Chapter 5 Quiz — Vector Spaces
Twelve quick checks on the big ideas. Try each before opening the answer. These are conceptual — no calculator needed, though a couple ask you to decide "is this a vector space?" by spotting a single counterexample.
Q1. In one phrase, what is a vector space?
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A **set of objects with an addition and a scalar multiplication that satisfy the eight axioms.** "Vector" names a *role* (anything you can add and scale lawfully), not a kind of object — arrows, polynomials, matrices, and functions all qualify.Q2. The eight axioms split into two families. Name them.
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(1)–(4) say **addition is well-behaved**: closure, commutativity, associativity, a zero vector, and additive inverses (together: a commutative group under $+$). (5)–(8) say **scalar multiplication cooperates with addition**: $1\mathbf{v}=\mathbf{v}$, $c(d\mathbf{v})=(cd)\mathbf{v}$, and the two distributive laws. Slogan: *you can take linear combinations freely and stay in the space.*Q3. Which axiom is the one to check first when asked "is this subset a vector space?", and why?
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**Closure (Axiom 0)** — closure under addition and under scalar multiplication. For a subset of a known vector space, the arithmetic axioms (1)–(2), (5)–(8) are inherited automatically; the usual way a set fails is that adding or (especially) scaling by a negative throws you out, or the set misses the zero vector. Probe closure, particularly with a negative scalar.Q4. True or false: the set $\{(x_1, x_2) \in \mathbb{R}^2 : x_1 \ge 0\}$ (the right half-plane) is a subspace of $\mathbb{R}^2$.
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**False.** It contains the origin and is closed under addition, but it is **not closed under multiplication by negative scalars**: $(1,0)$ is in the set, yet $(-1)(1,0)=(-1,0)$ is not. Equivalently, $(1,0)$ has no additive inverse inside the set. A single counterexample disqualifies it.Q5. Is the set of polynomials in $\mathbb{P}_2$ of degree exactly 2 a vector space?
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**No.** It fails closure (and misses the zero vector). The zero polynomial has no degree 2, so it isn't in the set, violating Axiom (3). And $(x^2 + x) + (-x^2) = x$ has degree 1, so the sum of two degree-2 polynomials can drop below degree 2 — closure under addition fails. (The polynomials of degree *at most* 2, $\mathbb{P}_2$, *are* a vector space — the "at most" is what saves it.)Q6. Why does the function space $\mathcal{F}[0,1]$ count as "infinite-dimensional"?
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Because no *finite* list of numbers (coordinates) can pin down an arbitrary function on $[0,1]$ — there are infinitely many points $x$ whose values you'd have to specify. It still satisfies all eight axioms, so it is a genuine vector space; it just needs infinitely many "axes." Sampling on an $n$-point grid gives a finite-dimensional shadow in $\mathbb{R}^n$.Q7. The chapter proves the zero vector is unique (Theorem 5.1). Why did that need a proof at all?
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Because the axioms only promise that **at least one** additive identity exists (Axiom 3); they never say there's only one. To be entitled to say "*the* zero vector," we must rule out a space having two different ones — which the proof does using only commutativity (Axiom 1) and the identity property (Axiom 3), so it holds in *every* vector space at once.Q8. What does the theorem $0\,\mathbf{v} = \mathbf{0}$ actually connect, and why isn't it just a definition?
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It connects two *different* zeros: the **scalar** $0$ (a number, from the field) on the left, and the **zero vector** $\mathbf{0}$ (an element of the space) on the right. Nothing in the axioms directly equates them — the link is *forced* by the distributive law (Axiom 8) plus additive inverses (Axiom 4), as the proof shows. It's a consequence, not a definition.Q9. A friend says "matrices form a vector space, so the additive inverse of $A$ is $A^{-1}$." What's wrong?
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The **additive inverse** of $A$ is $-A$ (negate every entry), the matrix that *adds* to $A$ to give the zero matrix — and it exists for every matrix. The **matrix inverse** $A^{-1}$ is about *multiplication* ($AA^{-1}=I$) and only exists for invertible matrices. The vector-space structure of $M_{2\times 2}$ uses only addition and scaling; matrix multiplication (Chapter 8) and the inverse (Chapter 9) are separate.Q10. Can the positive real numbers be a vector space? Under what operations?
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**Yes** — but not under ordinary $+$ and $\times$ (ordinary addition has no positive identity and no positive inverses). Define "addition" as ordinary *multiplication* ($x \oplus y = xy$) and "scalar multiplication" as *exponentiation* ($c \odot x = x^c$). Then the **zero vector is the number $1$**, the additive inverse of $x$ is $1/x$, and all eight axioms hold. The axioms constrain the *operations*, not the objects.Q11. Why is a qubit's state space a vector space over $\mathbb{C}$, not $\mathbb{R}$ — and does that change which theorems apply?
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The amplitudes $\alpha, \beta$ in $\alpha\mathbf{e}_0 + \beta\mathbf{e}_1$ are **complex numbers**, and the quantum interference effects that make the model physical require the field to be $\mathbb{C}$. It's a *complex* vector space. The eight axioms are identical (just with $\mathbb{C}$ as the scalars), so all the structural theorems of this chapter still apply — uniqueness of zero, $0\boldsymbol{\psi}=\mathbf{0}$, and so on — which is exactly why "the state is a vector, the gate is a matrix" requires no new mathematics.Q12. Fill in the blank: a constraint set equal to ____ tends to define a subspace, while the same constraint set equal to a nonzero value usually does not.