Examples

Computing a span by hand

Computing a span by hand

Consider the vectors $\mathbf{u}=(1,1,0,-1,2)$ and $\mathbf{v}=(0,2,3,1,-1)$.
Problem: Show that the set of vectors perpendicular to both $\mathbf{u}$ and $\mathbf{v}$ is a span.

Solution: Let $\mathbf{x}=(x_1,\dots ,x_5)$ be an arbitrary vector such that $\mathbf{x}\cdot \mathbf{u}=\mathbf{x}\cdot \mathbf{v}=0$. By writing these dot products using components, we find a system of two equations:

$$x_1+x_2-x_4+2x_5=0,2x_2+3x_3+x_4-x_5=0.$$

Solve for $x_1$ and $x_3$ in terms of the others:

$$x_1=-x_2+x_4-2x_5,x_3=-\frac{2}{3}{x}_2-\frac{1}{3}{x}_4+\frac{1}{3}{x}_5.$$

Now, we can let $x_2$, $x_4$, and $x_5$ take any value, and using these equations we can specify $x_1$ and $x_3$ to guarantee that the system of equations is valid, implying that $\mathbf{x}\cdot \mathbf{u}=\mathbf{x}\cdot \mathbf{v}=0$. Conversely, given values of $x_2$, $x_4$, and $x_5$, these equations fully determine the only possible values of $x_1$ and $x_3$. Therefore, the set of possible $\mathbf{x}$ is given by varying $x_2$, $x_4$, and $x_5$ in the vector:

$$\mathbf{x}=\left(\begin{array}{c}-x_2+x_4-2x_5\\ x_2\\ -\frac{2}{3}{x}_2-\frac{1}{3}{x}_4+\frac{1}{3}{x}_5\\ x_4\\ x_5\end{array}\right).$$

Observe that this vector can be written as the linear combination $x_2{\mathbf{w}}_2+x_4{\mathbf{w}}_4+x_5{\mathbf{w}}_5$, with

$${\mathbf{w}}_2=\left(\begin{array}{c}-1\\ 1\\ -2/3\\ 0\\ 0\end{array}\right),{\mathbf{w}}_4=\left(\begin{array}{c}1\\ 0\\ -1/3\\ 1\\ 0\end{array}\right),{\mathbf{w}}_5=\left(\begin{array}{c}-2\\ 0\\ 1/3\\ 0\\ 1\end{array}\right).$$

So indeed the set of vectors $\mathbf{x}$ perpendicular to $\mathbf{u}$ and $\mathbf{v}$ is the span $⟨{\mathbf{w}}_2,{\mathbf{w}}_4,{\mathbf{w}}_5⟩$.

Simple subspaces of n-dimensional space

Simple subspaces of ${\mathbb{R}}^n$

Let $W\subset {\mathbb{R}}^n$ be the collection of all possible vectors $\mathbf{w}=(w_1,\dots ,w_{n-1},0)$, meaning that the final term is always zero, and the other terms can be anything.

This set is a subspace because it contains $(0,0,\dots ,0)$, and any linear combination $a{\mathbf{w}}_1+b{\mathbf{w}}_2$ will still have a zero in the final component, so it lies in $W$.