# Orthogonal vectors

1. Mar 20, 2014

### negation

1. The problem statement, all variables and given/known data

Let S be the subspace of all vectors in R4 that are orthogonal to each of the vectors
(0, 4, 4, 2), (3, 4, -2, -4)
What is an example of a matrix for which S is the nullspace?

3. The attempt at a solution

I'm not sure how I should be intepreting the question:

[ 0 ,4 ,4 ,2 ;3, 4, -2, -4] = [ x , y ,z , t] = [ 0 , 0 ]

0x + 4y + 4z + 2t = 0

3x + 4y -2 z -4t = 0

from here I set up an augmented matrix and solve for the set of vectors x ,y ,z ,t? By definition if x, y, z, t results in each linear equation = 0, then, x ,y ,z ,t are vectors with properties of orthogonality?

2. Mar 20, 2014

### ehild

There is multiplication instead of equality in the first line? If yes, it looks all right, do it.

ehild

3. Mar 20, 2014

### negation

[ 0 ,4 ,4 ,2 ;3, 4, -2, -4] .[ x , y ,z , t] = [ 0 , 0 ]

Yes, it should be a multiplicative operation-dot product.
But I aren't exactly sure what should I be doing with the 2 linear equation. Do I perform reduce row echelon?

4. Mar 20, 2014

### ehild

Yes. But it is of that form already... You will have two free parameters.

ehild

5. Mar 20, 2014

### negation

I'm still unclear. Could you explain?

6. Mar 20, 2014

### ehild

There is the system of equations:

0x + 4y + 4z + 2t = 0

3x + 4y -2 z -4t = 0

What is the general solution?

ehild

7. Mar 20, 2014

### negation

It is done by setting up an augmented matrix:

0 4 4 2 | 0
3 4 -2 -4 | 0

Perform REF(Not RREF), then solve for the variables.

Edit: x1 = 2x3 + 2x4, x2 = -x3-0.5x4, x3, x4

Last edited: Mar 20, 2014
8. Mar 20, 2014

### ehild

Well, solve for the variables....

ehild

9. Mar 20, 2014

### negation

Post #7

10. Mar 20, 2014

### ehild

X3=a, x4=b a,b free parameters. So the S subspace consists of vectors ((2a+2b); (-a-0.5b); a; b).
And a matrix for which S is the nullspace?

ehild

11. Mar 20, 2014

### negation

Relooking at the question, I'm not sure if we interpreted it correctly.

I think the question is asking for the {SX=0|x1=0,4,4,2, x2=(3,4,-2,-4),S$\in R4$}

Edit: I think we were on the right track.

But this brings up a question. What if we're asked to solve S, given X where SX = 0?

Last edited: Mar 20, 2014
12. Mar 20, 2014

### Staff: Mentor

Let's untangle your notation. S is a subspace of R4 such that if v $\in$ S, then v $\cdot$ <0, 4, 4, 2> = 0 and v $\cdot$ <3, 4, -2, -4> = 0.

The problem asks that you find a matrix (call it A) for which S is the nullspace of A. IOW, if v $\in$ S, then Av = 0.

13. Mar 20, 2014

### negation

Isn't the vector [2z+2t; -z-0.5t;z;t] the answer?

14. Mar 20, 2014

### ehild

S is the nullspace of a linear transformation T, that is, Tv =0 for all vectors v$\in$S. Show a matrix of that transformation. (You have shown one in Post #3) .

ehild

15. Mar 20, 2014

### Staff: Mentor

See post #12. The "answer" should be a matrix, not a vector.

In addition, what you have above would be better written as two vectors - a basis for S.

16. Mar 20, 2014

### negation

Am I suppose to perform another one? I thought we have had all the solutions already. 2 are free variables and the corollary is that there are infinite solutions.

17. Mar 20, 2014

### ehild

The subspace S contains infinite number of vectors. You have found these vectors already.

The next question is
ehild

18. Mar 20, 2014

### negation

any matrix with coefficient $\in R$?

19. Mar 20, 2014

### negation

basis for S : z (2,-1,1,0) + t (2,-0.5,0,1)

20. Mar 20, 2014

### HallsofIvy

Your wording is peculiar. z (2,-1,1,0) + t (2,-0.5,0,1) is a general vector in S. The set of vectors {(2, -1, 1, 0), (2, -0.5, 0, 1)} is a basis.

The answer to this question is any matrix, A, such that A applied to each of those two vectors is 0.