Chapter 6 Quiz — Subspaces, Span, and Linear Independence
Twelve quick checks on subspaces, span, and linear independence. Try each before opening the answer. Most are conceptual; a couple need a little arithmetic.
Q1. What is the single fastest way to disqualify a subset of $\mathbb{R}^n$ from being a subspace?
Answer
**Check whether it contains the origin.** Every subspace must contain $\mathbf{0}$ (set $c = 0$ in closure under scalar multiplication). If $\mathbf{0} \notin W$, then $W$ is not a subspace and you can stop immediately — no need to test closure.Q2. Is the line $y = x + 3$ a subspace of $\mathbb{R}^2$? Why or why not?
Answer
**No.** It does not pass through the origin (at $x = 0$, $y = 3 \neq 0$), so it fails the zero-vector condition. Only lines *through the origin* are subspaces. This is the geometric cousin of the affine map $y = mx + b$ with $b \neq 0$ from Chapter 1.Q3. In one sentence, what is the span of a set of vectors?
Answer
**The set of all linear combinations of those vectors** — every vector reachable by scaling them and adding. Equivalently (and provably), it is the smallest subspace containing all of them.Q4. True or false: the span of any set of vectors is always a subspace.
Answer
**True.** This is the theorem proved in §6.8: the span is closed under addition and scalar multiplication (a combination of combinations is still a combination) and contains $\mathbf{0}$ (all-zero coefficients). Span is the machine that manufactures subspaces.Q5. What does it mean, in the precise definition, for vectors to be linearly independent?
Answer
The **only** linear combination of them that equals the zero vector is the trivial one (all coefficients zero): $c_1\mathbf{v}_1 + \dots + c_k\mathbf{v}_k = \mathbf{0} \Rightarrow c_1 = \dots = c_k = 0$. Geometrically: no vector is redundant — none lies in the span of the others.Q6. A set of vectors contains the zero vector. Are they independent or dependent?
Answer
**Dependent**, always. If $\mathbf{v}_1 = \mathbf{0}$, the combination $1\cdot\mathbf{v}_1 + 0\cdot\mathbf{v}_2 + \dots = \mathbf{0}$ is nontrivial (coefficient $1 \neq 0$) yet reaches $\mathbf{0}$. The zero vector is the ultimate redundant vector and can never be in an independent set.Q7. You stack $k$ vectors from $\mathbb{R}^n$ as the columns of a matrix and row reduce. What pivot condition tells you the vectors are independent? What condition tells you they span $\mathbb{R}^n$?
Answer
**Independent** $\iff$ a pivot in every *column* ($\operatorname{rank} = k$, no free variables). **Spans $\mathbb{R}^n$** $\iff$ a pivot in every *row* ($\operatorname{rank} = n$). They are a basis exactly when both hold, which forces $k = n$.Q8. Give a set of vectors that spans $\mathbb{R}^2$ but is not independent.
Answer
Any three (or more) vectors that include a spanning pair, e.g. $(1, 0),\ (0, 1),\ (1, 1)$. The first two already span $\mathbb{R}^2$, so the set spans; but three vectors in $\mathbb{R}^2$ cannot be independent (only two pivots are possible), and indeed $(1,1) = (1,0) + (0,1)$ is redundant. Rank $= 2 < 3$ columns.Q9. Are $(1, 1)$, $(2, 2)$, and $(3, 3)$ linearly independent? What is the dimension of their span?
Answer
**Dependent** — all three are multiples of $(1, 1)$, so they are parallel. Their span is a single **line** through the origin, which has **dimension 1**. The matrix of these columns has rank $1$.Q10. Two vectors in $\mathbb{R}^5$ are linearly independent. Can they span $\mathbb{R}^5$?
Answer
**No.** Their span has dimension at most $2$ (two vectors give at most two pivots), but $\mathbb{R}^5$ has dimension $5$. You need at least $5$ vectors to span $\mathbb{R}^5$. Independent-but-not-spanning is the classic "too few vectors" situation.Q11. Why does linear independence of a basis guarantee that every vector has unique coordinates?
Answer
If a vector had two representations $\sum a_i\mathbf{v}_i = \sum b_i\mathbf{v}_i$, subtracting gives $\sum(a_i - b_i)\mathbf{v}_i = \mathbf{0}$. By independence, the only combination reaching $\mathbf{0}$ is trivial, so $a_i = b_i$ for all $i$ — the two representations were identical. Dependence would destroy this uniqueness.Q12. Fill in the blank and explain: "A basis of a subspace is a set that is both ___ and ___."