Question

Let S be the subspace of R^3 spanned by the vectors x=(x1,x2,x3)^T and y=(y1,y2,y3)^T, let


A= x1 x2 x3
y1 y2 y3

show that orthagonal complement of S=N(A)

Answer 1

This is essentially the same problem you had yesterday.

Answer 2

no,for that one , we had to find basic; not sure what to do on this?

Answer 3

do you know what it means for a vector to be in the orthagonal complement of S

Answer 4

their dot product is 0?

Answer 5

yes...so you want all the vectors z such that when dotted with a vector from S you get the zero

Answer 6

some day I will say this correctly...

you want all the vectors z such that when dotted with any vector from S you get zero

Answer 7

so what about it being equal to N(A) part?

Answer 8

\[A=\left[\begin{matrix}x_1 & x_2 & x_3 \\ y_1 & y_2 & y_3 \end{matrix}\right]\]

\[\left[\begin{matrix}x_1 & x_2 & x_3 \\ y_1 & y_2 & y_3 \end{matrix}\right]\left[\begin{matrix}z_1\\ z_2 \\z_3\end{matrix}\right]\]
\[=\left[\begin{matrix}x_1z_1+x_2z_2+x_3z_3 \\ y_1z_1+y_2z_2+y_3z_3\end{matrix}\right]\]

\[=\left(\begin{matrix}\vec{x}\cdot \vec{z} \\ \vec{y}\cdot \vec{z}\end{matrix}\right)=\left(\begin{matrix}0 \\ 0\end{matrix}\right)\]


this is the null space of A

Answer 9

I have to go...hope this helps.

Answer 10

thanks so much