Examples from Spans

Lines and planes through the origin

For $v=(1,0,0)\in\mathbb R^3$,

which is the line $y=z=0$. The coefficient can be any real number, so the whole line is obtained; restricting it to an interval would give a bounded segment, which is not closed under scalar multiplication. Any plane through the origin containing this line is a larger subspace containing $v$.

Standard spanning sets

The set

spans $\mathbb R^2$, since every $(a,b)^T$ is

The set

spans $P_n$, since each polynomial of degree at most $n$ is a linear combination of these powers. The four matrices

span $M_{2\times2}(\mathbb R)$, because

A vector outside a span

Let

To test whether $(1,2,3)^T$ belongs to $\operatorname{span}(S)$, row-reduce

The last row is contradictory, so the system has no solution and

Three vectors spanning $\mathbb R^3$

For

row reduction gives

Thus $A c=b$ has a solution for every $b\in\mathbb R^3$. The columns

span $\mathbb R^3$.

Membership in a polynomial span

Let

and write

Then

The constant and linear coefficients require $a_0=-2\beta$ and $a_1=\beta$; the quadratic and cubic coefficients require $a_2=2\alpha$ and $a_3=\alpha$. Thus $p\in\operatorname{span}\{q_1,q_2\}$ exactly when

A set that does not span $P_2$

Consider

For

the equation

gives

The first two equations give $\gamma=a_0+2\beta$ and $\alpha=a_1-\beta$. Substituting into the third gives the condition

Conversely, when this condition holds, those choices of $\alpha$ and $\gamma$ solve all three equations for any $\beta$. The polynomial $p=x^2$ fails the condition, so these three polynomials do not span $P_2$.