Mathematics for Data Science II · Quiz 2 · Jan 2026

**1. Solve the nonhomogeneous equations:** The second equation is the negative of the first. From x+z=0, z=-x. Substitution into -x+y-z=1 gives y=1. Taking x=t, \[L=\{(t,1,-t):t\in\mathbb R\}=(0,1,0)+\operatorname{span}\{(1,0,-1)\}.\] **2. Identify the direction subspace:** The corresponding linear subspace consists of differences between solutions, or equivalently the solutions of Ax=0. Thus \(W=\operatorname{span}\{(1,0,-1)\}\). The affine offset (0,1,0) is not a direction vector. **Answer: B.** A: **Incorrect:** The extra vector (1,1,0) does not satisfy the homogeneous equations and adds a false direction. B: **Correct:** This is exactly the one free direction in the solution line. C: **Incorrect:** This adds another independent direction although the actual direction space is one-dimensional. D: **Incorrect:** (1,0,1) violates x+z=0.

**1. Use rank-nullity:** The domain has dimension 3 and rank(T)=2, so \[\dim\ker T=3-2=1.\] A nonzero kernel implies that distinct inputs can have the same output, so T is not injective. **2. Interpret idempotence:** T squared equals T, meaning that applying T again does not change vectors already in its image. It does not mean that T is the identity on every input. Its image is a two-dimensional plane, and that plane need not be a coordinate plane with a zero component in every vector. **Answer: C.** A: **Incorrect:** The identity on R3 has rank 3, not 2. B: **Incorrect:** A general image plane need not be a coordinate plane; it can contain vectors with all components nonzero. C: **Correct:** Nullity 1 proves failure of injectivity. D: **Incorrect:** The image has dimension 2 and therefore is not all of R3.

**1. Write the functional:** For fixed v, \(T_v(u)=\langle u,v\rangle\) maps R2 into R. **2. Prove surjectivity for v nonzero:** Given any scalar a, choose \(u=(a/\|v\|^2)v\). Then \(T_v(u)=a\), so every real number is attained. **3. Check injectivity:** When v is nonzero, the nonzero vector perpendicular to v maps to zero, giving a one-dimensional kernel. When v=0, every vector maps to zero. Therefore the map is never one-to-one, and no such map can be an isomorphism between spaces of dimensions 2 and 1. **Answer: B, C.** A: **Incorrect:** A perpendicular nonzero vector is in the kernel for every nonzero v. B: **Correct:** The displayed choice of u produces any requested scalar output. C: **Correct:** For v=0 the kernel is all R2; otherwise its dimension is 1. D: **Incorrect:** The domain and codomain have different dimensions, so a linear isomorphism is impossible.

**1. Recall matrix equivalence:** Two same-sized matrices are equivalent under invertible row and column operations exactly when they have the same rank. Equivalently, they can be connected by multiplication by invertible matrices on the left and right. **2. Use nullity:** Since A and B each have n columns, equal nullities imply equal ranks through rank+nullity=n. **3. Use the equivalence relation:** If A is equivalent to C and B is equivalent to C, symmetry and transitivity imply A is equivalent to B. Also, PA=BQ with P and Q invertible gives B=PAQ inverse, which directly supplies equivalence. **4. Reject insufficient conditions:** Equal determinants can both be zero while ranks differ; for example the 2-by-2 zero matrix and diag(1,0). Both column sets may likewise be dependent but have different ranks. **Answer: B, D, E.** A: **Incorrect:** Equal zero determinants do not force equal ranks. B: **Correct:** Equal nullity for the same number of columns forces equal rank. C: **Incorrect:** Linear dependence says only that rank is below n, not that the ranks match. D: **Correct:** Equivalence through a common matrix implies equivalence by transitivity. E: **Correct:** Invertible left and right transformations give the definition of equivalence.

**1. Map the spanning vectors:** U is spanned by u1=(1,0,1) and u2=(0,1,1). Their images are a=(1,2,1) and b=(0,1,1), which are independent: the first coordinate forces the coefficient of a to be zero, and then the coefficient of b is zero. **2. Determine dimension and injectivity:** Hence T(U)=span{a,b} has dimension 2. If T(su1+tu2)=sa+tb=0, independence gives s=t=0. Thus U meets ker(T) only at zero, and the restriction to U is injective. **3. Test the proposed vector:** For (1,-1,0)=sa+tb, the first coordinate gives s=1 and the second gives t=-3. The third coordinate would be 1-3=-2, not 0, so this vector does not belong to T(U). **Answer: A, C.** A: **Correct:** The two independent image vectors span a two-dimensional subspace. B: **Incorrect:** Its coordinates cannot be matched by a linear combination of the two image generators. C: **Correct:** The restriction has zero kernel because those images are independent. D: **Incorrect:** The restriction is injective, as its kernel calculation shows.

**1. Form the matrix:** \[A=\begin{pmatrix}1&1&0\\0&1&1\\1&k&1\end{pmatrix},\qquad\det A=(1-k)+1=2-k.\] Therefore T is injective when k is not 2, and rank is 3 in that case. At k=2 its rank is 2. **2. Test the proposed kernel vector:** \(T(1,-1,1)=(0,0,2-k)\), so it lies in the kernel when k=2. **3. Test fixed nonzero vectors:** T(x,y,z)=(x,y,z) implies y=0 from the first component and z=0 from the second. The third then implies x=0. Thus the only fixed vector is zero for every k. **Answer: D, E.** A: **Incorrect:** The exceptional value for injectivity is k=2, not k=-1. B: **Incorrect:** At k=-2, determinant is 4 and nullity is 0. C: **Incorrect:** The image is all R3 whenever k is not 2. D: **Correct:** The vector belongs to the kernel for k=2. E: **Correct:** Solving T(v)=v forces every component to be zero.

**1. Form the standard matrix:** From T(x,y,z)=(x-2y,y+z), \[A=\begin{pmatrix}1&-2&0\\0&1&1\end{pmatrix}.\] **2. Find rank:** Its first two columns have determinant 1 as a 2-by-2 minor, so rank is at least 2. There are only two rows, so rank is exactly 2. **3. Use equivalence:** Multiplication by invertible matrices and elementary row/column operations preserve rank. B is equivalent to A, hence rank(B)=2. **Answer: 2.**

**1. Write the kernel conditions:** T(a,b,c)=(a+2b-c,3a+b+c)=0 gives a+2b-c=0 and 3a+b+c=0. **2. Solve the relations:** The first gives c=a+2b. Substitute into the second: 4a+3b=0, so b=-4a/3 and c=-5a/3. **3. Use the additional constraint:** a+b+c=1 becomes -2a=1. Hence a=-1/2, b=2/3 and c=5/6. **4. Evaluate the expression:** \[a+2b+3c=-\frac12+\frac43+\frac{15}{6}=\frac{10}{3}=3.333\ldots.\] Rounded to two decimal places, the result is 3.33. **Answer: 3.33.**

**1. Use the two constraints:** L consists of points satisfying x+y+z=1 and x-y=0. Set x=y=t, which gives z=1-2t. **2. Transform the point:** \[T(t,t,1-2t)=(t+2t-(1-2t),3t+t+(1-2t))=(5t-1,2t+1).\] **3. Intersect with the y-axis:** On the y-axis, the first output coordinate is zero. Thus 5t-1=0 and t=1/5. The second coordinate is 2/5+1=7/5=1.4. The intersection is (0,1.4). **Answer: 1.4.**

**1. Apply the domain restriction:** Every vector in U satisfies x+y+z=0, so x=-y-z. **2. Substitute into T:** \[S(x,y,z)=T(x,y,z)=(0,y+2z,y+2z)=(y+2z)(0,1,1).\] Thus every output lies on the line spanned by (0,1,1). **3. Show the line is attained:** Choosing y=1,z=0,x=-1 gives the nonzero output (0,1,1). Therefore the image is exactly this line, and its dimension is 1. **Answer: 1.**

**1. Use the image formula on U:** For x+y+z=0, S(x,y,z)=(0,y+2z,y+2z). **2. Impose membership in span{(1,1,1)}:** Any vector in that span has equal coordinates. Since the first output coordinate is zero, all three must be zero. Hence y+2z=0. **3. Solve for v:** Combining y=-2z with x+y+z=0 gives x=z. Thus v=z(1,-2,1). **4. Use its squared norm:** \(\|v\|^2=z^2(1+4+1)=6z^2=6\). Therefore z=1 or -1, producing (1,-2,1) and (-1,2,-1). **Answer: A, C.** A: **Correct:** This lies in U, maps to zero and has squared norm 6. B: **Incorrect:** Its coordinates sum to 4, so it is not in U. C: **Correct:** It is the negative of the first valid vector and satisfies the same conditions. D: **Incorrect:** Its coordinates sum to -4, so it is not in U.

**1. Describe W:** The equations x+y=0 and z=0 give W={t(1,-1,0):t real}. Therefore dim(W)=1. **2. Describe orthogonality:** A vector (a,b,c) is perpendicular to (1,-1,0) when a-b=0. Thus W perpendicular consists of (a,a,c), which has two independent free parameters. **3. Give a basis:** (1,1,0) and (0,0,1) span this perpendicular space and are independent. Its dimension is 2, also agreeing with dim(W)+dim(W perpendicular)=3. **Answer: 2.**

**1. Find the line:** W=span{(1,-1,0)}. A basis of this one-dimensional space has exactly one nonzero vector on that line. **2. Normalize the generator:** Its length is \(\sqrt{1^2+(-1)^2}=\sqrt2\). Therefore \((1,-1,0)/\sqrt2\) is a unit vector. Its negative is equally valid. **3. Test the remaining candidates:** (1,-1,0) has length sqrt(2), not 1. (1,1,0)/sqrt(2) has length 1 but is not in W because its first two coordinates do not sum to zero. **Answer: A, B.** A: **Correct:** This is a unit vector spanning W. B: **Correct:** The negative of a unit basis vector is also an orthonormal basis of the same line. C: **Incorrect:** This vector spans W but is not normalized to length 1. D: **Incorrect:** This unit vector is perpendicular to the required line rather than lying in it.

**1. Use the projection formula:** Let u=(1,-1,0) span W and v=(1,1,2). The orthogonal projection is \[P_W(v)=\frac{v\cdot u}{u\cdot u}u.\] **2. Evaluate the dot products:** v dot u=1-1+0=0 and u dot u=1+1=2. Therefore P_W(v)=(0/2)u=(0,0,0). **3. Add the output coordinates:** a=b=c=0, so a+b+c=0. This also follows from the fact that v already lies in W perpendicular. **Answer: 0.**