Vector Spaces and Subspaces

Arrows in the plane, $3\times 3$ matrices, polynomials of degree at most 5, and real-valued functions on $[0,1]$ have almost nothing in common as objects. Each can be added to another of its kind, and each can be scaled by a number, and in every case those two operations obey the same eight rules.

That shared structure is what a vector space is. Naming it means a theorem proved from those rules alone holds for all four at once, which is why results in this subject transfer so far beyond the arrows they are usually drawn with.

The working question is narrower: given a subset of such a space, is it a space in its own right? Three checks settle it, and they fail independently. A set can pass two and fail the third, which is what makes the test worth doing rather than eyeballing.

The subspace test is short for a reason, and the reason explains which axioms can be skipped and which cannot.

What is inherited. Associativity, commutativity and both distributive laws are universally quantified equations: they say something holds for all elements of $V$. If it holds for all of $V$, it holds for the elements that happen to lie in $W$. Nothing about restricting attention to a subset can make a true equation false. The same applies to $1v=v$ and $a(bv)=(ab)v$.

What is not inherited. Two axioms assert that something exists in the set: an additive identity, and an additive inverse for each element. Existence claims do not survive restriction. The element may lie in $V$ and outside $W$. Neither does the requirement, implicit in calling them operations on $W$, that sums and multiples of members of $W$ land in $W$.

So exactly the existence-and-membership claims need checking, and the three conditions cover them:

Additive inverses come free from the third: $-v=(-1)v$, which closure under scaling puts in $W$.

On the redundancy. Condition 1 follows from condition 3 with $a=0$, provided $W$ is nonempty. It is kept as a separate step because it is the cheapest discriminator available: substituting zero into a defining condition takes seconds and settles every set defined by an inhomogeneous equation.

What the test does not address. Whether $W$ is interesting, finite-dimensional, spanned by something convenient, or equal to a space you already know. Those come later, and a set can be a subspace without any of them being apparent from its description.

In ${\mathbb{R}}^3$ the subspaces can be listed completely:

There is nothing else. No sphere, no half-space, no line that misses the origin, no bounded region.

The reason is scaling. Take any nonzero $v$ in a subspace; closure under scaling forces every multiple $av$ to be present, so the whole line through $v$ and the origin is in. A subspace therefore extends indefinitely in both directions along every one of its members, which rules out anything bounded at once, and rules out anything avoiding the origin, since $a=0$ produces it.

Add a second member off that line and closure under addition forces every combination $au+bv$, filling the plane those two span. So subspaces are built up in whole dimensions: a point, a line, a plane, everything.

Using this as a check. Before any algebra, ask whether the described set is a flat object through the origin extending to infinity. A set defined by $x+2y-z=5$ is a plane, but a shifted one: at the origin the left side is $0$ and the right is $5$. It is parallel to a subspace without being one.

The picture is a guide rather than a proof, and it runs out quickly, in spaces of matrices or polynomials there is nothing to draw. The three conditions are what carries over, and they are stated so they never depended on the picture.

Each is a subset of ${\mathbb{R}}^3$ or ${\mathbb{R}}^2$, tested by the same three conditions.

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$W_1=\{(x,y,z):x+2y-z=0\}$.

Zero. $0+2(0)-0=0$, so the origin is in.

Closure under addition. Take $u=(x_1,y_1,z_1)$ and $v=(x_2,y_2,z_2)$ in $W_1$, so $x_1+2y_1-z_1=0$ and $x_2+2y_2-z_2=0$. Then

so $u+v\in W_1$. The argument used arbitrary members, not examples.

Closure under scaling. For $a\in \mathbb{R}$ and $u\in W_1$,

Verdict: a subspace. Concretely, $u=(1,0,1)$ and $w=(0,1,2)$ are both in $W_1$; $u+w=(1,1,3)$ gives $1+2-3=0$, and $-2u=(-2,0,-2)$ gives $-2+0+2=0$. These check the general argument rather than substituting for it.

In fact $W_1=\mathrm{span}\{(1,0,1),(0,1,2)\}$: a combination $a(1,0,1)+b(0,1,2)=(a,b,a+2b)$ satisfies $a+2b-(a+2b)=0$, and conversely any $(x,y,z)$ with $z=x+2y$ is the combination with $a=x$, $b=y$. So $W_1$ is the plane through the origin spanned by those two vectors. One of the dimension-2 subspaces the previous block listed.

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$W_2=\{(x,y,z):x+2y-z=5\}$.

Zero. $0+2(0)-0=0\ne 5$. The origin is absent.

Verdict: not a subspace, and no further test is needed.

Had the zero test been skipped, addition would have caught it: $(5,0,0)$ and $(0,5,5)$ are both in $W_2$, and their sum $(5,5,5)$ gives $5+10-5=10\ne 5$. Any set defined by an inhomogeneous linear equation fails this way. The constant adds when the vectors do.

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$W_3=\{(x,y):xy=0\}$, the union of the two axes in ${\mathbb{R}}^2$.

Zero. $(0)(0)=0$, so the origin is in.

Closure under scaling. If $xy=0$ then $(ax)(ay)=a^{2}xy=0$, for every $a$. Checking: $7(1,0)=(7,0)$ has product $0$, and $-3(1,0)=(-3,0)$ has product $0$.

Closure under addition. $p=(1,0)$ has $xy=0$ and $q=(0,1)$ has $xy=0$, but

Verdict: not a subspace, failing closure under addition alone.

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$W_3$ passes the zero test and passes closure under scaling, and is still not a subspace. So the conditions cannot be collapsed: verifying one or two of them is not evidence about the others. Geometrically it is two lines through the origin, and the picture says why. A subspace containing two independent directions must contain the whole plane they span, and $W_3$ contains only the axes themselves.

Checking examples instead of arbitrary elements. "Take $(1,0,1)$ and $(0,1,2)$ in $W_1$; their sum is $(1,1,3)$, which satisfies the equation. So $W_1$ is closed under addition." Two members summing correctly is consistent with closure and does not establish it, $W_3$ above passes many such checks. Closure is a claim about every pair, so the argument must be carried out with symbols.

Concluding from one closure to the other. "Every multiple of a member stays in the set, so the set is closed." $W_3$ is the counterexample: closed under scaling, not under addition. The two conditions constrain different things and neither implies the other.

Treating the zero test as sufficient. "The origin is in the set, so it is a subspace." The origin lies in $W_3$, which is not one. The zero test is a fast way to rule sets out; it rules nothing in.

Mistaking a shifted flat for a subspace. " $x+2y-z=5$ describes a plane, and planes are subspaces." Only planes through the origin are. This one is parallel to $W_1$ and misses the origin, and it is a coset of a subspace rather than a subspace. The solution set of an inhomogeneous system, where the homogeneous version gives the subspace.

The common thread: each treats a necessary condition, or a verified instance, as though it settled a universal claim. The subspace test is three separate universal statements, and a verdict of yes requires all three argued in general, while a verdict of no requires only one explicit witness.

The definition earns its generality only if it is used away from coordinates. Four spaces where the same three conditions apply unchanged.

Matrices. $M_{2\times 2}(\mathbb{R})$ under entrywise addition and scaling. The symmetric matrices form a subspace: the zero matrix is symmetric, a sum of symmetric matrices is symmetric, and a multiple of one is symmetric. The invertible matrices do not. The zero matrix is not invertible, and is invertible while neither summand is, so closure fails in both directions at once.

Polynomials. $P_3$, the polynomials of degree at most 3. Those with $p(1)=0$ form a subspace: the zero polynomial vanishes at 1, and sums and multiples of polynomials vanishing at 1 vanish at 1. Those of degree exactly 3 do not. The zero polynomial is excluded, and $(x^3+1)+(-x^3)=1$ has degree 0.

Functions. Real-valued functions on $[0,1]$. The continuous ones form a subspace, since sums and multiples of continuous functions are continuous. So do the differentiable ones, and the solutions of $f^{″}+f=0$. The last being why linear differential equations have solution spaces whose members are combinations of a few basis solutions.

Sequences. Infinite real sequences, with the eventually-zero ones forming a subspace of them.

What the pattern shows. In every case the subspaces are the sets defined by conditions that are linear and homogeneous, vanishing at a point, equalling one's own transpose, satisfying a homogeneous differential equation. Conditions that are inhomogeneous ( $p(1)=2$), or nonlinear (degree exactly 3, invertibility, $xy=0$), fail. That is the same distinction as the ${\mathbb{R}}^3$ case, stated without coordinates.