the reduced matrix RREF(AA) span Col AA, but they are not necessarily the same vectors as those in Col AA itself, which is a subspace of R4\mathbb{R}^4.

Question 16

Consider the subspace TT of R4\mathbb{R}^4 defined by T={(a+b,a−b,2a+c,b−c):a,b,c∈R}T = \{(a+b, a-b, 2a+c, b-c) : a, b, c \in \mathbb{R}\}. Which of the following statements about a basis for TT is correct?

  1. {(1,1,2,0),(1,−1,0,1),(0,0,1,−1)}\{(1, 1, 2, 0), (1, -1, 0, 1), (0, 0, 1, -1)\} forms a basis and dim⁡(T)=3\dim(T) = 3 (correct answer)
  2. {(1,1,2,0),(1,−1,0,1),(0,0,1,1)}\{(1, 1, 2, 0), (1, -1, 0, 1), (0, 0, 1, 1)\} forms a basis and dim⁡(T)=3\dim(T) = 3
  3. {(2,0,4,0),(0,2,0,2)}\{(2, 0, 4, 0), (0, 2, 0, 2)\} forms a basis and dim⁡(T)=2\dim(T) = 2
  4. {(1,1,2,0),(1,−1,0,1)}\{(1, 1, 2, 0), (1, -1, 0, 1)\} forms a basis and dim⁡(T)=2\dim(T) = 2
Explanation: To find a basis for T, we express the general form: (a+b,a−b,2a+c,b−c)=a(1,1,2,0)+b(1,−1,0,1)+c(0,0,1,−1)(a+b, a-b, 2a+c, b-c) = a(1, 1, 2, 0) + b(1, -1, 0, 1) + c(0, 0, 1, -1). This shows that T is spanned by the three vectors {(1,1,2,0),(1,−1,0,1),(0,0,1,−1)}\{(1, 1, 2, 0), (1, -1, 0, 1), (0, 0, 1, -1)\}. To verify linear independence, we check if α(1,1,2,0)+β(1,−1,0,1)+γ(0,0,1,−1)=(0,0,0,0)\alpha(1, 1, 2, 0) + \beta(1, -1, 0, 1) + \gamma(0, 0, 1, -1) = (0, 0, 0, 0) only when α=β=γ=0\alpha = \beta = \gamma = 0. This gives the system: α+β=0\alpha + \beta = 0, α−β=0\alpha - \beta = 0, 2α+γ=02\alpha + \gamma = 0, β−γ=0\beta - \gamma = 0. From the first two equations, α=β=0\alpha = \beta = 0, and then from the third and fourth, γ=0\gamma = 0. Thus the vectors are linearly independent, so option A is correct. Option B has the wrong third vector, options C and D have insufficient vectors to span the 3-dimensional space.