Finding the basis for W perp

1. Mar 20, 2012

triucsd

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

Let W be a subspace of ℝ4 spanned by the vectors:

u1 = [1; -4; 0; 1], u2 = [7; -7; -4; 1]

Find an orthogonal basis for W by performing the Gram Schmidt proces to there vectors. Find a basis for W perp (W with the upside down T).

2. Relevant equations

Gram Schmidt

3. The attempt at a solution

I applied the Gram Schmidt process and found the basis for W which was {[1; -4; 0; 1], [5; 1; -4; -1].

Here's where I run into the problem. I'm trying to get the basis for W perp, but I don't really know what to do. I tried finding the basis for the transpose of W, I got x3 and x4 as the free variables, but the resulting vectors I got was wrong, where do I start from here?

Thanks.

2. Mar 20, 2012

Robert1986

I did the same thing and got the same free variables. Those are the correct variables. Are you sure you are getting the wrong vectors? Which ones are you getting?

3. Mar 20, 2012

triucsd

I'm getting x3[0; 4/21; 1; 0] and x4[-1; 6/21; 0; 1], but the answer is [4; 0; 6; -4], [0; 4; -3; 16]. I've done the calculation several times and I keep getting the same answers.

EDIT:

I got to those vectors by transposing W, resulting in:

[1, -4, 0, 1; 5 1 -4 -1]. After that I row reduced it and got my parametric equations.

Last edited: Mar 20, 2012
4. Mar 20, 2012

Robert1986

What is your row-reduced matrix? It looks like you might be making a mistake there.
I might suggest putting this into wolfam|alpha and seeing if you are messing up you row-reductions any.

Also, forget about what the book says the basis should be. I have found that in problems like this, book authors often put a technically correct answer in the back of the book, but perhaps not one that you would normally get. So, to check your answers, just take the dot product of the vectors you get with each of the vectors you were given. They should all be zero, right?

5. Mar 20, 2012

triucsd

Plugging in [1, -4, 0, 1; 5 1 -4 -1] in wolfram|alpha and having it row reduce the matrix gave me [1, 0, -16/21, -1/7; 0 1 -4/21, -2/7].

Doing the calculations again I get the free variables x3 and x4 with x3[16/21; 4/21; 1; 0] and x4[1/7; 2/7; 0; 1].

Scaling these vectors doesn't get me anywhere near [4; 0; 6; -4], [0; 4; -3; 16]. I suspect that I'm working with the wrong vectors while solving for the parametric equations, but I can't think of what other vectors I can use.

Last edited: Mar 20, 2012
6. Mar 20, 2012

Robert1986

OK, to make calculations simple, let x3 = 21 and x4 = 7. Then your two vectors are:

[16 4 21 0] and [1 2 0 7]

if I did my arithmetic correctly, both of these are orthogonal to both of the vectors you were given. So, these span W perp. Now, you were asked for A basis in the problem right? That is, not THE basis. The answer in the back of the book is also A basis for W perp, just not the same as the one you got. Your answer is correct. To convince yourself, show that the two vectors given in the book are in the linear span of the two vectors you computed.

7. Mar 20, 2012

triucsd

Is my basis different because there are infinitely many solutions for this problem and I just happened to find another one? Thanks, for all the help by the way!

EDIT:

To show if the two vectors are in the linear span of the two vectors do I just take:

c1[16; 4; 21; 0] + c2 [1; 2; 0; 7] = [1; -4; 0; 1]

c1[16; 4; 21; 0] + c2 [1; 2; 0; 7] = [-7; -7; -4; 1]

and solve for c1 and c2?

Last edited: Mar 20, 2012
8. Mar 20, 2012

Robert1986

Yes to both questions.

I'm not really sure how whoever wrote the solutions came up with that particular basis. I'd have to guess that he did it exactly the way we did it, and then took some odd linear combinations to come up with what is in the back of the book. Anyway, on problems like this, I wouldn't even look at the back of the book because it is really easy to check by yourself anyway. And, as you correctly pointed out, there are infinite solutions to this problem.