solutions

MATA22 Midterm — Solutions & Grading Rubric

MATA22 — Linear Algebra I · Summer 2026 · Total 100 pts (Part I 28 · Part II 72)

Grading key for graders. Part II point splits are the TA rubrics (author named per part), transcribed from Grading rubrics.md. LA 1 has no TA rubric yet — the split shown is provisional and marked as such.

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Part I — Short Questions (28 pts)

Graded from the student’s answer table. Each item is all-or-nothing; a multiple-answer item scores only for the exact correct set (no partial unless noted at grading meeting). ★ = correct option.

#	Type · pts	Answer	One-line reason
Q1	MCQ · 2	(a)	; .
Q2	MCQ · 2	(b) (S1) fails	; other seven axioms hold.
Q3	Multi · 3	(a),(c),(d),(e)	: dependent, span is a plane; . Not (b).
Q4	Multi · 3	(a),(c),(e)	Solution set ; homog. sols are the direction; .
Q5	MCQ · 2	(b)	.
Q6	Multi · 3	(a),(c),(e),(f)	Kernels of linear maps are subspaces; (b) affine, (d) not closed.
Q7	Numeric · 2		3 conditions, rank 2 (third first second); .
Q8	Numeric · 2		in ; sum .
Q9	MCQ · 2	(a)	Dependence of images pulls back through the basis to a nonzero kernel vector.
Q10	MCQ · 2	(c)	.
Q11	Numeric · 2		, rank ; .
Q12	Multi · 3	(a),(b),(c),(e)	(d) false: zero map kills independence.

Partial grades for short answer questions

Multiple choice

The partial grade was decided on the set symmetric difference between the posted answer and the correct answer. For example, answering abc in Q3 would yield a symmetric difference of 3 with the correct answer (incorrect b written down, correct d and e missed). For most questions the scale was where each next arrow signifies accruing one more mistake.

Q7 had a partial grade for the answer as it made sense given the problem statement. Some rare partial grades were awarded in other questions if it was apparent that partial progress has been achieved given the form of the answer.

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Part II — Long Answer (72 pts)

Each question is (a) compute · 9, (b) explain · 6, (c) prove · 9.

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Question 1 — Coordinates in an Abstract Vector Space (24 pts)

is -dim with basis ; , , .

(a) — 9 pts. Row-reduce the coordinate vectors as columns: rank dependent. Column relation , i.e. as the coordinate map is a linear bijection (). ; basis e.g. .

(b) — 6 pts. Classmate is wrong. The coordinate map is linear and injective, so a linear relation holds in iff the same relation holds among the coordinate vectors. Dependence (and ) is decided entirely by the coordinates; the actual never enter.

(c) — 9 pts. Show is a basis. Independence: regroups to ; independence of gives . Spanning: and independent vectors in a -dim space span (dimension shortcut) basis.

Rubric — LA 1. Not provided yet; the split below is provisional, pending the TA’s rubric. (a) [9] — [4] correctly determines dependence (row-reduce / exhibit relation); [2] explicit relation ; [3] and a valid basis of the span. (b) [6] — [4] coordinate map is linear & injective, so relations transfer between and ; [2] concludes classmate wrong / identity of irrelevant. (c) [9] — [3] set combination and regroup onto ; [3] use independence of to get the system and force ; [3] spanning via dimension shortcut and conclude basis.

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Question 2 — A Linear Map, Its Matrix, and Rank–Nullity (24 pts)

, .

(a) — 9 pts.

; two different kernel elements: and .

(b) — 6 pts. , , so . (i) Yes, onto: . (ii) Yes, not injective: ; e.g. and agree at . In words: three values don’t determine a cubic (a cubic needs four), so distinct cubics can match at .

(c) — 9 pts. Rank–nullity: (using ). So has a nonzero ; for any , has . Hence not injective.

Rubric — LA 2. (a) [9] — [4] matrix correct (columns are images of : rows ); [2] ; [3] two genuinely different polynomials in , e.g. and a nonzero multiple. (b) [6] — [2] rank–nullity with numbers placed (), not just named; [2] (i) YES onto, justified by ; [2] (ii) YES not injective, justified by + plain-words statement (two cubics can agree at ; three values don’t determine a cubic). (c) [9] — grade by the student’s approach:

• Case 1 (direct, rank–nullity): [3] correct use of rank–nullity + hypothesis toward ; [3] correct justification via ; [3] identifies not injective, i.e. the question’s definition holds when . Partial: [1.5] if it wrongly claims .
• Case 2 (contradiction, rank–nullity): [3] rank–nullity + hypothesis; [3] identifies injective ; [3] justification via giving a contradiction.
• Case 3 (transformation matrix): [2] correct matrix setup; [2] uses so free variables ; [2] from the matrix; [3] identifies not injective.
• Case 4 (rare — contradiction via independence): [3] correct setup of the contradiction; [3] uses independence of a basis of + injectivity to show images independent; [3] cardinality (independent set vs spanning set of ) contradiction.
• General partials: [1.5] right idea (rank–nullity) but incomplete/incorrect execution; [3] correct intuition + significant progress but largely incomplete; [1.5] correct intuition but no/irrelevant progress.

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Question 3 — A Subspace Cut Out by Conditions (24 pts)

.

(a) — 9 pts. and are linear, so is a subspace (, closed under and scalar mult). With : , . So ; basis , .

(b) — 6 pts. A single non-trivial condition removes exactly dimension. Two conditions remove a full only when the functionals are linearly independent; they remove fewer when dependent (one is a scalar multiple of / implied by the other). Dim- example: and — both non-trivial, but the same constraint, so has dimension .

(c) — 9 pts. () If and then , so . () : pick with , set . For any , has , so . Then , i.e. .

Rubric — LA 3. (a) [9] — [1] shows non-empty; [1.5] closed under addition; [1.5] closed under scalar multiplication; [1] set up the two equations from on general ; [1] solve / RREF; [1] found spanning set (as polynomials); [1] verify independence; [1] state . (b) [6] — [3] identifies the case: the two functionals are linearly dependent (one a scalar multiple of the other, or one the zero functional) — “one condition implied by the other”; [3] valid example: two non-trivial conditions on with common solution of dim . (c) [9] — () [6]: [1] ; [1] define ; [1] construct ; [1] show so ; [1] use to get ; [1] derive , conclude . () [3]: [1] take , ; [1] compute ; [1] conclude , so .