Find a basis for $\mathbf{W}^{\perp}$ given spanning vectors of $\mathbf{W}$ I am solving these problems in preparation to a midterm. Here's the problem I have solved before tackling the one I am asking help with:
I was given 1 basis vector for $\mathbf{W}$ - $\begin{bmatrix}
1\\ 
2\\ 
3\\ 
4\\
5
\end{bmatrix}$, and I had to find a basis for $\mathbf{W}^{\perp}$. I set up an equation $\begin{bmatrix}
1\\ 
2\\ 
3\\ 
4\\
5
\end{bmatrix}$ dotted with $\begin{bmatrix}
a & b & c & d & e 
\end{bmatrix}$ = 0 and found the kernel (it was 
$\begin{bmatrix}
-2b-3c-4d-5e\\ 
b\\ 
c\\ 
d\\ 
e
\end{bmatrix}$.)
Now, I know that Span of $\mathbb{W}$ = $\begin{bmatrix}
1\\ 
2\\ 
3\\ 
0
\end{bmatrix}$, $\begin{bmatrix}
1\\ 
2\\ 
3\\ 
4
\end{bmatrix}$. I need to find a basis for $\mathbf{W}^{\perp}$, or the sub-space that contains vectors such that if two vectors from $\mathbf{W}$ and $\mathbf{W}^{\perp}$ were dotted, the result would be 0. 
But I am not sure how to go about this similar problem when I am given two vectors. Thanks!
 A: Okay, first of all you can simplify your basis vectors a bit. You can write
$$W = \operatorname{span} \left\{ \begin{pmatrix} 1 \\ 2 \\ 3 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix} \right\}$$
In general you can apply Gram-Schmidt before to get an ON-basis for the subspace.
Call the vectors $v_1$ and $v_2$. Now, if $u \in W^\perp$, $\langle u, v_1 \rangle = \langle u, v_2 \rangle = 0$. Let $u = (a,b,c,d)^T$. Immidiately you get:
$$\langle u, v_2 \rangle = 0 \Leftrightarrow d = 0$$
And
$$\langle u, v_1 \rangle = 0 \Leftrightarrow a + 2b + 3c = 0$$
which gives a parametric solution (set $c = t$ and $b = s$, solve fo $a$):
$$u = s \begin{pmatrix} 3 \\ 0 \\ -1 \\ 0 \end{pmatrix} + t \begin{pmatrix} 2 \\ -1 \\ 0 \\ 0 \end{pmatrix}$$
$$W^\perp = \operatorname{span} \left\{ \begin{pmatrix} 3 \\ 0 \\ -1 \\ 0 \end{pmatrix}, \begin{pmatrix} 2 \\ -1 \\ 0 \\ 0 \end{pmatrix} \right\}$$
A: Your same ideas as before will work.
Take a vector
$\begin{bmatrix}
a\\ 
b\\ 
c\\ 
d
\end{bmatrix}$
You want this vector to be perpendicular to $\begin{bmatrix}
1\\ 
2\\ 
3\\ 
0
\end{bmatrix}$ and to $\begin{bmatrix}
1\\ 
2\\ 
3\\ 
4
\end{bmatrix}$
You get $a + 2b + 3c = 0$ and $a+ 2b + 3c + 4d = 0$.
From which you get $d = 0$ and $a+2b+3c = 0$.
Hence, $W^{\perp}$ is a $2D$ subspace, with $d=0$ and $a+2b+3c = 0$ i.e. $\begin{bmatrix}
-2b-3c\\ 
b\\ 
c\\ 
0
\end{bmatrix}$
Select two pairs $(a,b,c)$ such that the two are linearly independent to get a pair of basis for $W^{\perp}$.
An example for the two basis are $[3,0,-1,0]$ and $[2,-1,0,0]$
