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?
$\endgroup$3 Answers
$\begingroup$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.
$\endgroup$ 6 $\begingroup$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 $$ \begin{bmatrix} – & {w_1}^T & –\\ - & {w_2}^T & – \end{bmatrix} \begin{bmatrix} | \\ x \\ | \end{bmatrix} = \begin{bmatrix}0\\0\end{bmatrix} $$
$\endgroup$ 6 $\begingroup$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 $w_2\cdot x$.
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