Linear Algebra Quiz: Composition Of Transformations
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Composition Of TransformationsQuestion 1 of 17

Let T:R2→R3T: \mathbb{R}^2 \to \mathbb{R}^3 be a linear transformation with standard matrix A=(1−12003)A = \begin{pmatrix} 1 & -1 \\ 2 & 0 \\ 0 & 3 \end{pmatrix} and let S:R3→R2S: \mathbb{R}^3 \to \mathbb{R}^2 be a linear transformation with standard matrix B=(1010−12)B = \begin{pmatrix} 1 & 0 & 1 \\ 0 & -1 & 2 \end{pmatrix}. What is the standard matrix of the composite transformation S∘TS \circ T?

(12−26)\begin{pmatrix} 1 & 2 \\ -2 & 6 \end{pmatrix}
(11−12020−36)\begin{pmatrix} 1 & 1 & -1 \\ 2 & 0 & 2 \\ 0 & -3 & 6 \end{pmatrix}
(1226)\begin{pmatrix} 1 & 2 \\ 2 & 6 \end{pmatrix}
(2−12−1−15)\begin{pmatrix} 2 & -1 \\ 2 & -1 \\ -1 & 5 \end{pmatrix}
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Linear Algebra Quiz

Linear Algebra Quiz: Composition Of Transformations

Practice Composition Of Transformations 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 Composition Of Transformations, 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 T:R2→R3T: \mathbb{R}^2 \to \mathbb{R}^3 be a linear transformation with standard matrix A=(1−12003)A = \begin{pmatrix} 1 & -1 \\ 2 & 0 \\ 0 & 3 \end{pmatrix} and let S:R3→R2S: \mathbb{R}^3 \to \mathbb{R}^2 be a linear transformation with standard matrix B=(1010−12)B = \begin{pmatrix} 1 & 0 & 1 \\ 0 & -1 & 2 \end{pmatrix}. What is the standard matrix of the composite transformation S∘TS \circ T?

  1. (12−26)\begin{pmatrix} 1 & 2 \\ -2 & 6 \end{pmatrix} (correct answer)
  2. (11−12020−36)\begin{pmatrix} 1 & 1 & -1 \\ 2 & 0 & 2 \\ 0 & -3 & 6 \end{pmatrix}
  3. (1226)\begin{pmatrix} 1 & 2 \\ 2 & 6 \end{pmatrix}
  4. (2−12−1−15)\begin{pmatrix} 2 & -1 \\ 2 & -1 \\ -1 & 5 \end{pmatrix}
Explanation: The standard matrix of a composite transformation S∘TS \circ T is the product of the standard matrices of SS and TT, in that order. The matrix for SS is BB and the matrix for TT is AA. The resulting matrix is BABA. The product is computed as: $$BA = \begin{pmatrix} 1 & 0 & 1 \ 0 & -1 & 2 \end{pmatrix} \begin{pmatrix} 1 & -1 \ 2 & 0 \ 0 & 3 \end{pmatrix} = \begin{pmatrix} (1)(1)+(0)(2)+(1)(0) & (1)(-1)+(0)(0)+(1)(3) \ (0)(1)+(-1)(2)+(2)(0) & (0)(-1)+(-1)(0)+(2)(3) \end{pmatrix} = \begin{pmatrix} 1 & 2 \ -2 & 6 \end{pmatrix} **B is incorrect:** This is the result of computing the matrix product in the wrong order, $AB$. This would represent the transformation $T \circ S$, which maps from $\mathbb{R}^3$ to $\mathbb{R}^3$. **C is incorrect:** This matrix results from a sign error when computing the $(2,1)$ entry of the product $BA$: $(0)(1)+(-1)(2)+(2)(0)$ is $-2$, not $2$. **D is incorrect:** This matrix has incorrect dimensions. The composition $S \circ T$ maps from $\mathbb{R}^2$ to $\mathbb{R}^2$, so its standard matrix must be $2 \times 2$.

Question 2

Let T:R2→R3T: \mathbb{R}^2 \to \mathbb{R}^3 be defined by T(x1,x2)=(x1−x2,3x2,x1+x2)T(x_1, x_2) = (x_1 - x_2, 3x_2, x_1+x_2) and S:R3→R2S: \mathbb{R}^3 \to \mathbb{R}^2 be defined by S(y1,y2,y3)=(y1+y3,y2−y3)S(y_1, y_2, y_3) = (y_1+y_3, y_2-y_3). What is the formula for the composition (S∘T)(x1,x2)(S \circ T)(x_1, x_2)?

  1. (2x1,2x2−x1)(2x_1, 2x_2 - x_1) (correct answer)
  2. (−2x2,4x2)(-2x_2, 4x_2)
  3. (2x1,2x2+x1)(2x_1, 2x_2 + x_1)
  4. The composition is not defined.
Explanation: To find the formula for (S∘T)(x1,x2)(S \circ T)(x_1, x_2), we first apply TT to (x1,x2)(x_1, x_2) and then apply SS to the result. (S∘T)(x1,x2)=S(T(x1,x2))=S(x1−x2,3x2,x1+x2)(S \circ T)(x_1, x_2) = S(T(x_1, x_2)) = S(x_1 - x_2, 3x_2, x_1+x_2) Now we use the definition of SS with y1=x1−x2y_1 = x_1 - x_2, y2=3x2y_2 = 3x_2, and y3=x1+x2y_3 = x_1+x_2. S(y1,y2,y3)=(y1+y3,y2−y3)S(y_1, y_2, y_3) = (y_1+y_3, y_2-y_3) Substituting the expressions in terms of x1x_1 and x2x_2: ((x1−x2)+(x1+x2),(3x2)−(x1+x2))((x_1 - x_2) + (x_1+x_2), (3x_2) - (x_1+x_2)) Simplifying each component gives: (2x1,3x2−x1−x2)=(2x1,−x1+2x2)(2x_1, 3x_2 - x_1 - x_2) = (2x_1, -x_1 + 2x_2) The correct answer is (2x1,2x2−x1)(2x_1, 2x_2 - x_1). B is incorrect: This may result from calculation errors, for example, calculating the first component as (x1−x2)−(x1+x2)=−2x2(x_1 - x_2) - (x_1+x_2) = -2x_2. C is incorrect: This results from a sign error in the second component: (3x2)+(x1+x2)=x1+4x2(3x_2) + (x_1+x_2) = x_1+4x_2, which is also not the choice, but sign errors are common. This specific choice may come from (3x2)−(x1−x2)(3x_2)- (x_1-x_2) instead of (3x2)−(x1+x2)(3x_2)-(x_1+x_2). D is incorrect: The composition is well-defined because the codomain of TT (R3\mathbb{R}^3) matches the domain of SS (R3\mathbb{R}^3).

Question 3

Let T:R2→R2T: \mathbb{R}^2 \to \mathbb{R}^2 be a linear transformation with standard matrix AT=(1213)A_T = \begin{pmatrix} 1 & 2 \\ 1 & 3 \end{pmatrix}. Let L:R2→R2L: \mathbb{R}^2 \to \mathbb{R}^2 be the composite transformation L=S∘TL = S \circ T, with standard matrix AL=(2154)A_L = \begin{pmatrix} 2 & 1 \\ 5 & 4 \end{pmatrix}. If S:R2→R2S: \mathbb{R}^2 \to \mathbb{R}^2 is also a linear transformation, what is its standard matrix ASA_S?

  1. (−4−533)\begin{pmatrix} -4 & -5 \\ 3 & 3 \end{pmatrix}
  2. (37922)\begin{pmatrix} 3 & 7 \\ 9 & 22 \end{pmatrix}
  3. (1291713)\begin{pmatrix} 12 & 9 \\ 17 & 13 \end{pmatrix}
  4. (5−311−6)\begin{pmatrix} 5 & -3 \\ 11 & -6 \end{pmatrix} (correct answer)
Explanation: The standard matrix of the composition L=S∘TL = S \circ T is given by the product of the individual standard matrices: AL=ASATA_L = A_S A_T. To find the matrix ASA_S, we need to isolate it by right-multiplying by the inverse of ATA_T. AS=ALAT−1A_S = A_L A_T^{-1} First, we must calculate AT−1A_T^{-1}. The determinant of ATA_T is det⁡(AT)=(1)(3)−(2)(1)=1\det(A_T) = (1)(3) - (2)(1) = 1. Since the determinant is non-zero, the inverse exists. For a 2×22 \times 2 matrix (abcd)\begin{pmatrix} a & b \\ c & d \end{pmatrix}, the inverse is 1ad−bc(d−b−ca)\frac{1}{ad-bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}. So, AT−1=11(3−2−11)=(3−2−11)A_T^{-1} = \frac{1}{1} \begin{pmatrix} 3 & -2 \\ -1 & 1 \end{pmatrix} = \begin{pmatrix} 3 & -2 \\ -1 & 1 \end{pmatrix} Now, we compute the product AS=ALAT−1A_S = A_L A_T^{-1}: $$A_S = \begin{pmatrix} 2 & 1 \ 5 & 4 \end{pmatrix} \begin{pmatrix} 3 & -2 \ -1 & 1 \end{pmatrix} = \begin{pmatrix} (2)(3)+(1)(-1) & (2)(-2)+(1)(1) \ (5)(3)+(4)(-1) & (5)(-2)+(4)(1) \end{pmatrix} = \begin{pmatrix} 5 & -3 \ 11 & -6 \end{pmatrix} **A is incorrect:** This is the result of multiplying in the wrong order, $A_T^{-1} A_L$. Matrix multiplication is not commutative. **B is incorrect:** This is the result of calculating $A_L A_T$, which incorrectly assumes $A_S = A_L A_T$. **C is incorrect:** This is the result of calculating $A_T A_L$, another incorrect matrix multiplication order.

Question 4

Let R:R2→R2R: \mathbb{R}^2 \to \mathbb{R}^2 be a counterclockwise rotation by 45∘45^\circ, and let P:R2→R2P: \mathbb{R}^2 \to \mathbb{R}^2 be the orthogonal projection onto the line y=xy=x. What is the rank of the composite transformation P∘RP \circ R?

  1. 0
  2. 1 (correct answer)
  3. 2
  4. The rank depends on the vector it is applied to.
Explanation: The rank of a composite transformation S∘TS \circ T is bounded by the ranks of the individual transformations: rank(S∘T)≤min⁡(rank(S),rank(T))\mathrm{rank}(S \circ T) \le \min(\mathrm{rank}(S), \mathrm{rank}(T)). The transformation RR is a rotation, which is an invertible transformation. Its standard matrix is invertible, so rank(R)=2\mathrm{rank}(R) = 2. The transformation PP is a projection onto a one-dimensional subspace (a line) in R2\mathbb{R}^2. The image of this projection is the line itself, so its dimension is 1. Thus, rank(P)=1\mathrm{rank}(P) = 1. For the composition P∘RP \circ R, we have: rank(P∘R)≤min⁡(rank(P),rank(R))=min⁡(1,2)=1\mathrm{rank}(P \circ R) \le \min(\mathrm{rank}(P), \mathrm{rank}(R)) = \min(1, 2) = 1 This means the rank of P∘RP \circ R is either 0 or 1. The rank is 0 only if P∘RP \circ R is the zero transformation, meaning it maps every vector to the zero vector. This would happen if the image of RR was entirely in the kernel of PP. The image of the rotation RR is all of R2\mathbb{R}^2. The kernel of the projection PP is the line perpendicular to y=xy=x, which is y=−xy=-x. Since the image of RR is not a subset of the kernel of PP, the composition is not the zero transformation. Therefore, its rank must be 1. A is incorrect: The transformation is not the zero map, so its rank is not 0. C is incorrect: The rank cannot exceed the rank of the projection matrix PP, which is 1. D is incorrect: The rank is a property of the linear transformation itself, not of the specific vector it acts upon.

Question 5

Let T:U→VT: U \to V and S:V→WS: V \to W be linear transformations. Which of the following statements about the image (also known as range) of the composition S∘TS \circ T is always true?

  1. Im(S∘T)=Im(S)\mathrm{Im}(S \circ T) = \mathrm{Im}(S)
  2. Im(S∘T)=Im(T)\mathrm{Im}(S \circ T) = \mathrm{Im}(T)
  3. Im(S)⊆Im(S∘T)\mathrm{Im}(S) \subseteq \mathrm{Im}(S \circ T)
  4. Im(S∘T)⊆Im(S)\mathrm{Im}(S \circ T) \subseteq \mathrm{Im}(S) (correct answer)
Explanation: The image of the composite transformation S∘TS \circ T is the set of all outputs (S∘T)(u)(S \circ T)(\mathbf{u}) for u∈U\mathbf{u} \in U. This can be written as S(T(u))S(T(\mathbf{u})). The set of all vectors T(u)T(\mathbf{u}) is the image of TT, denoted Im(T)\mathrm{Im}(T). So, the image of the composition is S(Im(T))S(\mathrm{Im}(T)). By definition, the image of SS, Im(S)\mathrm{Im}(S), is the set of all outputs S(v)S(\mathbf{v}) for all v∈V\mathbf{v} \in V. Since Im(T)\mathrm{Im}(T) is a subspace of VV, the set of vectors S(Im(T))S(\mathrm{Im}(T)) must be a subset of the set of all possible outputs of SS. Therefore, Im(S∘T)⊆Im(S)\mathrm{Im}(S \circ T) \subseteq \mathrm{Im}(S). A is incorrect: This is not always true. If TT is not surjective, Im(T)\mathrm{Im}(T) might not be all of VV. Then SS is applied to a smaller set than its full domain, potentially resulting in a smaller image. For example, if TT is the zero map, Im(S∘T)={0W}\mathrm{Im}(S \circ T) = \{\mathbf{0}_W\}, which is generally not equal to Im(S)\mathrm{Im}(S). B is incorrect: The image of S∘TS \circ T is a subspace of WW, while the image of TT is a subspace of VV. They cannot be equal unless V=WV=W and SS is the identity on Im(T)\mathrm{Im}(T). C is incorrect: This reverses the correct inclusion. For a counterexample, let S:R2→R2S: \mathbb{R}^2 \to \mathbb{R}^2 be the identity map and T:R2→R2T: \mathbb{R}^2 \to \mathbb{R}^2 be the projection onto the x-axis. Then Im(S)=R2\mathrm{Im}(S) = \mathbb{R}^2 but Im(S∘T)\mathrm{Im}(S \circ T) is only the x-axis.

Question 6

Given the following linear transformations: T1:R3→R2T_1: \mathbb{R}^3 \to \mathbb{R}^2 T2:R2→R3T_2: \mathbb{R}^2 \to \mathbb{R}^3 T3:R3→R3T_3: \mathbb{R}^3 \to \mathbb{R}^3 T4:R2→R2T_4: \mathbb{R}^2 \to \mathbb{R}^2 Which of the following composite transformations is NOT well-defined?

  1. T1∘T3T_1 \circ T_3
  2. T2∘T1T_2 \circ T_1
  3. T3∘T4T_3 \circ T_4 (correct answer)
  4. T4∘T1T_4 \circ T_1
Explanation: For a composition S∘TS \circ T to be well-defined, the codomain of the inner transformation (TT) must be the same as the domain of the outer transformation (SS). We check each option: A) T1∘T3T_1 \circ T_3: The inner transformation is T3:R3→R3T_3: \mathbb{R}^3 \to \mathbb{R}^3. Its codomain is R3\mathbb{R}^3. The outer transformation is T1:R3→R2T_1: \mathbb{R}^3 \to \mathbb{R}^2. Its domain is R3\mathbb{R}^3. Since the codomain of T3T_3 matches the domain of T1T_1, this composition is well-defined. The resulting transformation maps R3→R2\mathbb{R}^3 \to \mathbb{R}^2. B) T2∘T1T_2 \circ T_1: The inner transformation is T1:R3→R2T_1: \mathbb{R}^3 \to \mathbb{R}^2. Its codomain is R2\mathbb{R}^2. The outer transformation is T2:R2→R3T_2: \mathbb{R}^2 \to \mathbb{R}^3. Its domain is R2\mathbb{R}^2. Since the codomain of T1T_1 matches the domain of T2T_2, this composition is well-defined. The resulting transformation maps R3→R3\mathbb{R}^3 \to \mathbb{R}^3. C) T3∘T4T_3 \circ T_4: The inner transformation is T4:R2→R2T_4: \mathbb{R}^2 \to \mathbb{R}^2. Its codomain is R2\mathbb{R}^2. The outer transformation is T3:R3→R3T_3: \mathbb{R}^3 \to \mathbb{R}^3. Its domain is R3\mathbb{R}^3. Since the codomain of T4T_4 (R2\mathbb{R}^2) does not match the domain of T3T_3 (R3\mathbb{R}^3), this composition is NOT well-defined. D) T4∘T1T_4 \circ T_1: The inner transformation is T1:R3→R2T_1: \mathbb{R}^3 \to \mathbb{R}^2. Its codomain is R2\mathbb{R}^2. The outer transformation is T4:R2→R2T_4: \mathbb{R}^2 \to \mathbb{R}^2. Its domain is R2\mathbb{R}^2. Since the codomain of T1T_1 matches the domain of T4T_4, this composition is well-defined. The resulting transformation maps R3→R2\mathbb{R}^3 \to \mathbb{R}^2.

Question 7

Let T:R2→R2T: \mathbb{R}^2 \to \mathbb{R}^2 be a horizontal shear that maps the standard basis vector e2\mathbf{e}_2 to e2−2e1\mathbf{e}_2 - 2\mathbf{e}_1 and leaves e1\mathbf{e}_1 fixed. What is the standard matrix for the transformation T3=T∘T∘TT^3 = T \circ T \circ T?

  1. (1−801)\begin{pmatrix} 1 & -8 \\ 0 & 1 \end{pmatrix}
  2. (1−601)\begin{pmatrix} 1 & -6 \\ 0 & 1 \end{pmatrix} (correct answer)
  3. (3−603)\begin{pmatrix} 3 & -6 \\ 0 & 3 \end{pmatrix}
  4. (1−201)\begin{pmatrix} 1 & -2 \\ 0 & 1 \end{pmatrix}
Explanation: The standard matrix AA of a linear transformation TT has columns that are the images of the standard basis vectors. We are given T(e1)=e1=(10)T(\mathbf{e}_1) = \mathbf{e}_1 = \begin{pmatrix} 1 \\ 0 \end{pmatrix} and T(e2)=e2−2e1=(01)−2(10)=(−21)T(\mathbf{e}_2) = \mathbf{e}_2 - 2\mathbf{e}_1 = \begin{pmatrix} 0 \\ 1 \end{pmatrix} - 2\begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} -2 \\ 1 \end{pmatrix}. Thus, the standard matrix for TT is A=(1−201)A = \begin{pmatrix} 1 & -2 \\ 0 & 1 \end{pmatrix}. The matrix for T3T^3 is A3A^3. We compute this by repeated multiplication: A2=AA=(1−201)(1−201)=(1−2−201)=(1−401)A^2 = A A = \begin{pmatrix} 1 & -2 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & -2 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & -2-2 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & -4 \\ 0 & 1 \end{pmatrix} A3=A2A=(1−401)(1−201)=(1−2−401)=(1−601)A^3 = A^2 A = \begin{pmatrix} 1 & -4 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & -2 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & -2-4 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & -6 \\ 0 & 1 \end{pmatrix} A general pattern for this shear matrix is Ak=(1−2k01)A^k = \begin{pmatrix} 1 & -2k \\ 0 & 1 \end{pmatrix}. A is incorrect: This matrix results from incorrectly cubing the entries of the original matrix AA, specifically (−2)3=−8(-2)^3 = -8. Matrix exponentiation is not element-wise. C is incorrect: This is the matrix for 3T3T, which corresponds to scaling the transformation. This is not the same as applying the transformation three times. D is incorrect: This is the matrix for a single application of TT, not three.

Question 8

In R2\mathbb{R}^2, let RR be the linear transformation that rotates points 90∘90^\circ counterclockwise about the origin, and let FF be the linear transformation that reflects points across the line y=xy=x. Which matrix represents the composite transformation F∘RF \circ R?

  1. (−1001)\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}
  2. (100−1)\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} (correct answer)
  3. (0110)\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}
  4. (0−110)\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}
Explanation: First, we find the standard matrices for RR and FF. The matrix for a counterclockwise rotation by angle θ\theta is (cos⁡θ−sin⁡θsin⁡θcos⁡θ)\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}. For θ=90∘\theta = 90^\circ, the matrix for RR is AR=(0−110)A_R = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}. The matrix for reflection across the line y=xy=x swaps the coordinates, so its matrix is AF=(0110)A_F = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}. The matrix for the composition F∘RF \circ R is the product AFARA_F A_R. $$A_F A_R = \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} \begin{pmatrix} 0 & -1 \ 1 & 0 \end{pmatrix} = \begin{pmatrix} (0)(0)+(1)(1) & (0)(-1)+(1)(0) \ (1)(0)+(0)(1) & (1)(-1)+(0)(0) \end{pmatrix} = \begin{pmatrix} 1 & 0 \ 0 & -1 \end{pmatrix} **A is incorrect:** This is the matrix for $R \circ F$, which is $A_R A_F = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}$. This shows that composition of these transformations is not commutative. **C is incorrect:** This is the matrix for the reflection $F$ alone. **D is incorrect:** This is the matrix for the rotation $R$ alone.

Question 9

Let P1\mathcal{P}_1 be the vector space of polynomials of degree at most 1. Let T:P1→R2T: \mathcal{P}_1 \to \mathbb{R}^2 be defined by T(p(x))=(p(0)p′(0))T(p(x)) = \begin{pmatrix} p(0) \\ p'(0) \end{pmatrix} and S:R2→P1S: \mathbb{R}^2 \to \mathcal{P}_1 be defined by S((ab))=(a+b)+(a−b)xS\left(\begin{pmatrix} a \\ b \end{pmatrix}\right) = (a+b) + (a-b)x. What is (S∘T)(3−2x)(S \circ T)(3-2x)?

  1. 3+3x3+3x
  2. 5+x5+x
  3. 1+5x1+5x (correct answer)
  4. 3−2x3-2x
Explanation: We evaluate the composition (S∘T)(3−2x)(S \circ T)(3-2x) by first applying TT to the polynomial p(x)=3−2xp(x) = 3-2x, and then applying SS to the resulting vector. Step 1: Apply TT. We need to find p(0)p(0) and p′(0)p'(0). p(x)=3−2xp(x) = 3-2x, so p(0)=3−2(0)=3p(0) = 3-2(0) = 3. The derivative is p′(x)=−2p'(x) = -2, so p′(0)=−2p'(0) = -2. Therefore, T(3−2x)=(3−2)T(3-2x) = \begin{pmatrix} 3 \\ -2 \end{pmatrix}. Step 2: Apply SS to this vector. The definition of SS is S((ab))=(a+b)+(a−b)xS\left(\begin{pmatrix} a \\ b \end{pmatrix}\right) = (a+b) + (a-b)x. Here, a=3a=3 and b=−2b=-2. Plugging these values in: S((3−2))=(3+(−2))+(3−(−2))x=(1)+(5)x=1+5xS\left(\begin{pmatrix} 3 \\ -2 \end{pmatrix}\right) = (3 + (-2)) + (3 - (-2))x = (1) + (5)x = 1+5x Thus, (S∘T)(3−2x)=1+5x(S \circ T)(3-2x) = 1+5x. A is incorrect: This result would be obtained if the derivative p′(x)=−2p'(x)=-2 was incorrectly calculated as p′(x)=0p'(x)=0, leading to T(p(x))=(3,0)TT(p(x)) = (3,0)^T and S(3,0)=3+3xS(3,0) = 3+3x. B is incorrect: This result would be obtained if the roles of aa and bb were swapped in the second part of the formula for SS, i.e., using (a+b)+(b−a)x(a+b) + (b-a)x. This would give (3−2)+(−2−3)x=1−5x(3-2) + (-2-3)x = 1-5x. Another possibility is mis-assigning the components of the vector to the polynomial formula for SS. For example, if one computed (a−b)+(a+b)x(a-b)+(a+b)x, it would yield (3−(−2))+(3+(−2))x=5+x(3-(-2))+(3+(-2))x = 5+x. D is incorrect: This is the original polynomial. It would be the result if S∘TS \circ T were the identity transformation, which it is not.

Question 10

Let T:R5→R3T: \mathbb{R}^5 \to \mathbb{R}^3 and S:R3→R5S: \mathbb{R}^3 \to \mathbb{R}^5 be linear transformations. Consider the composite transformation L=S∘T:R5→R5L = S \circ T: \mathbb{R}^5 \to \mathbb{R}^5. Which of the following statements about LL must be true?

  1. LL is injective.
  2. LL is surjective.
  3. The nullity of LL is at least 2. (correct answer)
  4. The rank of LL is exactly 3.
Explanation: The rank of a linear transformation is at most the minimum of the dimensions of its domain and codomain. Thus, rank(T)≤min⁡(5,3)=3\mathrm{rank}(T) \le \min(5, 3) = 3 and rank(S)≤min⁡(3,5)=3\mathrm{rank}(S) \le \min(3, 5) = 3. The rank of a composite transformation S∘TS \circ T is at most the minimum of the ranks of SS and TT. Therefore, rank(L)=rank(S∘T)≤min⁡(rank(S),rank(T))≤3\mathrm{rank}(L) = \mathrm{rank}(S \circ T) \le \min(\mathrm{rank}(S), \mathrm{rank}(T)) \le 3. The domain of LL is R5\mathbb{R}^5. By the Rank-Nullity Theorem, rank(L)+nullity(L)=dim⁡(domain)=5\mathrm{rank}(L) + \mathrm{nullity}(L) = \dim(\text{domain}) = 5. Since rank(L)≤3\mathrm{rank}(L) \le 3, we can find a lower bound for the nullity: nullity(L)=5−rank(L)≥5−3=2\mathrm{nullity}(L) = 5 - \mathrm{rank}(L) \ge 5 - 3 = 2. So, the nullity of LL must be at least 2. A is incorrect: LL is injective if and only if its nullity is 0. Since the nullity is at least 2, LL cannot be injective. B is incorrect: For a transformation from R5\mathbb{R}^5 to R5\mathbb{R}^5, being surjective is equivalent to being injective. Since LL is not injective, it cannot be surjective. Alternatively, a surjective map from R5\mathbb{R}^5 to R5\mathbb{R}^5 must have rank 5, but we know rank(L)≤3\mathrm{rank}(L) \le 3. D is incorrect: The rank of LL can be less than 3. For example, if TT is the zero map, then LL is also the zero map, and its rank is 0.

Question 11

Consider linear transformations A:Rn→RnA: \mathbb{R}^n \to \mathbb{R}^n and B:Rn→RnB: \mathbb{R}^n \to \mathbb{R}^n where A2=AA^2 = A (AA is idempotent) and B2=IB^2 = I (BB is involutory). If C=B∘A∘BC = B \circ A \circ B, which property must CC satisfy?

  1. C2=CC^2 = C (CC is idempotent) (correct answer)
  2. C2=IC^2 = I (CC is involutory)
  3. C3=IC^3 = I (CC has order 3)
  4. C2=−IC^2 = -I (CC is a square root of −I-I)
Explanation: We need to compute C2C^2 where C=B∘A∘BC = B \circ A \circ B. Using the given properties A2=AA^2 = A and B2=IB^2 = I: C2=(B∘A∘B)∘(B∘A∘B)=B∘A∘B∘B∘A∘B=B∘A∘(B∘B)∘A∘B=B∘A∘I∘A∘B=B∘A∘A∘B=B∘A2∘B=B∘A∘B=CC^2 = (B \circ A \circ B) \circ (B \circ A \circ B) = B \circ A \circ B \circ B \circ A \circ B = B \circ A \circ (B \circ B) \circ A \circ B = B \circ A \circ I \circ A \circ B = B \circ A \circ A \circ B = B \circ A^2 \circ B = B \circ A \circ B = C. Therefore, C2=CC^2 = C, making CC idempotent. This shows that the conjugation of an idempotent by an involution preserves the idempotent property. Choice B would require CC to be involutory, but we showed C2=C≠IC^2 = C \neq I in general. Choice C suggests a different order structure. Choice D would make CC a complex structure, which doesn't follow from the given conditions.

Question 12

Consider the linear transformations S:R3→R3S: \mathbb{R}^3 \to \mathbb{R}^3 defined by S(x,y,z)=(y,z,x)S(x,y,z) = (y,z,x) and T:R3→R3T: \mathbb{R}^3 \to \mathbb{R}^3 defined by T(x,y,z)=(x,z,y)T(x,y,z) = (x,z,y). The transformation S∘T∘S−1S \circ T \circ S^{-1} has the effect of:

  1. Swapping the first and third coordinates while keeping the second coordinate fixed
  2. Swapping the second and third coordinates while keeping the first coordinate fixed
  3. Performing a cyclic permutation (x,y,z)↦(z,x,y)(x,y,z) \mapsto (z,x,y)
  4. Swapping the first and second coordinates while keeping the third coordinate fixed (correct answer)
Explanation: First, SS performs the cyclic permutation (x,y,z)↦(y,z,x)(x,y,z) \mapsto (y,z,x), so S−1S^{-1} performs (x,y,z)↦(z,x,y)(x,y,z) \mapsto (z,x,y). For any vector (x,y,z)(x,y,z): Step 1: S−1(x,y,z)=(z,x,y)S^{-1}(x,y,z) = (z,x,y). Step 2: T(S−1(x,y,z))=T(z,x,y)=(z,y,x)T(S^{-1}(x,y,z)) = T(z,x,y) = (z,y,x). Step 3: S(T(S−1(x,y,z)))=S(z,y,x)=(y,x,z)S(T(S^{-1}(x,y,z))) = S(z,y,x) = (y,x,z). Therefore, S∘T∘S−1(x,y,z)=(y,x,z)S \circ T \circ S^{-1}(x,y,z) = (y,x,z), which swaps the first and second coordinates while keeping the third fixed. Choice A incorrectly identifies which coordinates are swapped. Choice B corresponds to the transformation TT alone. Choice C corresponds to S2S^2. The key insight is recognizing that this is a conjugation that transforms TT's coordinate swapping via the change of basis induced by SS.

Question 13

Suppose P1P_1 and P2P_2 are orthogonal projections from R4\mathbb{R}^4 to R4\mathbb{R}^4 such that rank(P1)=2\text{rank}(P_1) = 2, rank(P2)=2\text{rank}(P_2) = 2, and P1∘P2=0P_1 \circ P_2 = 0. If Q=P1+P2Q = P_1 + P_2, what is the rank of QQ?

  1. rank(Q)=2\text{rank}(Q) = 2
  2. rank(Q)=3\text{rank}(Q) = 3
  3. rank(Q)=4\text{rank}(Q) = 4 (correct answer)
  4. rank(Q)=1\text{rank}(Q) = 1
Explanation: Since P1P_1 and P2P_2 are orthogonal projections with P1∘P2=0P_1 \circ P_2 = 0, their ranges are orthogonal subspaces. Let V1=im(P1)V_1 = \text{im}(P_1) and V2=im(P2)V_2 = \text{im}(P_2). We have dim(V1)=dim(V2)=2\text{dim}(V_1) = \text{dim}(V_2) = 2, and V1⊥V2V_1 \perp V_2 (since P1∘P2=0P_1 \circ P_2 = 0 implies ⟨P1(v),P2(w)⟩=0\langle P_1(v), P_2(w) \rangle = 0 for all v,wv,w). For any x∈R4x \in \mathbb{R}^4, we have Q(x)=P1(x)+P2(x)Q(x) = P_1(x) + P_2(x). Since P1(x)∈V1P_1(x) \in V_1, P2(x)∈V2P_2(x) \in V_2, and V1⊥V2V_1 \perp V_2, we get im(Q)=V1⊕V2\text{im}(Q) = V_1 \oplus V_2. Therefore, rank(Q)=dim(V1⊕V2)=dim(V1)+dim(V2)=2+2=4\text{rank}(Q) = \text{dim}(V_1 \oplus V_2) = \text{dim}(V_1) + \text{dim}(V_2) = 2 + 2 = 4. Choice A incorrectly assumes the ranges overlap significantly. Choice B assumes one dimension of overlap. Choice D severely underestimates by assuming near-complete overlap.

Question 14

Consider linear transformations F:R2→R2F: \mathbb{R}^2 \to \mathbb{R}^2 and G:R2→R2G: \mathbb{R}^2 \to \mathbb{R}^2 with matrix representations [F]=(1201)[F] = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} and [G]=(2013)[G] = \begin{pmatrix} 2 & 0 \\ 1 & 3 \end{pmatrix} . If H=F∘G∘F−1H = F \circ G \circ F^{-1}, what is the trace of the matrix representation of HH?

  1. tr(H)=5\text{tr}(H) = 5 (correct answer)
  2. tr(H)=6\text{tr}(H) = 6
  3. tr(H)=7\text{tr}(H) = 7
  4. tr(H)=4\text{tr}(H) = 4
Explanation: The matrix of H=F∘G∘F−1H = F \circ G \circ F^{-1} is [F][G][F]−1[F][G][F]^{-1}. Since [F]=(1201)[F] = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} , we have [F]−1=(1−201)[F]^{-1} = \begin{pmatrix} 1 & -2 \\ 0 & 1 \end{pmatrix} . The key insight is that HH represents a similarity transformation, so tr(H)=tr([F][G][F]−1)=tr([G])=2+3=5\text{tr}(H) = \text{tr}([F][G][F]^{-1}) = \text{tr}([G]) = 2 + 3 = 5. This follows from the cyclic property of trace: tr(ABC)=tr(CAB)\text{tr}(ABC) = \text{tr}(CAB), so tr([F][G][F]−1)=tr([G][F]−1[F])=tr([G]I)=tr([G])\text{tr}([F][G][F]^{-1}) = \text{tr}([G][F]^{-1}[F]) = \text{tr}([G]I) = \text{tr}([G]). Choice B incorrectly adds all matrix entries. Choice C results from computational errors in matrix multiplication. Choice D incorrectly computes the trace as the determinant of [G][G].

Question 15

Let α:R2→R3\alpha: \mathbb{R}^2 \to \mathbb{R}^3 be defined by α(x,y)=(x,x+y,2y)\alpha(x,y) = (x, x+y, 2y) and β:R3→R2\beta: \mathbb{R}^3 \to \mathbb{R}^2 be defined by β(a,b,c)=(a−b,c)\beta(a,b,c) = (a-b, c). If γ=α∘β∘α\gamma = \alpha \circ \beta \circ \alpha, then the nullity of γ\gamma is:

  1. 00
  2. 11 (correct answer)
  3. 22
  4. The transformation γ\gamma is not well-defined
Explanation: First, let's compute γ=α∘β∘α\gamma = \alpha \circ \beta \circ \alpha. For (x,y)∈R2(x,y) \in \mathbb{R}^2: Step 1: α(x,y)=(x,x+y,2y)\alpha(x,y) = (x, x+y, 2y). Step 2: β(α(x,y))=β(x,x+y,2y)=(x−(x+y),2y)=(−y,2y)\beta(\alpha(x,y)) = \beta(x, x+y, 2y) = (x-(x+y), 2y) = (-y, 2y). Step 3: γ(x,y)=α(β(α(x,y)))=α(−y,2y)=(−y,−y+2y,2(2y))=(−y,y,4y)\gamma(x,y) = \alpha(\beta(\alpha(x,y))) = \alpha(-y, 2y) = (-y, -y + 2y, 2(2y)) = (-y, y, 4y). So γ(x,y)=(−y,y,4y)\gamma(x,y) = (-y, y, 4y). To find the nullity, we solve γ(x,y)=(0,0,0)\gamma(x,y) = (0,0,0): −y=0-y = 0, y=0y = 0, 4y=04y = 0. All equations give y=0y = 0, with xx arbitrary. Therefore, ker⁡(γ)={(x,0):x∈R}=span{(1,0)}\ker(\gamma) = \{(x,0) : x \in \mathbb{R}\} = \text{span}\{(1,0)\}, which has dimension 1. Hence, the nullity is 1. Choice A would mean γ\gamma is injective. Choice C would mean γ=0\gamma = 0. Choice D incorrectly suggests a domain/codomain mismatch.

Question 16

For which of the following pairs of linear transformations T,S:R2→R2T, S: \mathbb{R}^2 \to \mathbb{R}^2 does the property S∘T=T∘SS \circ T = T \circ S hold (i.e., the transformations commute)?

  1. TT is a reflection across the x-axis, and SS is a rotation by 90∘90^\circ counterclockwise.
  2. TT is a projection onto the x-axis, and SS is a rotation by 90∘90^\circ counterclockwise.
  3. TT is a horizontal shear with factor 1, and SS is a rotation by 90∘90^\circ counterclockwise.
  4. TT is a reflection across the x-axis, and SS is a reflection across the y-axis. (correct answer)
Explanation: Two transformations SS and TT commute if their standard matrices ASA_S and ATA_T commute, i.e., ASAT=ATASA_S A_T = A_T A_S. Let's check the matrices for each option. A: AT=(100−1)A_T = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}, AS=(0−110)A_S = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}. ASAT=(0110)A_S A_T = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, but ATAS=(0−1−10)A_T A_S = \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}. They do not commute. B: AT=(1000)A_T = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, AS=(0−110)A_S = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}. ASAT=(0010)A_S A_T = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}, but ATAS=(0−100)A_T A_S = \begin{pmatrix} 0 & -1 \\ 0 & 0 \end{pmatrix}. They do not commute. C: AT=(1101)A_T = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}, AS=(0−110)A_S = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}. ASAT=(0−111)A_S A_T = \begin{pmatrix} 0 & -1 \\ 1 & 1 \end{pmatrix}, but ATAS=(1−110)A_T A_S = \begin{pmatrix} 1 & -1 \\ 1 & 0 \end{pmatrix}. They do not commute. D: AT=(100−1)A_T = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} (reflection across x-axis), AS=(−1001)A_S = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} (reflection across y-axis). Both matrices are diagonal. ASAT=(−1001)(100−1)=(−100−1)A_S A_T = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} ATAS=(100−1)(−1001)=(−100−1)A_T A_S = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} Since ASAT=ATASA_S A_T = A_T A_S, the transformations commute. The composition results in a rotation by 180∘180^\circ.

Question 17

Let T:U→VT: U \to V and S:V→WS: V \to W be linear transformations between vector spaces. Which statement regarding the kernel of the composition S∘TS \circ T is always true?

  1. ker⁡(S∘T)⊆ker⁡(T)\ker(S \circ T) \subseteq \ker(T)
  2. ker⁡(T)⊆ker⁡(S∘T)\ker(T) \subseteq \ker(S \circ T) (correct answer)
  3. ker⁡(S∘T)=ker⁡(S)\ker(S \circ T) = \ker(S)
  4. Im(T)⊆ker⁡(S)\mathrm{Im}(T) \subseteq \ker(S)
Explanation: The kernel of a transformation consists of all vectors in the domain that map to the zero vector in the codomain. Let's prove that ker⁡(T)⊆ker⁡(S∘T)\ker(T) \subseteq \ker(S \circ T). Let u\mathbf{u} be an arbitrary vector in ker⁡(T)\ker(T). By definition of the kernel, T(u)=0VT(\mathbf{u}) = \mathbf{0}_V, where 0V\mathbf{0}_V is the zero vector in VV. Now, let's apply the composite transformation S∘TS \circ T to u\mathbf{u}: (S∘T)(u)=S(T(u))=S(0V)(S \circ T)(\mathbf{u}) = S(T(\mathbf{u})) = S(\mathbf{0}_V) Since SS is a linear transformation, it maps the zero vector to the zero vector, so S(0V)=0WS(\mathbf{0}_V) = \mathbf{0}_W. Therefore, (S∘T)(u)=0W(S \circ T)(\mathbf{u}) = \mathbf{0}_W. This means that u\mathbf{u} is in the kernel of S∘TS \circ T. Since this holds for any u∈ker⁡(T)\mathbf{u} \in \ker(T), we have ker⁡(T)⊆ker⁡(S∘T)\ker(T) \subseteq \ker(S \circ T). A is incorrect: This reverses the correct inclusion. For a counterexample, consider SS being a non-injective map. Vectors outside ker⁡(T)\ker(T) can be mapped by TT into ker⁡(S)\ker(S), thus becoming part of ker⁡(S∘T)\ker(S \circ T). C is incorrect: This is not true in general as ker⁡(S∘T)\ker(S \circ T) is a subspace of UU while ker⁡(S)\ker(S) is a subspace of VV. The statement is dimensionally and conceptually flawed. D is incorrect: This statement implies that for every vector u∈U\mathbf{u} \in U, S(T(u))=0WS(T(\mathbf{u})) = \mathbf{0}_W. This would mean S∘TS \circ T is the zero transformation, which is not always the case.