Mathematics for Data Science II · Quiz 2 · May 2026

**1. Distinguish equivalence and similarity:** Matrix equivalence allows independent invertible changes on the left and right: B=PAQ. Similarity is the more restrictive relation B=P inverse AP for a square matrix. **2. Derive the implication:** A similarity is a particular equivalence, since both P inverse and P are invertible. Therefore similar matrices are equivalent. **3. Reject the converse:** Equivalent matrices need only have the same rank. Similar matrices must also share characteristic polynomial, eigenvalues and trace. For example, the one-by-one matrices [1] and [2] are equivalent but not similar. Independent changes of domain and codomain bases give equivalence, not necessarily similarity. **Answer: B.** A: **Incorrect:** Equivalence only fixes rank and does not imply similarity. B: **Correct:** The two invertible factors in a similarity also give an equivalence. C: **Incorrect:** Same determinant is neither the full criterion for equivalence nor sufficient when both determinants are zero. D: **Incorrect:** Independent basis changes in domain and codomain need not be the same similarity basis change.

**1. Use the dimension bound:** In a vector space of dimension n, a largest independent set has n vectors. Therefore every set with more than n vectors is dependent. **2. Check related facts:** Exactly n independent vectors do form a basis. An onto transformation may have a nonzero kernel if its domain has greater dimension than its codomain. Row space and column space have the same dimension, but they need not be the same sets and may even lie in different ambient spaces. **Answer: C.** A: **Incorrect:** Exactly n independent vectors in dimension n necessarily form a basis. B: **Incorrect:** The onto map (x,y) maps to x has a nonzero kernel. C: **Correct:** More than n vectors exceed the maximum possible independent-set size. D: **Incorrect:** Equal dimension of row and column spaces does not make those spaces identical.

**1. Expand using the inner product:** \[\|u+v\|^2=\|u\|^2+2(u\cdot v)+\|v\|^2.\] The sum-of-squares equality holds precisely when u dot v=0. Likewise, \[\|u-v\|^2=\|u\|^2+\|v\|^2-2(u\cdot v).\] **2. Recall affine spaces:** A nonempty affine subspace is a translate v+U of a linear subspace U. **3. Test nonhomogeneous solution sets:** Ax=b can be inconsistent. If it is consistent, its solutions form x0+ker(A), usually a translated subspace. Unless b=0, the solution set need not contain zero and therefore need not be a vector subspace. **Answer: A, B, C.** A: **Correct:** The cross term vanishes exactly when u and v are orthogonal. B: **Correct:** This is the standard inner-product expansion of the squared difference. C: **Correct:** A nonempty affine subspace is a translate of a linear subspace. D: **Incorrect:** A nonhomogeneous system can have no solutions or an affine solution set that excludes zero.

**1. Solve for the kernel:** T(x,y,z)=(x+y+z,x-y+z)=0. Subtract the equations to get 2y=0, so y=0 and z=-x. Hence \[\ker T=\operatorname{span}\{(1,0,-1)\},\qquad\dim\ker T=1.\] **2. Determine the range:** The matrix has independent rows (1,1,1) and (1,-1,1), so rank is 2. Since the codomain is R2, the map is onto. **3. Test the claimed basis:** (-1,0,1) is the negative of (1,0,-1). Both lie in the kernel, but the two-vector set is dependent and is therefore not a basis. **Answer: A, C, D.** A: **Correct:** The computed kernel has one free scalar parameter. B: **Incorrect:** The proposed two vectors are scalar multiples, so the set is not a basis. C: **Correct:** Rank 2 equals the codomain dimension and proves onto. D: **Correct:** For example, the nonzero vector (1,0,-1) maps to zero.

**1. Write the matrices:** \[A=\begin{pmatrix}2&1\\1&1\end{pmatrix},\qquad B=\begin{pmatrix}1&c\\1&2\end{pmatrix}.\] They both have trace 3, but det(A)=1 and det(B)=2-c. For c=2 the determinants differ, so similarity is impossible. **2. Prove similarity at c=1:** Choose \(P=\begin{pmatrix}1&2\\1&1\end{pmatrix}\), whose determinant is -1. Direct multiplication gives AP=PB=[[3,5],[2,3]]. Thus P inverse AP=B, proving P is true. For the other candidate P=[[1,0],[1,2]], AP=[[3,2],[2,2]] whereas PB=[[1,1],[3,5]], so R is false and S is true. **3. Show infinitely many changes of basis:** Every nonzero scalar multiple tP is invertible and satisfies (tP) inverse A(tP)=B. Therefore T is true. **Answer: A, D, E.** A: **Correct:** Similarity holds at c=1, while unequal determinants rule it out at c=2. B: **Incorrect:** The assertion for c=2 is false. C: **Incorrect:** The first proposed P does not satisfy AP=PB. D: **Correct:** The direct multiplications show R false and S true. E: **Correct:** Nonzero scalar multiples of a valid P give infinitely many valid matrices.

**1. Find rank:** The second row of A is twice the first. The first and third rows, (1,2,1,3) and (1,1,0,2), are independent. Thus rank(A)=2. **2. Use rank-nullity:** The domain is R4, so nullity=4-2=2. The image has dimension 2 and cannot equal R3. **3. Test the column-space vector:** 5 column 1 +4 column 3 gives 5(1,2,1)+4(1,2,0)=(9,18,5), establishing membership. **4. Apply dimension theorems:** For equal finite dimensions, injective implies full rank and therefore onto. If the domain dimension exceeds the codomain dimension, nullity is positive, preventing injectivity. **Answer: A, C, D, E.** A: **Correct:** Rank-nullity gives a two-dimensional kernel. B: **Incorrect:** Rank 2 is below the codomain dimension 3. C: **Correct:** Injective means rank equals the common dimension, so onto follows. D: **Correct:** The explicit combination 5c1+4c3 equals (9,18,5). E: **Correct:** The rank cannot exceed the smaller codomain dimension, forcing a nonzero kernel.

**1. Reduce the equations:** A has rows (1,1,1,1) and twice that row; b has matching entries 3 and 6. Thus the system is consistent and equivalent to x1+x2+x3+x4=3. **2. Count free coordinates:** Choose x2, x3 and x4 freely, then x1=3-x2-x3-x4. This gives three independent directions. **3. Apply the affine-dimension formula:** The solution space is a translate of ker(A), so dim(L)=4-rank(A)=4-1=3. **Answer: 3.**

**1. Substitute the specified kernel vector:** T(1,b,c,d)=0 gives \[1+b+c=0,\qquad b+c+d=0,\qquad 1+2b+2c+d=0.\] **2. Solve the required combinations:** From the first equation b+c=-1. The second then gives d=1. These also satisfy the third because 1+2(-1)+1=0. **3. Evaluate only what is asked:** b+c+d=-1+1=0. The separate values of b and c are not unique, but their requested sum is. **Answer: 0.**

**1. Identify the column-space condition:** The second row of A=( (1,2,3),(2,4,6),(1,3,4) ) is twice the first. Therefore every vector (a,b,c) in its column space satisfies b=2a. The first and third rows are independent, so this condition describes the entire two-dimensional column space. **2. Apply the proposed vector:** For (x,-x+1,2x+1), require -x+1=2x. Therefore 3x=1 and x=1/3. **3. Check the decimal:** One-third is approximately 0.3333333333, which lies within the accepted numerical interval 0.33 to 0.34. **Answer: 0.3333333333.**

**1. Compute rank:** For the given 3-by-3 matrix, row 2 is twice row 1, while row 3 is independent of row 1. Hence rank(A)=2. **2. Test one-to-one:** Rank-nullity gives nullity=3-2=1, so a nonzero kernel prevents injectivity. **3. Test onto:** The image has dimension 2, below the codomain dimension 3, so it is not onto. Thus neither property holds. **Answer: D.** A: **Incorrect:** The positive nullity means it is not one-to-one. B: **Incorrect:** The deficient image dimension means it is not onto. C: **Incorrect:** Rank 2 in dimension 3 rules out both properties. D: **Correct:** The kernel is nontrivial and the image is a proper subspace of R3.

**1. Form the matrix rows:** The three output expressions give rows r1=(1,1,-1,1), r2=(2,3,-1,2), and r3=(1,2,0,1). **2. Identify dependence and independence:** r2=r1+r3, so rank is at most 2. Rows r1 and r3 are independent; for example, the determinant of their first two columns is 1. Thus r=2. **3. Apply rank-nullity:** The domain is R4, so k=4-r=2. Therefore 3r+2k=3(2)+2(2)=6+4=10. **Answer: 10.**

**1. Reuse the rank calculation for this transformation:** Its rows satisfy r2=r1+r3, and r1,r3 are independent. Therefore rank(A)=2. **2. Use the transpose theorem:** Row rank and column rank are equal. Transposing exchanges rows and columns, so rank(A transpose)=rank(A)=2. The number of rows after transposition does not alter this invariant. **Answer: 2.**

**1. Use orthogonal spanning vectors:** Let u1=(2,1,0,1), u2=(1,-2,2,0) and v=(7,5,6,4). Their dot product is 2-2=0, so projection onto their span is the sum of the two independent orthogonal projections. **2. Compute coefficients:** \[u_1\cdot u_1=6,\quad v\cdot u_1=23,\quad u_2\cdot u_2=9,\quad v\cdot u_2=9.\] Thus \(\operatorname{proj}_W(v)=(23/6)u_1+u_2\). **3. Add coordinates efficiently:** The coordinate sums of u1 and u2 are 4 and 1. Therefore the sum of projected coordinates is \[\frac{23}{6}\times4+1=\frac{92}{6}+1=\frac{49}{3}=16.333333\ldots.\] This lies in the given accepted interval 16.3 to 16.4. **Answer: 16.3333333333.**

**1. Normalize the first basis vector:** For v1=(1,2,2), the norm is 3, so w1=(1,2,2)/3. **2. Remove the first projection from v2:** With v2=(-1,0,2), v2 dot w1=(-1+4)/3=1. The second orthogonal vector is \[u_2=v_2-(v_2\cdot w_1)w_1=(-1,0,2)-(1,2,2)/3=(-4,-2,4)/3.\] **3. Normalize:** The length of u2 is sqrt(16+4+16)/3=2. Thus w2=(-2,-1,2)/3. **4. Match the required form:** Comparing with (a/3,b/3,c/3) gives a=-2,b=-1,c=2. Hence |a+b+c|=|-1|=1. **Answer: 1.**