4.1 Vector spaces and subspaces

Big idea

Everything done so far used columns of numbers. But the arguments never used the numbers, only the rules: you can add two of these things, you can scale one by a number, and the arithmetic behaves the way you expect. Any set with those two operations obeying those rules is a vector space, and every theorem proved from the rules alone applies to it.

The payoff is reach. Polynomials of degree at most two add and scale exactly like vectors in $\mathbb{R}^3$. So do $2 \times 3$ matrices, and continuous functions on an interval, and solutions of a differential equation. Span, independence, basis and dimension all carry over unchanged, so one round of work covers many settings.

The second half of the section is about spaces inside spaces. A subspace is a subset that is a vector space on its own, using the same two operations. Checking the full list of axioms for a subset is wasted effort: everything except three conditions is inherited from the parent. Those three conditions, contains the zero vector and is closed under addition and scalar multiplication, become the standard test.

Decoder

A subspace of a vector space is a subset that contains the zero vector and is closed under addition and scalar multiplication.

“Closed under addition” means: take any two things in the subset, add them, and the result is still in the subset. Not usually, always. “Closed under scalar multiplication” means the same for scaling by any real number, including negative ones and zero.

The zero condition is not redundant in practice, even though closure under scaling by $0$ would give it. Requiring the zero vector rules out the empty set and gives you a one-line disproof: if the candidate subset misses the origin, it is not a subspace and you can stop. That check costs nothing and kills most of the bad candidates you will meet.

Definitions and results

Vector space. A vector space is a nonempty set $V$ with an addition and a scalar multiplication satisfying ten rules, for all $\mathbf{u}, \mathbf{v}, \mathbf{w}$ in $V$ and all scalars $c, d$. Both operations must land back in $V$: $\mathbf{u} + \mathbf{v}$ is in $V$ and $c\mathbf{u}$ is in $V$. Addition is commutative and associative. There is a zero vector with $\mathbf{u} + \mathbf{0} = \mathbf{u}$, and each $\mathbf{u}$ has a negative with $\mathbf{u} + (-\mathbf{u}) = \mathbf{0}$. Scalars distribute over vector sums, sums of scalars distribute over a vector, scalar multiplication is associative, and $1\mathbf{u} = \mathbf{u}$.

Immediate consequences. From the axioms alone: the zero vector is unique, each negative is unique, $0\mathbf{u} = \mathbf{0}$, $c\mathbf{0} = \mathbf{0}$, and $-\mathbf{u} = (-1)\mathbf{u}$. These are proved once and then used in every vector space without reproof.

Standard examples. $\mathbb{R}^n$ with entrywise operations. The set $\mathbb{P}_n$ of polynomials of degree at most $n$, added and scaled coefficient by coefficient; the zero vector is the zero polynomial. The set of all $m \times n$ matrices. The set of all real-valued functions on a set, added pointwise. The set of doubly infinite sequences of real numbers.

Degree is the wrong bookkeeping. $\mathbb{P}_n$ collects polynomials of degree at most $n$, not exactly $n$. Polynomials of degree exactly $n$ do not form a vector space: the zero polynomial is missing, and adding $t^2$ to $-t^2 + t$ drops the degree.

Subspace. A subset $H$ of a vector space $V$ is a subspace when the zero vector of $V$ is in $H$, the sum of any two vectors of $H$ is in $H$, and any scalar multiple of a vector of $H$ is in $H$. A subspace is itself a vector space; the other seven axioms hold in $H$ because they already hold in $V$.

Extremes. Every vector space has at least two subspaces: the set containing only the zero vector, and the whole space. The first is called the zero subspace.

Spans are subspaces. If $\mathbf{v}_1, \dots, \mathbf{v}_p$ are in $V$, then $\text{Span}\{\mathbf{v}_1, \dots, \mathbf{v}_p\}$ is a subspace of $V$. All three conditions are immediate: the zero vector uses all-zero weights, the sum of two combinations is a combination, and a multiple of a combination is a combination. So the fastest way to prove a set is a subspace is to exhibit it as a span.

How subsets fail. A subset defined by an inequality is usually not closed under scaling by a negative number. A subset defined by a nonzero constant on the right-hand side usually misses the origin. A subset defined by a product being zero is usually not closed under addition. Each failure is disproved by one explicit pair of vectors, not by a general argument.

Worked examples

A plane through the origin, recognised as a span

Let $H$ be the set of all vectors in $\mathbb{R}^3$ of the form

$$ \begin{bmatrix} a \\ b \\ a + 2b \end{bmatrix} $$

where $a$ and $b$ range over all real numbers. Split the description into the two free parameters:

$$ \begin{bmatrix} a \\ b \\ a + 2b \end{bmatrix} = a\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} + b\begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix} $$

So $H$ is the span of those two vectors, and spans are subspaces. Done, with no closure checks.

Check the description against a concrete member. Take $a = 3$, $b = -1$: the vector is $(3, -1, 1)$, and indeed $3 + 2(-1) = 1$. Its double $(6, -2, 2)$ satisfies the rule too, since $6 + 2(-2) = 2$.

Two subsets that fail

First let $W$ be the set of vectors $(x_1, x_2)$ in $\mathbb{R}^2$ with $x_1 x_2 = 0$, the union of the two axes. The zero vector is in $W$, and every scalar multiple of a member stays in $W$. But addition fails: $(1, 0)$ and $(0, 1)$ are both in $W$, while their sum $(1, 1)$ has product $1$. So $W$ is not a subspace, and one pair of vectors settled it.

Now let $K$ be the set of vectors $(x_1, x_2)$ with $x_1 + x_2 = 1$. The zero vector gives $0 + 0 = 0$, not $1$, so $\mathbf{0}$ is not in $K$ and $K$ is not a subspace. Geometrically $K$ is a line that misses the origin; only lines through the origin are subspaces of $\mathbb{R}^2$.

A subspace of polynomials

Work inside $\mathbb{P}_2$, the polynomials $p(t) = a_0 + a_1 t + a_2 t^2$. Let $H$ be the set of those with $p(0) = 0$.

The zero polynomial has value $0$ at $t = 0$, so $\mathbf{0}$ is in $H$. If $p(0) = 0$ and $q(0) = 0$ then $(p + q)(0) = 0 + 0 = 0$, so $H$ is closed under addition. If $p(0) = 0$ then $(cp)(0) = c \cdot 0 = 0$, so $H$ is closed under scaling. All three hold, so $H$ is a subspace.

Check what $H$ actually contains. Since $p(0) = a_0$, the condition says $a_0 = 0$, so $H$ is the set of polynomials $a_1 t + a_2 t^2$, which is $\text{Span}\{t, t^2\}$. That is the second proof, and it matches.

Contrast the set with $p(0) = 1$. The zero polynomial is not in it, so it is not a subspace, for the same reason the line $x_1 + x_2 = 1$ was not.

A subspace of matrices

Inside the space of all $2 \times 2$ matrices, let $S$ be the symmetric ones, those equal to their own transpose. The zero matrix is symmetric. If $A^T = A$ and $B^T = B$ then $(A + B)^T = A^T + B^T = A + B$, and $(cA)^T = cA^T = cA$. So $S$ is a subspace, and every member looks like $\begin{bmatrix} a & b \\ b & d \end{bmatrix}$, which is the span of three fixed matrices.

Practice

First the axioms and the standard examples: decide whether a given set with given operations is a vector space, and name the zero vector when it is.

Then the subspace test: check the zero vector and the two closure conditions, or exhibit the set as a span.

Videos

Watch the closing argument about why the axioms are stated the way they are. The point to take away is that a result proved from the axioms is automatically true for functions and polynomials, not only for arrows.

Quiz

Five items on the axioms, the standard spaces, and deciding whether a subset is a subspace.