Some subsets of a space are worlds of their own: adding two of their vectors, or scaling one, never takes you outside. Those are the subsets worth a name.

The whole numbers are closed under addition, since adding two of them never produces anything else. Halving one does leave them, so they are not closed under everything.

A subset $W$ of a vector space $V$ is a subspace when it contains the zero vector and stays inside itself under both operations. All three conditions are asked for, and the first rules out the empty set. A subspace is a vector space in its own right: the eight rules hold for its vectors because they hold for all of $V$. In $\mathbb{R}^2$ a line through the origin is a subspace; the same line shifted away from the origin is not.

• The span of vectors $\mathbf{v}_1,\ldots,\mathbf{v}_k$ of $V$, the set of all their combinations, is written $$, as in .

Both are lines. The blue one passes through the origin, and adding or scaling its vectors never leaves it, so it is a subspace. The amber one misses the origin: scaling one of its vectors by zero leaves it immediately, and the first condition already fails.

All three conditions are read straight off the shape of a linear combination. Every weight zero gives the zero vector; adding two combinations of the same list gives another, because the weights on each vector add; and scaling one multiplies every weight, which leaves a combination of the same list.

Taking every weight to be zero shows that $\mathbf{0}$ belongs to the span; each $0\mathbf{v}_j$ is $\mathbf{0}$, by the theorem of the lesson on vector spaces.

Add $\mathbf{x}=\sum c_j\mathbf{v}_j$ and $\mathbf{y}=\sum d_j\mathbf{v}_j$ and collect the terms on each : the weights add, and the result is again a combination of the same list.

Scaling a combination multiplies each weight, which is again a combination of the same list.

All three conditions hold, which is the whole of the definition.

A line, but $(0,0)$ does not satisfy $y=x+1$, so it is not a subspace.

The conclusion names $W$, not the space around it.

Still a combination of the same two vectors, so still in the span.

True of every subspace $W$.

The three conditions make sense in any vector space: contain $\mathbf{0}$, stay inside under addition, stay inside under scaling. A subspace is then a vector space in its own right, because the rules hold for its vectors as they do for all the others.

This is a subspace of $\mathbb{P}_2$: the zero polynomial is in it, and sums and multiples of polynomials with value $0$ at $0$ still have value $0$ there. Those with $p(0)=1$ are not: the zero polynomial is missing.

For an $m\times n$ matrix $A$, the solutions of $A\mathbf{x}=\mathbf{0}$ form a subspace of $\mathbb{R}^n$, called the null space of and written .

• A subset $W$ of a vector space is a subspace when it contains $\mathbf{0}$ and is closed under addition and scaling.
• A span is always a subspace, and so is the null space of a matrix.
• Intersections of subspaces are subspaces; unions usually are not.
• One condition failing is enough to rule a subset out.

Great work! You now know how to:

• test a subset of any vector space with the three conditions;
• recognise spans and null spaces as subspaces;
• find the condition that fails when a subset is not a subspace;
• use intersections, and distrust unions.