How Do You Find a Basis for the Orthogonal Complement of Given Vectors in ℝ5?

[ichabodgrant]

Let u = [1, 2, 3, -1, 2]^T, v = [2, 4, 7, 2, -1]^T in ℝ⁵.
Find a basis of a space W such that w ⊥ u and w ⊥ v for all w ∈ W.

I think the question is quite easy. Given this vector w in the space W is orthogonal to both u and v. I can only think of w being a zero vector. But would this be too trivial?

w^Tu = w^Tv = 0

w^T(u - v) = 0

I believe there is something wrong. Things above are what I can compute...
Is there any other way to solve this question?

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Oh, I've thought of another way.

Set a augmented matrix
[1 2 3 -1 2 | 0]
[2 4 7 2 -1 | 0]

~

[1 2 3 -1 2 | 0]
[0 2 4 3 -3 | 0]

Let x₅ = t, x₄ = s, x₃ = z where t, s, z ∈ ℝ.
∴ x₂ = 1.5t - 1.5s - 2z
x₁ = -5t + 4s + z

Therefore, the solution set is {t[-5 1.5 2 0 0 1]^T + s[4 -1.5 0 1 0]^T + z[1 -2 1 0 0]^T}.
The basis is {[-5 1.5 2 0 0 1]^T, [4 -1.5 0 1 0]^T, [1 -2 1 0 0]^T}.Is this correct?

 

[Svein]

wTu = wTv = 0

You have started OK, but then you have hurried along too fast. Let w = [w1, w2, w3, w4, w5]. Now apply your conditions (written as scalar products: $\left\langle w, u \right\rangle = 0$ and $\left\langle w, v \right\rangle = 0$ ). You will not find a complete solution (after all, there are 5 variables and two equations), but you will have some constraints on the coordinates.

 

[ichabodgrant]

So my second try is correct?
Better to let w1, w2, w3, w4, w5 instead of x...?

 

[Mark44]

So my second try is correct?

No.
The first vector in your set has too many components (typo?). Each vector in your set has to be perpendicular to u and v, which you can easily check by doing dot products.

Better to let w1, w2, w3, w4, w5 instead of x...?

 

[ichabodgrant]

Yeah typo...sorry
the 2 is redundant

 

[WWGD]

Seems like W would be the intersection of the ortho complements of u,v .

 

[HallsofIvy]

I would do it this way- write the general vector in your subspace as (a, b, c, d, e). Such a vector is perpendicular to (1, 2, 3, -1, 1) if and only if a+ 2b+ 3c- d+ e= 0. It is perpendicular to (2, 4, 7, 2, -1) if and only if 2a+ 4b+ 7c+ 2d- e= 0. Adding those two equations together, 3a+ 6b+ 10c- d= 0 so that d= 3a+ 6b+ 10c. Putting that into either or the first equations let's you solve for, say, e in terms of a, b, c, and d. Replace d and e in (a, b, c, d, e) by those and then write the result as a( ...)+ b(...)+ c(...).

 

[Ahmad Kishki]

for your first answer that w=0, you should understand that a set of vectors with the zero vector is never linearly independent. Think about it for a bit. You can easily prove it yourself. And if you don't have a linearly independent set then you don't have a basis. A basis is defined to be a linearly independent spanning list afterall.

 

[WWGD]

for your first answer that w=0, you should understand that a set of vectors with the zero vector is never linearly independent. Think about it for a bit. You can easily prove it yourself. .

A 1-line proof: take any non-zero c , then c.0=0 is a non-trivial combination that gives you 0.