Composition of Transformations - Linear Algebra

1. Let $T: U \to V$ and $S: V \to W$ be linear transformations between vector spaces. Which statement regarding the kernel of the composition $S \circ T$ is always true?
   Answer: $\ker(T) \subseteq \ker(S \circ T)$ 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 $\ker(T) \subseteq \ker(S \circ T)$. Let $\mathbf{u}$ be an arbitrary vector in $\ker(T)$. By definition of the kernel, $T(\mathbf{u}) = \mathbf{0}_V$, where $\mathbf{0}_V$ is the zero vector in $V$. Now, let's apply the composite transformation $S \circ T$ to $\mathbf{u}$: $(S \circ T)(\mathbf{u}) = S(T(\mathbf{u})) = S(\mathbf{0}_V)$ Since $S$ is a linear transformation, it maps the zero vector to the zero vector, so $S(\mathbf{0}_V) = \mathbf{0}_W$. Therefore, $(S \circ T)(\mathbf{u}) = \mathbf{0}_W$. This means that $\mathbf{u}$ is in the kernel of $S \circ T$. Since this holds for any $\mathbf{u} \in \ker(T)$, we have $\ker(T) \subseteq \ker(S \circ T)$. A is incorrect: This reverses the correct inclusion. For a counterexample, consider $S$ being a non-injective map. Vectors outside $\ker(T)$ can be mapped by $T$ into $\ker(S)$, thus becoming part of $\ker(S \circ T)$. C is incorrect: This is not true in general as $\ker(S \circ T)$ is a subspace of $U$ while $\ker(S)$ is a subspace of $V$. The statement is dimensionally and conceptually flawed. D is incorrect: This statement implies that for every vector $\mathbf{u} \in U$, $S(T(\mathbf{u})) = \mathbf{0}_W$. This would mean $S \circ T$ is the zero transformation, which is not always the case.
2. Consider linear transformations $A: \mathbb{R}^n \to \mathbb{R}^n$ and $B: \mathbb{R}^n \to \mathbb{R}^n$ where $A^2 = A$ ($A$ is idempotent) and $B^2 = I$ ($B$ is involutory). If $C = B \circ A \circ B$, which property must $C$ satisfy?
   Answer: $C^2 = C$ ($C$ is idempotent) We need to compute $C^2$ where $C = B \circ A \circ B$. Using the given properties $A^2 = A$ and $B^2 = I$: $C^2 = (B \circ A \circ B) \circ (B \circ A \circ B) = B \circ A \circ B \circ B \circ A \circ B = B \circ A \circ (B \circ B) \circ A \circ B = B \circ A \circ I \circ A \circ B = B \circ A \circ A \circ B = B \circ A^2 \circ B = B \circ A \circ B = C$. Therefore, $C^2 = C$, making $C$ idempotent. This shows that the conjugation of an idempotent by an involution preserves the idempotent property. Choice B would require $C$ to be involutory, but we showed $C^2 = C \neq I$ in general. Choice C suggests a different order structure. Choice D would make $C$ a complex structure, which doesn't follow from the given conditions.
3. Let $T: U \to V$ and $S: V \to W$ be linear transformations. Which of the following statements about the image (also known as range) of the composition $S \circ T$ is always true?
   Answer: $\mathrm{Im}(S \circ T) \subseteq \mathrm{Im}(S)$ The image of the composite transformation $S \circ T$ is the set of all outputs $(S \circ T)(\mathbf{u})$ for $\mathbf{u} \in U$. This can be written as $S(T(\mathbf{u}))$. The set of all vectors $T(\mathbf{u})$ is the image of $T$, denoted $\mathrm{Im}(T)$. So, the image of the composition is $S(\mathrm{Im}(T))$. By definition, the image of $S$, $\mathrm{Im}(S)$, is the set of all outputs $S(\mathbf{v})$ for all $\mathbf{v} \in V$. Since $\mathrm{Im}(T)$ is a subspace of $V$, the set of vectors $S(\mathrm{Im}(T))$ must be a subset of the set of all possible outputs of $S$. Therefore, $\mathrm{Im}(S \circ T) \subseteq \mathrm{Im}(S)$. A is incorrect: This is not always true. If $T$ is not surjective, $\mathrm{Im}(T)$ might not be all of $V$. Then $S$ is applied to a smaller set than its full domain, potentially resulting in a smaller image. For example, if $T$ is the zero map, $\mathrm{Im}(S \circ T) = \{\mathbf{0}_W\}$, which is generally not equal to $\mathrm{Im}(S)$. B is incorrect: The image of $S \circ T$ is a subspace of $W$, while the image of $T$ is a subspace of $V$. They cannot be equal unless $V=W$ and $S$ is the identity on $\mathrm{Im}(T)$. C is incorrect: This reverses the correct inclusion. For a counterexample, let $S: \mathbb{R}^2 \to \mathbb{R}^2$ be the identity map and $T: \mathbb{R}^2 \to \mathbb{R}^2$ be the projection onto the x-axis. Then $\mathrm{Im}(S) = \mathbb{R}^2$ but $\mathrm{Im}(S \circ T)$ is only the x-axis.
4. For which of the following pairs of linear transformations $T, S: \mathbb{R}^2 \to \mathbb{R}^2$ does the property $S \circ T = T \circ S$ hold (i.e., the transformations commute)?
   Answer: $T$ is a reflection across the x-axis, and $S$ is a reflection across the y-axis. Two transformations $S$ and $T$ commute if their standard matrices $A_S$ and $A_T$ commute, i.e., $A_S A_T = A_T A_S$. Let's check the matrices for each option. A: $A_T = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$, $A_S = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$. $A_S A_T = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$, but $A_T A_S = \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}$. They do not commute. B: $A_T = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$, $A_S = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$. $A_S A_T = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}$, but $A_T A_S = \begin{pmatrix} 0 & -1 \\ 0 & 0 \end{pmatrix}$. They do not commute. C: $A_T = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$, $A_S = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$. $A_S A_T = \begin{pmatrix} 0 & -1 \\ 1 & 1 \end{pmatrix}$, but $A_T A_S = \begin{pmatrix} 1 & -1 \\ 1 & 0 \end{pmatrix}$. They do not commute. D: $A_T = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$ (reflection across x-axis), $A_S = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}$ (reflection across y-axis). Both matrices are diagonal. $A_S A_T = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}$ $A_T A_S = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}$ Since $A_S A_T = A_T A_S$, the transformations commute. The composition results in a rotation by $180^\circ$.
5. Let $T: \mathbb{R}^2 \to \mathbb{R}^2$ be a linear transformation with standard matrix $A_T = \begin{pmatrix} 1 & 2 \\ 1 & 3 \end{pmatrix}$. Let $L: \mathbb{R}^2 \to \mathbb{R}^2$ be the composite transformation $L = S \circ T$, with standard matrix $A_L = \begin{pmatrix} 2 & 1 \\ 5 & 4 \end{pmatrix}$. If $S: \mathbb{R}^2 \to \mathbb{R}^2$ is also a linear transformation, what is its standard matrix $A_S$?
   Answer: $\begin{pmatrix} 5 & -3 \\ 11 & -6 \end{pmatrix}$ The standard matrix of the composition $L = S \circ T$ is given by the product of the individual standard matrices: $A_L = A_S A_T$. To find the matrix $A_S$, we need to isolate it by right-multiplying by the inverse of $A_T$. $A_S = A_L A_T^{-1}$ First, we must calculate $A_T^{-1}$. The determinant of $A_T$ is $\det(A_T) = (1)(3) - (2)(1) = 1$. Since the determinant is non-zero, the inverse exists. For a $2 \times 2$ matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$, the inverse is $\frac{1}{ad-bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$. So, $A_T^{-1} = \frac{1}{1} \begin{pmatrix} 3 & -2 \\ -1 & 1 \end{pmatrix} = \begin{pmatrix} 3 & -2 \\ -1 & 1 \end{pmatrix}$ Now, we compute the product $A_S = A_L A_T^{-1}$: $A_S = \begin{pmatrix} 2 & 1 \\ 5 & 4 \end{pmatrix} \begin{pmatrix} 3 & -2 \\ -1 & 1 \end{pmatrix} = \begin{pmatrix} (2)(3)+(1)(-1) & (2)(-2)+(1)(1) \\ (5)(3)+(4)(-1) & (5)(-2)+(4)(1) \end{pmatrix} = \begin{pmatrix} 5 & -3 \\ 11 & -6 \end{pmatrix}$ 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.
6. Given the following linear transformations: $T_1: \mathbb{R}^3 \to \mathbb{R}^2$ $T_2: \mathbb{R}^2 \to \mathbb{R}^3$ $T_3: \mathbb{R}^3 \to \mathbb{R}^3$ $T_4: \mathbb{R}^2 \to \mathbb{R}^2$ Which of the following composite transformations is NOT well-defined?
   Answer: $T_3 \circ T_4$ For a composition $S \circ T$ to be well-defined, the codomain of the inner transformation ($T$) must be the same as the domain of the outer transformation ($S$). We check each option: **A) $T_1 \circ T_3$:** The inner transformation is $T_3: \mathbb{R}^3 \to \mathbb{R}^3$. Its codomain is $\mathbb{R}^3$. The outer transformation is $T_1: \mathbb{R}^3 \to \mathbb{R}^2$. Its domain is $\mathbb{R}^3$. Since the codomain of $T_3$ matches the domain of $T_1$, this composition is well-defined. The resulting transformation maps $\mathbb{R}^3 \to \mathbb{R}^2$. **B) $T_2 \circ T_1$:** The inner transformation is $T_1: \mathbb{R}^3 \to \mathbb{R}^2$. Its codomain is $\mathbb{R}^2$. The outer transformation is $T_2: \mathbb{R}^2 \to \mathbb{R}^3$. Its domain is $\mathbb{R}^2$. Since the codomain of $T_1$ matches the domain of $T_2$, this composition is well-defined. The resulting transformation maps $\mathbb{R}^3 \to \mathbb{R}^3$. **C) $T_3 \circ T_4$:** The inner transformation is $T_4: \mathbb{R}^2 \to \mathbb{R}^2$. Its codomain is $\mathbb{R}^2$. The outer transformation is $T_3: \mathbb{R}^3 \to \mathbb{R}^3$. Its domain is $\mathbb{R}^3$. Since the codomain of $T_4$ ($\mathbb{R}^2$) does not match the domain of $T_3$ ($\mathbb{R}^3$), this composition is NOT well-defined. **D) $T_4 \circ T_1$:** The inner transformation is $T_1: \mathbb{R}^3 \to \mathbb{R}^2$. Its codomain is $\mathbb{R}^2$. The outer transformation is $T_4: \mathbb{R}^2 \to \mathbb{R}^2$. Its domain is $\mathbb{R}^2$. Since the codomain of $T_1$ matches the domain of $T_4$, this composition is well-defined. The resulting transformation maps $\mathbb{R}^3 \to \mathbb{R}^2$.
7. Let $\alpha: \mathbb{R}^2 \to \mathbb{R}^3$ be defined by $\alpha(x,y) = (x, x+y, 2y)$ and $\beta: \mathbb{R}^3 \to \mathbb{R}^2$ be defined by $\beta(a,b,c) = (a-b, c)$. If $\gamma = \alpha \circ \beta \circ \alpha$, then the nullity of $\gamma$ is:
   Answer: $1$ First, let's compute $\gamma = \alpha \circ \beta \circ \alpha$. For $(x,y) \in \mathbb{R}^2$: Step 1: $\alpha(x,y) = (x, x+y, 2y)$. Step 2: $\beta(\alpha(x,y)) = \beta(x, x+y, 2y) = (x-(x+y), 2y) = (-y, 2y)$. Step 3: $\gamma(x,y) = \alpha(\beta(\alpha(x,y))) = \alpha(-y, 2y) = (-y, -y + 2y, 2(2y)) = (-y, y, 4y)$. So $\gamma(x,y) = (-y, y, 4y)$. To find the nullity, we solve $\gamma(x,y) = (0,0,0)$: $-y = 0$, $y = 0$, $4y = 0$. All equations give $y = 0$, with $x$ arbitrary. Therefore, $\ker(\gamma) = \{(x,0) : x \in \mathbb{R}\} = \text{span}\{(1,0)\}$, which has dimension 1. Hence, the nullity is 1. Choice A would mean $\gamma$ is injective. Choice C would mean $\gamma = 0$. Choice D incorrectly suggests a domain/codomain mismatch.
8. Let $T: \mathbb{R}^2 \to \mathbb{R}^3$ be a linear transformation with standard matrix $A = \begin{pmatrix} 1 & -1 \\ 2 & 0 \\ 0 & 3 \end{pmatrix}$ and let $S: \mathbb{R}^3 \to \mathbb{R}^2$ be a linear transformation with standard matrix $B = \begin{pmatrix} 1 & 0 & 1 \\ 0 & -1 & 2 \end{pmatrix}$. What is the standard matrix of the composite transformation $S \circ T$?
   Answer: $\begin{pmatrix} 1 & 2 \\ -2 & 6 \end{pmatrix}$ The standard matrix of a composite transformation $S \circ T$ is the product of the standard matrices of $S$ and $T$, in that order. The matrix for $S$ is $B$ and the matrix for $T$ is $A$. The resulting matrix is $BA$. 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 $2 \times 2$.
9. Let $T: \mathbb{R}^2 \to \mathbb{R}^2$ be a horizontal shear that maps the standard basis vector $\mathbf{e}_2$ to $\mathbf{e}_2 - 2\mathbf{e}_1$ and leaves $\mathbf{e}_1$ fixed. What is the standard matrix for the transformation $T^3 = T \circ T \circ T$?
   Answer: $\begin{pmatrix} 1 & -6 \\ 0 & 1 \end{pmatrix}$ The standard matrix $A$ of a linear transformation $T$ has columns that are the images of the standard basis vectors. We are given $T(\mathbf{e}_1) = \mathbf{e}_1 = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ and $T(\mathbf{e}_2) = \mathbf{e}_2 - 2\mathbf{e}_1 = \begin{pmatrix} 0 \\ 1 \end{pmatrix} - 2\begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} -2 \\ 1 \end{pmatrix}$. Thus, the standard matrix for $T$ is $A = \begin{pmatrix} 1 & -2 \\ 0 & 1 \end{pmatrix}$. The matrix for $T^3$ is $A^3$. We compute this by repeated multiplication: $A^2 = A A = \begin{pmatrix} 1 & -2 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & -2 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & -2-2 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & -4 \\ 0 & 1 \end{pmatrix}$ $A^3 = A^2 A = \begin{pmatrix} 1 & -4 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & -2 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & -2-4 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & -6 \\ 0 & 1 \end{pmatrix}$ A general pattern for this shear matrix is $A^k = \begin{pmatrix} 1 & -2k \\ 0 & 1 \end{pmatrix}$. A is incorrect: This matrix results from incorrectly cubing the entries of the original matrix $A$, specifically $(-2)^3 = -8$. Matrix exponentiation is not element-wise. C is incorrect: This is the matrix for $3T$, 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 $T$, not three.
10. Consider the linear transformations $S: \mathbb{R}^3 \to \mathbb{R}^3$ defined by $S(x,y,z) = (y,z,x)$ and $T: \mathbb{R}^3 \to \mathbb{R}^3$ defined by $T(x,y,z) = (x,z,y)$. The transformation $S \circ T \circ S^{-1}$ has the effect of:
    Answer: Swapping the first and second coordinates while keeping the third coordinate fixed First, $S$ performs the cyclic permutation $(x,y,z) \mapsto (y,z,x)$, so $S^{-1}$ performs $(x,y,z) \mapsto (z,x,y)$. For any vector $(x,y,z)$: Step 1: $S^{-1}(x,y,z) = (z,x,y)$. Step 2: $T(S^{-1}(x,y,z)) = T(z,x,y) = (z,y,x)$. Step 3: $S(T(S^{-1}(x,y,z))) = S(z,y,x) = (y,x,z)$. Therefore, $S \circ T \circ S^{-1}(x,y,z) = (y,x,z)$, 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 $T$ alone. Choice C corresponds to $S^2$. The key insight is recognizing that this is a conjugation that transforms $T$'s coordinate swapping via the change of basis induced by $S$.
11. Consider linear transformations $F: \mathbb{R}^2 \to \mathbb{R}^2$ and $G: \mathbb{R}^2 \to \mathbb{R}^2$ with matrix representations $[F] = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}$ and $[G] = \begin{pmatrix} 2 & 0 \\ 1 & 3 \end{pmatrix}$. If $H = F \circ G \circ F^{-1}$, what is the trace of the matrix representation of $H$?
    Answer: $\text{tr}(H) = 5$ The matrix of $H = F \circ G \circ F^{-1}$ is $[F][G][F]^{-1}$. Since $[F] = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}$, we have $[F]^{-1} = \begin{pmatrix} 1 & -2 \\ 0 & 1 \end{pmatrix}$. The key insight is that $H$ represents a similarity transformation, so $\text{tr}(H) = \text{tr}([F][G][F]^{-1}) = \text{tr}([G]) = 2 + 3 = 5$. This follows from the cyclic property of trace: $\text{tr}(ABC) = \text{tr}(CAB)$, so $\text{tr}([F][G][F]^{-1}) = \text{tr}([G][F]^{-1}[F]) = \text{tr}([G]I) = \text{tr}([G])$. 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 $[G]$.
12. In $\mathbb{R}^2$, let $R$ be the linear transformation that rotates points $90^\circ$ counterclockwise about the origin, and let $F$ be the linear transformation that reflects points across the line $y=x$. Which matrix represents the composite transformation $F \circ R$?
    Answer: $\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$ First, we find the standard matrices for $R$ and $F$. The matrix for a counterclockwise rotation by angle $\theta$ is $\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$. For $\theta = 90^\circ$, the matrix for $R$ is $A_R = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$. The matrix for reflection across the line $y=x$ swaps the coordinates, so its matrix is $A_F = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$. The matrix for the composition $F \circ R$ is the product $A_F A_R$. $A_F A_R = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} (0)(0)+(1)(1) & (0)(-1)+(1)(0) \\ (1)(0)+(0)(1) & (1)(-1)+(0)(0) \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$ This resulting transformation is a reflection across the x-axis. 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.
13. Let $T: \mathbb{R}^2 \to \mathbb{R}^3$ be defined by $T(x_1, x_2) = (x_1 - x_2, 3x_2, x_1+x_2)$ and $S: \mathbb{R}^3 \to \mathbb{R}^2$ be defined by $S(y_1, y_2, y_3) = (y_1+y_3, y_2-y_3)$. What is the formula for the composition $(S \circ T)(x_1, x_2)$?
    Answer: $(2x_1, 2x_2 - x_1)$ To find the formula for $(S \circ T)(x_1, x_2)$, we first apply $T$ to $(x_1, x_2)$ and then apply $S$ to the result. $(S \circ T)(x_1, x_2) = S(T(x_1, x_2)) = S(x_1 - x_2, 3x_2, x_1+x_2)$ Now we use the definition of $S$ with $y_1 = x_1 - x_2$, $y_2 = 3x_2$, and $y_3 = x_1+x_2$. $S(y_1, y_2, y_3) = (y_1+y_3, y_2-y_3)$ Substituting the expressions in terms of $x_1$ and $x_2$: $((x_1 - x_2) + (x_1+x_2), (3x_2) - (x_1+x_2))$ Simplifying each component gives: $(2x_1, 3x_2 - x_1 - x_2) = (2x_1, -x_1 + 2x_2)$ The correct answer is $(2x_1, 2x_2 - x_1)$. B is incorrect: This may result from calculation errors, for example, calculating the first component as $(x_1 - x_2) - (x_1+x_2) = -2x_2$. C is incorrect: This results from a sign error in the second component: $(3x_2) + (x_1+x_2) = x_1+4x_2$, which is also not the choice, but sign errors are common. This specific choice may come from $(3x_2)- (x_1-x_2)$ instead of $(3x_2)-(x_1+x_2)$. D is incorrect: The composition is well-defined because the codomain of $T$ ($\mathbb{R}^3$) matches the domain of $S$ ($\mathbb{R}^3$).
14. Let $\mathcal{P}_1$ be the vector space of polynomials of degree at most 1. Let $T: \mathcal{P}_1 \to \mathbb{R}^2$ be defined by $T(p(x)) = \begin{pmatrix} p(0) \\ p'(0) \end{pmatrix}$ and $S: \mathbb{R}^2 \to \mathcal{P}_1$ be defined by $S\left(\begin{pmatrix} a \\ b \end{pmatrix}\right) = (a+b) + (a-b)x$. What is $(S \circ T)(3-2x)$?
    Answer: $1+5x$ We evaluate the composition $(S \circ T)(3-2x)$ by first applying $T$ to the polynomial $p(x) = 3-2x$, and then applying $S$ to the resulting vector. Step 1: Apply $T$. We need to find $p(0)$ and $p'(0)$. $p(x) = 3-2x$, so $p(0) = 3-2(0) = 3$. The derivative is $p'(x) = -2$, so $p'(0) = -2$. Therefore, $T(3-2x) = \begin{pmatrix} 3 \\ -2 \end{pmatrix}$. Step 2: Apply $S$ to this vector. The definition of $S$ is $S\left(\begin{pmatrix} a \\ b \end{pmatrix}\right) = (a+b) + (a-b)x$. Here, $a=3$ and $b=-2$. Plugging these values in: $S\left(\begin{pmatrix} 3 \\ -2 \end{pmatrix}\right) = (3 + (-2)) + (3 - (-2))x = (1) + (5)x = 1+5x$ Thus, $(S \circ T)(3-2x) = 1+5x$. A is incorrect: This result would be obtained if the derivative $p'(x)=-2$ was incorrectly calculated as $p'(x)=0$, leading to $T(p(x)) = (3,0)^T$ and $S(3,0) = 3+3x$. B is incorrect: This result would be obtained if the roles of $a$ and $b$ were swapped in the second part of the formula for $S$, i.e., using $(a+b) + (b-a)x$. This would give $(3-2) + (-2-3)x = 1-5x$. Another possibility is mis-assigning the components of the vector to the polynomial formula for $S$. For example, if one computed $(a-b)+(a+b)x$, it would yield $(3-(-2))+(3+(-2))x = 5+x$. D is incorrect: This is the original polynomial. It would be the result if $S \circ T$ were the identity transformation, which it is not.
15. Suppose $P_1$ and $P_2$ are orthogonal projections from $\mathbb{R}^4$ to $\mathbb{R}^4$ such that $\text{rank}(P_1) = 2$, $\text{rank}(P_2) = 2$, and $P_1 \circ P_2 = 0$. If $Q = P_1 + P_2$, what is the rank of $Q$?
    Answer: $\text{rank}(Q) = 4$ Since $P_1$ and $P_2$ are orthogonal projections with $P_1 \circ P_2 = 0$, their ranges are orthogonal subspaces. Let $V_1 = \text{im}(P_1)$ and $V_2 = \text{im}(P_2)$. We have $\text{dim}(V_1) = \text{dim}(V_2) = 2$, and $V_1 \perp V_2$ (since $P_1 \circ P_2 = 0$ implies $\langle P_1(v), P_2(w) \rangle = 0$ for all $v,w$). For any $x \in \mathbb{R}^4$, we have $Q(x) = P_1(x) + P_2(x)$. Since $P_1(x) \in V_1$, $P_2(x) \in V_2$, and $V_1 \perp V_2$, we get $\text{im}(Q) = V_1 \oplus V_2$. Therefore, $\text{rank}(Q) = \text{dim}(V_1 \oplus V_2) = \text{dim}(V_1) + \text{dim}(V_2) = 2 + 2 = 4$. Choice A incorrectly assumes the ranges overlap significantly. Choice B assumes one dimension of overlap. Choice D severely underestimates by assuming near-complete overlap.
16. Let $T: \mathbb{R}^5 \to \mathbb{R}^3$ and $S: \mathbb{R}^3 \to \mathbb{R}^5$ be linear transformations. Consider the composite transformation $L = S \circ T: \mathbb{R}^5 \to \mathbb{R}^5$. Which of the following statements about $L$ must be true?
    Answer: The nullity of $L$ is at least 2. The rank of a linear transformation is at most the minimum of the dimensions of its domain and codomain. Thus, $\mathrm{rank}(T) \le \min(5, 3) = 3$ and $\mathrm{rank}(S) \le \min(3, 5) = 3$. The rank of a composite transformation $S \circ T$ is at most the minimum of the ranks of $S$ and $T$. Therefore, $\mathrm{rank}(L) = \mathrm{rank}(S \circ T) \le \min(\mathrm{rank}(S), \mathrm{rank}(T)) \le 3$. The domain of $L$ is $\mathbb{R}^5$. By the Rank-Nullity Theorem, $\mathrm{rank}(L) + \mathrm{nullity}(L) = \dim(\text{domain}) = 5$. Since $\mathrm{rank}(L) \le 3$, we can find a lower bound for the nullity: $\mathrm{nullity}(L) = 5 - \mathrm{rank}(L) \ge 5 - 3 = 2$. So, the nullity of $L$ must be at least 2. A is incorrect: $L$ is injective if and only if its nullity is 0. Since the nullity is at least 2, $L$ cannot be injective. B is incorrect: For a transformation from $\mathbb{R}^5$ to $\mathbb{R}^5$, being surjective is equivalent to being injective. Since $L$ is not injective, it cannot be surjective. Alternatively, a surjective map from $\mathbb{R}^5$ to $\mathbb{R}^5$ must have rank 5, but we know $\mathrm{rank}(L) \le 3$. D is incorrect: The rank of $L$ can be less than 3. For example, if $T$ is the zero map, then $L$ is also the zero map, and its rank is 0.
17. Let $R: \mathbb{R}^2 \to \mathbb{R}^2$ be a counterclockwise rotation by $45^\circ$, and let $P: \mathbb{R}^2 \to \mathbb{R}^2$ be the orthogonal projection onto the line $y=x$. What is the rank of the composite transformation $P \circ R$?
    Answer: 1 The rank of a composite transformation $S \circ T$ is bounded by the ranks of the individual transformations: $\mathrm{rank}(S \circ T) \le \min(\mathrm{rank}(S), \mathrm{rank}(T))$. The transformation $R$ is a rotation, which is an invertible transformation. Its standard matrix is invertible, so $\mathrm{rank}(R) = 2$. The transformation $P$ is a projection onto a one-dimensional subspace (a line) in $\mathbb{R}^2$. The image of this projection is the line itself, so its dimension is 1. Thus, $\mathrm{rank}(P) = 1$. For the composition $P \circ R$, we have: $\mathrm{rank}(P \circ R) \le \min(\mathrm{rank}(P), \mathrm{rank}(R)) = \min(1, 2) = 1$ This means the rank of $P \circ R$ is either 0 or 1. The rank is 0 only if $P \circ R$ is the zero transformation, meaning it maps every vector to the zero vector. This would happen if the image of $R$ was entirely in the kernel of $P$. The image of the rotation $R$ is all of $\mathbb{R}^2$. The kernel of the projection $P$ is the line perpendicular to $y=x$, which is $y=-x$. Since the image of $R$ is not a subset of the kernel of $P$, 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 $P$, which is 1. D is incorrect: The rank is a property of the linear transformation itself, not of the specific vector it acts upon.

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