# finding a basis for $W^\perp$ and understanding it.

Given

$$w_1 = \begin{bmatrix} 1 \\ -1 \\ 1 \\ 1 \end{bmatrix},w_2= \begin{bmatrix} 0 \\ 1\\2\\3 \end{bmatrix}$$ let $W$ be the subspace spanned by the given vectors. Find a basis for $W^\perp$ Now my problem is, how do envision this? They do the following:

They use the vectors as rows. Then they say that W is the row space of A, and so it holds that $W^\perp = null(A)$ . and we thus solve for $Ax=0$

Now my problem is: how do i envision this? Why is $W^\perp = null(A)$ I dont like learning these kinds fo things, is there a way to understand this? WHY is this the case, why do they specifically let A use $w_1$ and $w_2$ as the rows?

Take a vector ${\bf x}=(x_1,x_2,x_3,x_4)$. You want to find the solutions to the system $$\begin{cases}{\bf x}\cdot {\bf w}_1=0\\{\bf x}\cdot {\bf w}_2=0\end{cases}$$ See? Then find a basis for this.

• why not two variables? why are we not taking them as columns? Thats moreover my question, though I get your intuition Commented Jul 9, 2013 at 20:04
• What do you mean by "two variables"? We want $x$ to be simultaneously orthogonal to $w_1$ and $w_2$, so it will be orthogonal to all the elements of their span.
– Pedro
Commented Jul 9, 2013 at 20:10
• I mean, why are they using $w_1$ and $w_2$ as rows? Why not as columns? Thats my main pickle with this. Commented Jul 9, 2013 at 20:11
• ohhhh, of course. how could it not be $x_1,x_2,x_3,x_3$, else we couldnt even do the dot product. d'oh Commented Jul 9, 2013 at 20:14
• @WiseStrawberry You can do it with columns, as long as you start using row vectors as inputs, so that the matrix dimensions are compatible. Commented Jul 9, 2013 at 20:18

To elaborate somewhat on Peter's post, we note that for any two column vectors of real entries: $$v\cdot w=v^Tw$$ So our system is simply $$\begin{cases} {w_1}^Tx = 0\\ {w_2}^Tx = 0 \end{cases}$$

That is,

$$\begin{bmatrix} {w_1}^Tx \\ {w_2}^Tx \end{bmatrix}= \begin{bmatrix}0\\0\end{bmatrix}$$ And in case you need me to tell you where this is going, that is $$\begin{bmatrix} – & {w_1}^T & –\\ - & {w_2}^T & – \end{bmatrix} \begin{bmatrix} | \\ x \\ | \end{bmatrix} = \begin{bmatrix}0\\0\end{bmatrix}$$

• damn. that does explain a lot. Commented Jul 9, 2013 at 20:12
• Dear @omno... Oop, you beat me to it, but since I spent so much time typing I hope you don't mind if I just vote for yours and leave my answer up. Commented Jul 9, 2013 at 20:16
• @rschwieb the more the merrier, and the quickest answer is not generally the best. The upvote is appreciated also :) Commented Jul 9, 2013 at 20:18
• whats up with the last matrix? what are the '-' on the left and right of $w_1$ and $w_2$? Commented Jul 9, 2013 at 20:21
• emphasizing that it's a row vector in each row. Thought it looks better Commented Jul 9, 2013 at 20:22

Why is $W^\perp =null(A)$?

Well, $W^\perp=\{x\in V\mid w_1\cdot x=w_2\cdot x=0\}$

and

$null(A)=\left\{x\in V \mid Ax^\top=\begin{bmatrix}0\\0\end{bmatrix} \right\}$

The connection is that $Ax^\top$ computes exactly $w_1\cdot x$ and $w_2\cdot x$.