# Finding a basis of a space

Let u = [1, 2, 3, -1, 2]T, v = [2, 4, 7, 2, -1]T in 5.
Find a basis of a space W such that wu and wv for all wW.

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?

wTu = wTv = 0

wT(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?

Last edited:

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.

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

Mark44
Mentor
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.
ichabodgrant said:
Better to let w1, w2, w3, w4, w5 instead of x...?

Yeah typo...sorry
the 2 is redundant

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

HallsofIvy
Homework Helper
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 lets 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(...).

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 dont have a linearly independent set then you dont have a basis. A basis is defined to be a linearly independent spanning list afterall.

WWGD
Gold Member
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.