Mathematics for Data Science II · Quiz 2 · Sep 2025

**1. Recall the matrix property of an orthogonal linear transformation:** A linear transformation $T: \mathbb{R}^3 \to \mathbb{R}^3$ is an **orthogonal transformation** if it preserves inner products ($\langle Tv, Tw \rangle = \langle v, w \rangle$). With respect to the standard orthonormal basis of $\mathbb{R}^3$, its matrix representation $A$ is an **orthogonal matrix**, meaning its transpose equals its inverse: $$A^T = A^{-1} \iff A A^T = A^T A = I$$ **Answer: Option C** A: **Incorrect:** For an orthogonal matrix, $A^T = A^{-1}$, not $-A^{-1}$. B: **Incorrect:** For any square matrix, $\det(A^T) = \det(A)$, not $-\det(A)$. C: **Correct:** The matrix representation $A$ of an orthogonal transformation with respect to the standard orthonormal basis satisfies $A A^T = I$. D: **Incorrect:** The columns of $A$ are mutually orthogonal to each other (and rows are mutually orthogonal to each other), not columns to rows.

**1. Define an affine subspace vs. a vector subspace:** - A **vector subspace** $V$ must contain the zero vector $\mathbf{0}$ and be closed under addition and scalar multiplication (the solution set of a homogeneous linear system $Ax = \mathbf{0}$). - An **affine subspace** is a translate of a vector subspace, $L = v_0 + V$, which is the solution set of a consistent linear system $Ax = b$. It is an affine subspace **but not a vector subspace** if and only if $b \ne \mathbf{0}$ (so $\mathbf{0} \notin L$). **2. Check each option:** - **Option A:** Contains the non-linear term $wz$, so it is neither a vector subspace nor an affine subspace. - **Option B:** The solution space of the homogeneous linear system $x + y = 0,\; y + z = 0$ is a vector subspace (contains $(0, 0, 0)$). - **Option C:** The union of two lines is not closed under affine combinations, so it is neither a vector subspace nor an affine subspace. - **Option D:** The solution set of the non-homogeneous linear system $9x - 3y + 10z = 6,\; 13y + 11z = 7$ is a non-empty translate of the null space of the coefficient matrix, and $(0, 0, 0)$ is not a solution (since $0 \ne 6$ and $0 \ne 7$). Thus, it is an **affine subspace but not a vector subspace**. **Answer: Option D** A: **Incorrect:** Due to the non-linear product $wz$, $L$ is not an affine subspace. B: **Incorrect:** This homogeneous linear system defines a vector subspace containing the origin $(0,0,0)$. C: **Incorrect:** The union of two distinct lines is not an affine subspace. D: **Correct:** The solution set of a consistent non-homogeneous system $Ax = b$ ($b \ne \mathbf{0}$) is an affine subspace that does not contain the origin, so it is not a vector subspace.

**1. Evaluate each statement for a linear transformation $T: \mathbb{R}^3 \to \mathbb{R}^3$:** - **Option A:** By linearity, $T(x + y, z, 0) = T((x, z, 0) + (y, 0, 0)) = T(x, z, 0) + T(y, 0, 0)$, which is not equal to $T(x, z, 0) + T(y, z, 0)$ unless $T(0, z, 0) = \mathbf{0}$. (**Incorrect**) - **Option B:** A linear transformation is injective (one-to-one) if and only if its kernel (null space) is trivial, i.e., $\ker(T) = \{(0, 0, 0)\}$. (**Correct**) - **Option C:** Since $(x + y, z, 0) = (x, 0, 0) + (y, z, 0)$, linearity of $T$ directly gives: $$T(x + y, z, 0) = T(x, 0, 0) + T(y, z, 0)$$ (**Correct**) - **Option D:** By the Rank-Nullity Theorem for $T: \mathbb{R}^3 \to \mathbb{R}^3$, $\text{Nullity}(T) + \text{Rank}(T) = \dim(\mathbb{R}^3) = 3$, not $< 3$. (**Incorrect**) - **Option E:** For a linear operator $T: V \to V$ on a finite-dimensional vector space $V$ ($\dim V = 3$), $T$ is surjective if and only if $\text{Rank}(T) = 3 \iff \text{Nullity}(T) = 0 \iff T$ is injective, so $T$ is bijective. (**Correct**) **Answer: Options B, C, E** A: **Incorrect:** $(x, z, 0) + (y, z, 0) = (x + y, 2z, 0) \ne (x + y, z, 0)$ for $z \ne 0$. B: **Correct:** A linear map is injective if and only if $\ker(T) = \{(0, 0, 0)\}$. C: **Correct:** Since $(x, 0, 0) + (y, z, 0) = (x + y, z, 0)$, additivity of $T$ gives $T(x+y, z, 0) = T(x, 0, 0) + T(y, z, 0)$. D: **Incorrect:** By the Rank-Nullity Theorem, $\text{Nullity}(T) + \text{Rank}(T) = 3$. E: **Correct:** A linear map from a finite-dimensional space $\mathbb{R}^3$ to itself is surjective if and only if it is bijective.

**1. Determine for which non-negative integers $k$ the function $T(x, y, z) = \big(k^2x,\; (z + y)^{\sqrt{k}}\big)$ is linear:** - For $T$ to be a linear map, we must have $T(0, 0, 0) = (0, 0)$ and the second component $(z + y)^{\sqrt{k}}$ must be homogeneous of degree $1$ and additive in $(y, z)$. - **If $k = 0$:** $(z + y)^0 = 1$, so $T(0, 0, 0) = (0, 1) \ne (0, 0)$. Thus, for $k = 0$, $T$ is **not** a linear map (**Option C is correct**). - **If $k > 0$:** Homogeneity $T(cx, cy, cz) = cT(x, y, z)$ requires $(c(z + y))^{\sqrt{k}} = c(z + y)^{\sqrt{k}}$ for all $c, y, z \in \mathbb{R}$, which holds if and only if $\sqrt{k} = 1 \iff k = 1$. - Thus, $T$ is linear **only when $k = 1$** (so **Options A and D are incorrect**). **2. Find the null space of $T$ when $k = 1$:** For $k = 1$, $T(x, y, z) = (x, z + y)$. Setting $T(x, y, z) = (0, 0)$ gives: $$x = 0 \quad \text{and} \quad y + z = 0 \implies z = -y$$ Thus, any vector in $\ker(T)$ has the form $(0, y, -y) = y(0, 1, -1)$, so $\{(0, 1, -1)\}$ is a basis of the null space of $T$ (**Option B is correct**). **Answer: Options B, C** A: **Incorrect:** If $T$ is linear, the exponent $\sqrt{k}$ must equal $1$, so $k = 1$, not $k = 2$. B: **Correct:** When $k = 1$, $T(x, y, z) = (x, y + z) = (0, 0) \iff x = 0, z = -y$, so $\{(0, 1, -1)\}$ is a basis for $\ker(T)$. C: **Correct:** When $k = 0$, $T(0, 0, 0) = (0, 1) \ne (0, 0)$, so $T$ is not a linear map. D: **Incorrect:** When $k = 4$, $\sqrt{k} = 2$, so the second component is $(z + y)^2$, which is quadratic and not linear.

**1. Write the matrix representation $A$ of $T(x, y) = (-3x + 4y, 6x - ky)$ with respect to the standard basis:** $$A = \begin{bmatrix} -3 & 4 \\ 6 & -k \end{bmatrix}$$ The determinant of $A$ is: $$\det(A) = (-3)(-k) - (4)(6) = 3k - 24 = 3(k - 8)$$ **2. Check each option:** - **Option A:** When $k \ne 8$, $\det(A) \ne 0$ so $\text{rank}(T) = 2$ and $\text{nullity}(T) = 0$. When $k = 8$, $\det(A) = 0$ so $\text{rank}(T) = 1$ and $\text{nullity}(T) = 1$. Thus rank and nullity depend on $k$. (**Incorrect**) - **Option B:** For $k = -8$, $\det(A) = -48 \ne 0$, so $\ker(T) = \{(0, 0)\}$ (in fact, $(4, 3)$ spans $\ker(T)$ when $k = +8$, not $-8$). (**Incorrect**) - **Option C:** Since $\det(A) = 3(k - 8) \ne 0$ for all $k \ne 8$, the rank of $T$ is $2$ if $k \ne 8$. (**Correct**) - **Option D:** For $k = 0$, $A = \begin{bmatrix} -3 & 4 \\ 6 & 0 \end{bmatrix}$ has $\text{tr}(A) = -3$, whereas $B = \begin{bmatrix} 1 & 1 \\ -2 & 1 \end{bmatrix}$ has $\text{tr}(B) = 2 \ne -3$, so they are not similar. (**Incorrect**) - **Option E:** The columns of $A$ are $c_1 = (-3, 6)$ and $c_2 = (4, -k)$. Their dot product is: $$c_1 \cdot c_2 = (-3)(4) + (6)(-k) = -12 - 6k$$ Thus, $c_1 \cdot c_2 = 0 \iff -12 - 6k = 0 \iff k = -2$. (**Correct**) **Answer: Options C, E** A: **Incorrect:** At $k = 8$, $\text{rank}(T) = 1$, whereas for $k \ne 8$, $\text{rank}(T) = 2$. B: **Incorrect:** $(4, 3)$ is in $\ker(T)$ when $k = 8$, whereas for $k = -8$, $\ker(T) = \{(0, 0)\}$. C: **Correct:** Since $\det(A) = 3k - 24$, $\det(A) \ne 0$ and $\text{rank}(T) = 2$ whenever $k \ne 8$. D: **Incorrect:** For $k = 0$, $\text{tr}(A) = -3 \ne \text{tr}\begin{pmatrix} 1 & 1 \\ -2 & 1 \end{pmatrix} = 2$, so they cannot be similar. E: **Correct:** The dot product of the columns $(-3, 6)$ and $(4, -k)$ is $-12 - 6k = 0 \iff k = -2$.

**1. Understand how the matrix representation changes when changing the domain basis:** We are given $T: \mathbb{R}^2 \to \mathbb{R}^2$ with standard matrix: $$A = \begin{pmatrix} 3 & -8 \\ -1 & 0 \end{pmatrix}$$ So for any vector $(x, y) \in \mathbb{R}^2$, its image in the standard co-domain basis is: $$T(x, y) = \begin{pmatrix} 3 & -8 \\ -1 & 0 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 3x - 8y \\ -x \end{pmatrix}$$ **2. Evaluate $T$ on the domain basis $\{(1, 1), (0, 1)\}$ (Options A and C):** - $T(1, 1) = \begin{pmatrix} 3(1) - 8(1) \\ -1 \end{pmatrix} = \begin{pmatrix} -5 \\ -1 \end{pmatrix}$ - $T(0, 1) = \begin{pmatrix} 3(0) - 8(1) \\ 0 \end{pmatrix} = \begin{pmatrix} -8 \\ 0 \end{pmatrix}$ Putting these as columns gives $\begin{pmatrix} -5 & -8 \\ -1 & 0 \end{pmatrix}$. (**Option C is correct**, Option A is incorrect.) **3. Evaluate $T$ on the domain basis $\{(1, 0), (1, 1)\}$ (Options B and D):** - $T(1, 0) = \begin{pmatrix} 3 \\ -1 \end{pmatrix}$ - $T(1, 1) = \begin{pmatrix} -5 \\ -1 \end{pmatrix}$ Putting these as columns gives $\begin{pmatrix} 3 & -5 \\ -1 & -1 \end{pmatrix}$. (**Option B is correct**, Option D is incorrect.) **Answer: Options B, C** A: **Incorrect:** $T(1, 1) = (3 - 8, -1) = (-5, -1)$, not $(5, 1)$. B: **Correct:** The images of $(1, 0)$ and $(1, 1)$ are $(3, -1)$ and $(-5, -1)$, forming columns $\begin{pmatrix} 3 & -5 \\ -1 & -1 \end{pmatrix}$. C: **Correct:** The images of $(1, 1)$ and $(0, 1)$ are $(-5, -1)$ and $(-8, 0)$, forming columns $\begin{pmatrix} -5 & -8 \\ -1 & 0 \end{pmatrix}$. D: **Incorrect:** The signs of the entries do not match $T(1, 0) = (3, -1)$ and $T(1, 1) = (-5, -1)$.

**1. Set the dot product of $v_1$ and $v_2$ equal to zero:** Wait — notice that $v_1 = (2, 4, -1, 0)$ and $v_2 = (-3, 0, a, 1)$ are 4-component vectors (with the usual dot product): $$v_1 \cdot v_2 = (2)(-3) + (4)(0) + (-1)(a) + (0)(1) = 0$$ **2. Solve for $a$:** $$-6 + 0 - a + 0 = 0 \implies a = -6$$ **Answer: -6**

**1. Use linearity of $T$ to express $T(-1, 1, 2)$ in terms of the standard basis images:** Since $(-1, 1, 2) = -1(1, 0, 0) + 1(0, 1, 0) + 2(0, 0, 1)$, if $T$ is a linear transformation we must have: $$T(-1, 1, 2) = -T(1, 0, 0) + T(0, 1, 0) + 2T(0, 0, 1)$$ Substitute the given values $T(1, 0, 0) = (k, 1, 0)$, $T(0, 1, 0) = (0, 0, 1)$, and $T(0, 0, 1) = (1, k, 0)$: $$T(-1, 1, 2) = -(k, 1, 0) + (0, 0, 1) + 2(1, k, 0) = (2 - k,\; 2k - 1,\; 1)$$ **2. Equate with the given expression $T(-1, 1, 2) = (k^2 + 2,\; k^3 + 2k,\; 1)$:** 1. First component: $$k^2 + 2 = 2 - k \implies k^2 + k = 0 \implies k(k + 1) = 0 \implies k = 0 \text{ or } k = -1$$ 2. Second component: $$k^3 + 2k = 2k - 1 \implies k^3 = -1 \implies k = -1$$ Both components are satisfied simultaneously if and only if **$k = -1$**. **Answer: -1**

**1. Write the matrix $A$ of the linear map $T: \mathbb{R}^4 \to \mathbb{R}^4$ and find its rank $\beta$:** Given $T(x, y, z, w) = (3x + 4y + w,\; x - y,\; y + 3z,\; 4x + 4y + 3z + w)$, the rows of its matrix representation are: - $R_1 = (3, 4, 0, 1)$ - $R_2 = (1, -1, 0, 0)$ - $R_3 = (0, 1, 3, 0)$ - $R_4 = (4, 4, 3, 1)$ Notice that: $$R_1 + R_2 + R_3 = (3 + 1 + 0,\; 4 - 1 + 1,\; 0 + 0 + 3,\; 1 + 0 + 0) = (4, 4, 3, 1) = R_4$$ Thus, $R_4$ is a linear combination of $R_1, R_2, R_3$. Moreover, $R_1, R_2, R_3$ are clearly linearly independent (since only $R_1$ has a non-zero 4th entry, only $R_3$ has a non-zero 3rd entry, and $R_2 \ne 0$). Therefore, the rank is $\beta = 3$. **2. Find the nullity $\eta$ and compute $4\eta - 2\beta$:** By the Rank-Nullity Theorem on $\mathbb{R}^4$: $$\eta + \beta = 4 \implies \eta = 4 - 3 = 1$$ Hence: $$4\eta - 2\beta = 4(1) - 2(3) = 4 - 6 = -2$$ **Answer: -2**

**1. Compute the orthogonal projection $w$ of $v_1 = (-1, 4, 3)$ onto $v_2 = (2, 0, 1)$:** The formula for the projection of $v_1$ onto $v_2$ is: $$w = \text{proj}_{v_2}(v_1) = \frac{v_1 \cdot v_2}{\|v_2\|^2} v_2$$ - Dot product $v_1 \cdot v_2 = (-1)(2) + (4)(0) + (3)(1) = -2 + 0 + 3 = 1$. - Squared norm $\|v_2\|^2 = 2^2 + 0^2 + 1^2 = 5$. Therefore: $$w = (k_1, k_2, k_3) = \frac{1}{5}(2, 0, 1) = \left(\frac{2}{5},\; 0,\; \frac{1}{5}\right)$$ **2. Compute $5k_1 + k_2 + 5k_3$:** $$5k_1 + k_2 + 5k_3 = 5\left(\frac{2}{5}\right) + 0 + 5\left(\frac{1}{5}\right) = 2 + 0 + 1 = 3$$ **Answer: 3**

**1. Determine the set $S_1$ for which $f(v, w) = (5 - k)v \cdot w$ is an inner product:** Since the standard dot product $v \cdot w$ is an inner product on $\mathbb{R}^3$, a scalar multiple $c(v \cdot w)$ satisfies all inner product axioms (specifically positive-definiteness $f(v, v) = (5 - k)\|v\|^2 > 0$ for all $v \ne \mathbf{0}$) if and only if the scalar is strictly positive: $$5 - k > 0 \iff k < 5$$ For $k \in \mathbb{Z}$, $S_1 = \{k \in \mathbb{Z} \mid k \le 4\}$. **2. Determine the set $S_2$ for which $g(v) = (6 + k)\sqrt{v_1^2 + v_2^2 + v_3^2}$ defines a norm:** Since $\|v\|_2 = \sqrt{v_1^2 + v_2^2 + v_3^2}$ is the standard Euclidean norm on $\mathbb{R}^3$, a scalar multiple $(6 + k)\|v\|_2$ defines a valid norm (satisfying positivity $g(v) > 0$ for $v \ne \mathbf{0}$ and positive homogeneity) if and only if: $$6 + k > 0 \iff k > -6$$ For $k \in \mathbb{Z}$, $S_2 = \{k \in \mathbb{Z} \mid k \ge -5\}$. **3. Find the cardinality of $S_1 \cap S_2$:** $$S_1 \cap S_2 = \{k \in \mathbb{Z} \mid -5 \le k \le 4\} = \{-5, -4, -3, -2, -1, 0, 1, 2, 3, 4\}$$ The number of integers from $-5$ to $4$ inclusive is $4 - (-5) + 1 = 10$. **Answer: 10**