# Thread: Proving orthogonal vectors are a subspace

1. ## Proving orthogonal vectors are a subspace

Problem:
For a set of vectors { $u_1,....,u_k$} $\subset R^n$ show that the set of vectors $X \subset R^n$ that are orthogonal to each $u_i$ is a subspace of $R^n$

Solution:
Alright, I'm thinking that since the set of vectors X is orthogonal to $u_i$ then I can say something like
{ $u_1x_1, u_2x_2,....,u_kx_k$} = 0
therefore,
$u_1x_1$ = 0 , and $u_2x_2$= 0
so
$u_1x_1$ + $u_2x_2$ = 0

Am I thinking this through properly?

2. Think about what it means to be a subspace.

You're notation is confusing. You should be dealing with an inner product, and show that when pairing $u_i$ with a linear combination of vectors in the orthogonal complement, that you still get 0.

3. Originally Posted by MattMan
Think about what it means to be a subspace.

You're notation is confusing. You should be dealing with an inner product, and show that when pairing $u_i$ with a linear combination of vectors in the orthogonal complement, that you still get 0.
So you're saying it should look more like this?

$Xu_1$ = 0 and $Xu_2$ = 0

$X(u_1 + u_2)$ = 0

$Xu_1 + Xu_2$ = 0

4. ***Reply got posted twice***

5. If we let $U=\{u_1,...,u_k\}\subset\mathbb{R}^n$
Clearly $0 \in U^{\perp}$ since $(0,u_i)=0,\: 1
So $0 \in U^{\perp}\neq \emptyset$
Now let $x_1,x_2 \in U^{\perp}; \alpha,\beta \in \mathbb{R}$
Now consider $(\alpha x_1+\beta x_2,u_i)=\alpha (x_1,u_i)+\beta (x_2,u_i)=0+0=0, \text{ for } 1, hence $\alpha x_1 + \beta x_2 \in U^{\perp}$. So $U^{\perp}$ is closed under linear combinations and hence a subspace of $\mathbb{R}^n$.

6. Thank you very much for clearing that up!