Question bank — Abstract vector spaces — axioms, examples beyond ℝⁿ

True. Addition and scalar multiplication are entrywise, closure holds, the zero matrix is $0$, and all 8 axioms are inherited from $\mathbb{R}$; its dimension is $4$.

True. Axiom 3 forces the existence of a zero vector $0$, so no vector space can be empty — the smallest possible one is $\{0\}$.

False. It is not closed: $(x^3+x)+(-x^3)=x$ has degree $1$, not $3$, and the zero polynomial (degree undefined) is missing, so Axioms 3 and closure fail.

False. Closure is necessary but not sufficient; you still need the zero vector, inverses, and the distributive/compatibility laws — e.g. weird custom operations can be closed yet break Axiom 8.

False. Dimension depends on the field: $\mathbb{C}$ has dimension $2$ over $\mathbb{R}$ (basis $\{1,i\}$) but dimension $1$ over $\mathbb{C}$ (basis $\{1\}$) — same set, different scalars. See Fields and scalars.

True. The homogeneous equation is closed under linear combinations and contains the zero function; the inhomogeneous one fails closure ($y_1+y_2$ solves $y^{\prime \prime }+y=6\ne 3$) and lacks $0$.

False. Being infinite-dimensional is a property, not a disqualifier; $C[0,1]$ satisfies all 8 axioms and is a perfectly valid vector space. See Function spaces and Fourier series.

True. It is a $1$-dimensional vector space over itself with basis $\{1\}$; "vectors" and "scalars" happen to be the same objects, which is legal.

False. It has no zero vector ($0\notin {\mathbb{R}}_{>0}$) and no additive inverses ($-v<0$), so Axioms 3 and 4 fail outright.

Error: it confuses the scalar $0\in \mathbb{F}$ with the vector $0\in V$. The theorem says scaling any vector by the scalar $0$ produces $0$; it does not equate the two objects, which may live in totally different worlds (e.g. functions vs. numbers).

Error: "cancelling" is not an axiom. You must legitimise it by adding the additive inverse $-(0v)$ (Axiom 4) to both sides and using associativity; only then does the term vanish.

Error: "flips its sign" assumes number-arithmetic, but $v$ may be a function or matrix. The real proof uses $v+(-1)v=(1+(-1))v=0v=0$, so $(-1)v$ is the additive inverse of $v$, which is what $-v$ denotes.

Error: a subspace must contain $0=(0,0)$, but $(0,0)$ does not satisfy $y=x+1$. It also fails scaling: $2\cdot (1,2)=(2,4)$ is off the line. Only lines through the origin qualify. See Subspaces and the subspace test.

Error: dimension counts basis vectors, not roots. The basis $\{1,x,x^2\}$ has three elements, so $\dim P_2=3$; roots of a polynomial are irrelevant to the dimension of the space of polynomials.

Error: a basis must be linearly independent and spanning. Infinitely many functions can still be dependent (e.g. $\{1,x,2x,x^2,\dots \}$), so being infinite is not enough. See Linear independence, basis and dimension.

Because closure is baked into the type of the operations $+:V\times V\to V$ and $\cdot :\mathbb{F}\times V\to V$; the outputs must land back in $V$. Since it lives in the operation signatures, it's often not counted, yet it's the cheapest thing to check and the one beginners forget.

Because they use only the abstract axioms, which every vector space obeys regardless of what its elements "look like". That's the whole payoff of abstraction: one proof works simultaneously for arrows, functions, matrices, and ODE solutions.

Because scalar multiplication only makes sense once you fix which scalars are allowed. The same set can be a vector space over different fields with different dimensions, so "vector space" is incomplete without naming $\mathbb{F}$.

Because its elements are ordinary functions, and functions already obey all 8 axioms via pointwise operations. Closure is the only thing that could fail, so checking that linear combinations of solutions are still solutions is enough.

Because a vector space needs $+$ and a compatible scalar action linking it to a field, plus a zero, inverses, and the distributive laws. A lone $+$ gives at most a group or monoid, not the full linear-algebra structure.

With complex scalars, one basis vector $\{1\}$ reaches every complex number via $z\cdot 1$, so dimension $1$. With only real scalars you need two, $\{1,i\}$, since no real multiple of $1$ produces $i$ — hence dimension $2$.

Yes — the trivial/zero vector space. It satisfies all 8 axioms ($0+0=0$, $a0=0$) and has dimension $0$ with the empty set as its basis.

Yes. $0$ is whatever element satisfies $v+0=v$: it is the zero function in $C[0,1]$, the zero matrix in $M_{2\times 2}$, and the zero polynomial in $P_n$ — none of which is the number $0$.

No. It fails closure under real scalar multiplication: $\sqrt{2}\cdot 1=\sqrt{2}\notin \mathbb{Q}$. It is a vector space over $\mathbb{Q}$ itself, showing again that the field choice is decisive.

Yes, if you accept the standard axioms of set theory, but for infinite-dimensional spaces like "all polynomials" the basis is infinite and for $C[0,1]$ the basis is not something you can write down explicitly. Existence and constructibility are different things.

That element must be $0$ (Axiom 3), and every axiom collapses to the trivially true statement $0=0$. This confirms $\{0\}$ is the smallest legal vector space and that a vector space is never empty.