A basis is a list that is large enough to reach every vector of a space and small enough that nothing in it is wasted. Those two demands pull in opposite directions, and a basis is where they meet.

A set of weights for a balance should be able to measure every amount, and should contain no weight that the others could already make. Both conditions together decide the set.

A list is a basis of a subspace $W$ when it spans $W$ and is linearly independent. Spanning alone allows waste; independence alone allows gaps. A basis is a list that is exactly the right size, and a space has many different bases.

• The standard vectors $\mathbf{e}_1,\ldots,\mathbf{e}_n$ form a basis of $\mathbb{R}^n$, called the standard basis.

Spanning gives at least one set of weights. Suppose two sets both produced the same vector; subtracting one expression from the other gives a combination equal to the zero vector, whose weights are the differences. Independence forces each of those differences to be zero, so the two sets of weights were the same all along.

The list spans $W$, so at least one set of weights builds $\mathbf{x}$.

Suppose a second set of weights builds the same vector.

Subtract one expression from the other; the differences of the weights are left on the same list.

The list is independent, so the only weights giving $\mathbf{0}$ are all zero.

The two sets of weights agree throughout, so there was only ever one description.

$(1,0),(0,1),(1,1)$ spans $\mathbb{R}^2$

But it is dependent, so it is not a basis.

So the three together are dependent, though they still span $\mathbb{R}^2$.

$\operatorname{span}\{(1,0)\}$ is a line

Independent, but far from all of $\mathbb{R}^2$.

And in the standard basis there is no other way to write it.

$1,x,\ldots,x^n$ is a basis of $\mathbb{P}_n$: every polynomial of degree at most $n$ is a combination of them, in one way only. The matrices , with a single in position , are a basis of .

Solve $A\mathbf{x}=\mathbf{0}$ in parametric form. The vectors that multiply the free variables span the solutions, and they are independent, because each has a $1$ where its own free variable sits and $0$ where the others do.

So $(-1,1,0)$ and $(1,0,1)$ are a basis of that plane.

• $1,x,\ldots,x^n$ is a basis of $\mathbb{P}_n$, and the $E_{}$ are a basis of .

Great work! You now know how to:

• test a list for both halves of being a basis;
• use the standard bases of $\mathbb{R}^n$, $\mathbb{P}_n$ and $\mathbb{R}^{m\times n}$;