Q1 — Numerical Answer · 1 pt
A robot moves on a 3D factory floor. Starting at the origin, it can travel in exactly two directions:
To reach the target
Does this system have a solution? If yes, enter
Answer:
Solution. The system is:
Adding equations 1 and 3:
, so , then from equation 3. Checking equation 2: ✓. The target lies in
— the plane through the origin containing both direction vectors. When the target happens to lie on this plane, the overdetermined system is consistent. Answer:
.
Q2 — Multiple Choice (1 correct answer) · 1 pt
Two students start from the same augmented matrix and row-reduce it using different sequences of valid elementary row operations. Both arrive at a matrix in reduced row echelon form (RREF). Can their final matrices differ?
- (a) ★ No — RREF is unique. Any two sequences of valid row operations that terminate at RREF must produce the same result.
- (b) Yes — the final RREF depends on the order in which you perform the operations.
- (c) It depends on whether the matrix is square.
- (d) Yes — but only if the matrix has free variables.
Solution. The uniqueness of RREF was stated in Lecture 6: “the RREF of a given matrix is the same regardless of the sequence of row operations used — uniqueness is not obvious, and is proved carefully in Hefferon.” Different paths to RREF always arrive at the same destination. Options (b) and (d) are exactly the misconception this question targets: students who think of row reduction as a procedure that might depend on the choices made along the way. It doesn’t — the RREF is a property of the matrix, not of the reduction procedure.
Q3 — Multiple Choice (1 correct answer) · 1 pt
Two students are debating what happens when you add more vectors to a spanning set.
- Student A: “If
spans , then for any . Adding vectors cannot shrink the span.” - Student B: “It could shrink. If
is proportional to a vector already in , the new vectors might start cancelling each other out in linear combinations.”
Which student is correct, and why?
- (a) Student B — linear combinations can cancel, so adding proportional vectors reduces the span.
- (b) ★ Student A — any linear combination of
is also a linear combination of (just set the new coefficient to ), so . Since , the span cannot grow either. Hence . - (c) Both are partly right: the span grows if
, and shrinks if is proportional to some element of . - (d) Student B is correct specifically when
for some .
Solution. Student A is correct. The key is the subset relationship: every linear combination using only vectors from
is still valid in (with coefficient on ). So always holds — adding vectors cannot decrease the span. Student B’s error is thinking that adding
introduces new constraints on what combinations are allowed. It doesn’t. Linear span is not a vector subspace of some other set that “sees” redundancy — it is the set of all finite linear combinations, and more combinations are only ever more, not fewer. Option (c) is wrong in its second clause: if
is proportional to an element of , the span stays the same (not shrinks). Option (d) is wrong for the same reason: since already.
Q4 — Multiple Choice (1 correct answer) · 1 pt
A linear system
- (a) The columns of
are linearly independent. - (b) ★ The columns of
are linearly dependent: the vector is nonzero and satisfies , giving a nontrivial combination of the columns equal to . - (c) Nothing can be concluded about the columns — non-uniqueness of solutions is a property of the right-hand side
, not of itself. - (d) The columns of
must span (where is the number of equations).
Solution. Since
, subtracting gives . Because , the vector is nonzero — so the homogeneous system has a nontrivial solution. Writing , this says with not all — a non-trivial dependence relation among the columns of . This connects two key themes: the solution structure of
(from Lectures 4–5) and the dependence lemma (from Lecture 6). Non-uniqueness of solutions is never just about ; it is a property of — specifically, that the homogeneous system has non-trivial solutions. Option (a) reverses the implication. Options (c) and (d) both miss the structural connection.
Q5 — Numerical Answer · 1 pt
For exactly one value of
Answer:
Solution. Two vectors span
iff they are linearly independent (not proportional). Proportionality requires ; the first component forces , the second then forces . Equivalently: set up the coefficient matrix
and row-reduce for a general right-hand side. A zero row appears in the RREF iff .
Q6 — Multiple Choice (1 correct answer) · 1 pt
The vectors
The redundancy lemma from Lecture 6 seems to guarantee that at least one vector can be removed without changing
- (a) Only
can be removed — it is the one explicitly expressed in terms of the others. - (b) ★ Any one of the three can be removed: the dependence relation
has nonzero coefficient on each vector, so each can be solved for in terms of the other two. - (c) Only
and can be removed, because has the largest entries. - (d) None can be removed — removing any vector always reduces the span.
Solution. The redundancy lemma says: if
with , then is a linear combination of the others and can be removed. The relation here is , which has nonzero coefficient on all three vectors. Therefore all three are in the span of the other two: Option (a) is the most common error: the “obvious” redundant vector is the one that was expressed in the problem statement. But redundancy is symmetric whenever all coefficients in the dependence relation are nonzero. Contrast: in the dependent set
, the dependence involves only the first two vectors (coefficient on the third), so the third cannot be removed while preserving the span. Whether a vector is removable depends on the specific dependence relation, not on which vector “looks” most complex.
Q7 — Multiple Answers (select all that apply) · 2 pts
Let
Which of the following are bases of
- (a) ★
- (b) ★
- (c)
- (d)
Solution.
- (a) ★ The standard basis of
— immediately a basis. ✓ - (b) ★ Express each polynomial as a coordinate vector in terms of
: , , . Solve : the equations , , give , , then . So — independent. Three independent vectors in a 3-dimensional space automatically span (dimension shortcut). ✓ - (c) ✗ Four polynomials in a 3-dimensional space must be linearly dependent: by the Steinitz lemma, no linearly independent set can have more vectors than
. - (d) ✗ The three polynomials are linearly dependent:
, so the third is a linear combination of the first two. A dependent set cannot be a basis.
Q8 — Numerical Answer · 1 pt
Let
Answer:
Solution. Row-reduce the coefficient matrix:
Two pivot rows → two independent constraints →
. The third equation is redundant: subtracting the second from the first gives
, which is exactly the third. The three equations look independent — they describe different relationships among the variables — but the algebraic structure reveals the redundancy. A student who counts “3 equations in 4 unknowns” and concludes
is wrong. The number of independent constraints, not the number of equations written down, determines the dimension.
Q9 — Numerical Answer · 1 pt
In
Enter the value of
Answer:
Solution. The coordinate vector means
. Evaluating at :
Q10 — Multiple Answers (select all that apply) · 2 pts
Let
- (a) ★
- (b) ★
- (c)
implies - (d) ★
Solution.
- (a) ★ TRUE.
. ✓ - (b) ★ TRUE.
. ✓ - (c) FALSE. This holds only if
implies (i.e., the null space is trivial). In general, with is possible whenever the homogeneous system has non-trivial solutions. - (d) ★ TRUE. Set
in (b): . Alternatively, by (a), then subtract from both sides. ✓ Properties (a) and (b) together say that the map
is linear: it preserves addition and scalar multiplication. This will be the defining property of linear maps in the next part of the course.
Q11 — Multiple Choice (1 correct answer) · 1 pt
Lecture 6 gave two proofs that all bases of a finite-dimensional vector space have the same number of vectors: a coordinate proof (expressing one basis in terms of another, setting up a coefficient matrix, deriving a contradiction via RREF) and the Steinitz exchange proof (iteratively swapping vectors between two sets while preserving spanning).
A student says: “The coordinate proof is more rigorous because it uses explicit matrices and calculations, while the Steinitz proof just waves its hands at geometry.”
Which of the following best responds to this assessment?
- (a) The student is correct: proofs that use matrices and RREF are always more rigorous than coordinate-free arguments.
- (b) The student has the comparison backwards: the Steinitz proof is more rigorous precisely because it avoids coordinates.
- (c) ★ Both proofs are equally rigorous — neither depends on unproven steps. The difference is explanatory power: the coordinate proof translates the geometric statement into a rank condition and derives a contradiction mechanistically, without revealing why the result is true. The Steinitz proof stays geometric throughout — you run out of vectors to swap — and the conclusion mirrors the intuition directly.
- (d) The student is correct, but only because the coordinate proof also works for abstract vector spaces, whereas the Steinitz proof is limited to
.
Solution. As the lecture noted: “Neither is ‘more rigorous’ than the other, but they differ enormously in what they reveal.” Rigour is not the same as computation. The coordinate proof is short and mechanical, but the reader is left asking: why did that work? The answer — you were counting pivots — has no geometric content. The Steinitz proof is longer but self-explanatory: the exchange argument fails the moment you try to fit more independent vectors than spanning vectors, because you run out of spanning vectors to replace. Option (d) is false: both proofs work for any finite-dimensional vector space over
.
Q12 — Multiple Choice (1 correct answer) · 1 pt
The standard sequence vectors
Which of the following best explains why
- (a) There are infinitely many
‘s, and a spanning set must be finite. - (b) The
‘s are not linearly independent inside . - (c) ★ A linear combination is always a finite sum. The sequence
belongs to but would need infinitely many nonzero coefficients to be expressed in terms of the ‘s — and no such finite expression exists. - (d)
has uncountably many elements, while the index set is countable, so the latter cannot span the former.
Solution.
- (a) ✗ A spanning set can be infinite —
already spans despite being an infinite set. The issue is not the size of the spanning set but what “span” means. - (b) ✗ The
‘s are linearly independent in any space containing them: a finite combination forces each by looking at the -th component. - (c) ★ Span means the set of all finite linear combinations — a sum
with finitely many nonzero terms. Any such combination is eventually zero (it is zero past position ). But is never zero — it cannot be produced by any finite linear combination of the ‘s. The word “infinite series” describes a different object (a limit), and that is not what “linear combination” means. ✓ - (d) ✗ Cardinality is a red herring:
is also “small” (sequences with finite support), yet the same countable set spans it exactly. The obstruction is the finiteness built into the definition of span, not the countability of the index set.