# Thread: basis of the subspace of R4 consisting of all vectors perpendicular to v.

1. ## basis of the subspace of R4 consisting of all vectors perpendicular to v.

v=[3 -5 7 9]^T

Find a basis of the subspace of R4 consisting of all vectors perpendicular to v.

I have no clue how to do this...

Thank you,
Diggidy

2. Originally Posted by Diggidy
v=[3 -5 7 9]^T

Find a basis of the subspace of R4 consisting of all vectors perpendicular to v.

I have no clue how to do this...

Thank you,
Diggidy
Here is a hint Find the "plane" with $\displaystyle v$ as a normal vector that passes though the origin. It is

$\displaystyle 3x_1-5x_2+7x_3+9x_4=0$

Now solve this under-determined system to find three vectors perpendicular to $\displaystyle v$

3. Same thing, really: a vector <w, x , y, z> is perpendicular to <3, -5, 7, 9> if and only if there dot product, 3w- 5x+ 7y+ 9z= 0.

You can solve for any one of those variables in terms of the other 3, then write <w, x, y, z> in terms of those three.

Rather than do that problem for you consider the space of all vector perpendicular to <-2, 4, 6, 5>. That would consist of all vectors <w, x, y, z> such that -2w+ 4x+ 6y+ 5z= 0. Solving for w, w= 2x+ 3y+ (5/2)z. Then, <w, x, y, z>= <2x+ 3y+ (5/2)z, x, y, z>= <2x, x, 0, 0> + <3y, 0, y, 0> + <(5/2)z, 0, 0, z>= x<2, 1, 0, 0>+ y<3, 0, 1, 0> + z<5/2, 0, 0, 1>.

And now, a basis is obvious.

If you don't like fractions, that last one can be written (z/2)<5, 0, 0, 2>