Basis for orthognal vector

• Apr 6th 2011, 07:10 PM
Diggidy
Basis for orthognal vector
In R3 let S=span((1,2,4)^T)

What is a basis for S^(perpendicular)?

I know that 0=x*y=1y1+2y2+4y3

and that y2 and y3 are free variables so y1=-2y2-4y3

i have then tried to solve for each of the vectors this way but it is not correct.

Thank You,
Diggidy
• Apr 6th 2011, 07:47 PM
TheEmptySet
Quote:

Originally Posted by Diggidy
In R3 let S=span((1,2,4)^T)

What is a basis for S^(perpendicular)?

I know that 0=x*y=1y1+2y2+4y3

and that y2 and y3 are free variables so y1=-2y2-4y3

i have then tried to solve for each of the vectors this way but it is not correct.

Thank You,
Diggidy

You have not solved the system correctly.

$y_1+2y_2+4y_3=0$

Now let $y_3=s,y_2=t,y_1=-2t-4s$

Now we have the vector

$\begin{pmatrix} -2t-4s \\ t \\s \end{pmatrix}=t\begin{pmatrix} -2 \\ 1 \\ 0 \end{pmatrix}+s\begin{pmatrix} -4 \\ 0 \\ 1 \end{pmatrix}$

These two vectors will do the job.
• Apr 6th 2011, 07:51 PM
Diggidy
Quote:

Originally Posted by TheEmptySet
You have not solved the system correctly.

$y_1+2y_2+4y_3=0$

Now let $y_3=s,y_2=t,y_1=-2t-4s$

Now we have the vector

$\begin{pmatrix} -2t-4s \\ t \\s \end{pmatrix}=t\begin{pmatrix} -2 \\ 1 \\ 0 \end{pmatrix}+s\begin{pmatrix} -4 \\ 0 \\ 1 \end{pmatrix}$

These two vectors will do the job.

O, i see, Thank you very much