# Homework Help: Orthogonal spanning set

1. Apr 14, 2014

### mpittma1

1. The problem statement, all variables and given/known data
Let V be an inner product space. Show that if w is orthogonal to each of the vectors
u1,u2,...,ur, then it is orthogonal to every vector in the span{u1,u2,...,ur}.

2. Relevant equations

3. The attempt at a solution
Not sure how to show this, if w is orthogonal to every vector u1,u2,...,ur, wouldn't w already be orthogonal to the spanning set?

The spanning set is a subspace of the subspace of the inner product space of V.

Am I suppose to show it symbolically?

2. Apr 14, 2014

### jbunniii

Two things to consider:

(1) What is the definition of orthogonality?
(2) What does a general element of $\text{span}\{u_1, \ldots,u_r\}$ look like?

3. Apr 14, 2014

### mpittma1

So for orthogonality two vectors in an inner product space are orthogonal if (in this case) the vectors <u, w> = 0.

A spanning set is just the set of all linear combinations of the vectors in V.

So for the spanning set w = k1u1 + k2u2,...,krur

which becomes

0 = k1u1 + k2u2,...,krur

so the only difference between w being orthogonal with u1,u2,...,ur is that there is some scalar k thats been thrown into the mix?

4. Apr 14, 2014

### jbunniii

OK.

Well, the right hand side is a general element of the span, so that's good. But we want to check whether $w$ is orthogonal to this element, we don't want to set $w$ equal to this element. So to avoid confusion, let's give the general element a new name, say $v = k_1 u_1 + \ldots + k_r u_r$. Now you want to check whether $w$ is orthogonal to $v$. So what is $\langle w, v \rangle$?

5. Apr 14, 2014

### mpittma1

$\langle w, v \rangle$ = ku1w+ku2w+...+kurw =0

then after saying that, if k was the zero vector it proves that to be true

6. Apr 14, 2014

### jbunniii

No, you don't need $k$ to be the zero vector. Let's take a step back.
$$\langle w,v \rangle = \langle w, k_1 u_1 + \ldots + k_r u_r\rangle = ?$$
What properties of the inner product do you know that can help to simplify this? The goal is to get an expression involving $\langle w, u_i\rangle$ for $i=1, \ldots, r$.

7. Apr 14, 2014

### mpittma1

Properties of inner products say the scalar k can be brought out as such k<w,u>

so we could say k<w, ui> = 0 for k ε R and for i = 1, 2, ... r

then at this point the value of k doesnt matter because i have already said that <w, ui> = 0 for i = 1, 2, ... r.

Please correct me if Im wrong, I think im starting to see it.

I

8. Apr 14, 2014

### jbunniii

Yes, that's right. You also need one more fact, namely:
$$\langle w, k_1 u_1 + \ldots + k_r u_r \rangle = \langle w, k_1 u_1 \rangle + \ldots + \langle w, k_r u_r \rangle$$
Now you can apply your step of sliding the $k_i$'s out of the inner product, and use the fact that $\langle w, u_i \rangle = 0$ as you said.

One small detail: in case you happen to be working with complex numbers, the $k_i$'s get conjugated if you move them from the second position in the inner product. So:
$$\langle w, k_i u_i \rangle = \overline{k_i} \langle w, u_i\rangle$$
But if you move them from the first position then there is no conjugate:
$$\langle k_i u_i, w \rangle = k_i \langle u_i, w \rangle$$
If the $k_i$'s are real numbers, then it doesn't make any difference, since the conjugate of $k_i$ is just $k_i$.

9. Apr 14, 2014

### mpittma1

This is a basic linear algebra course so we only work with real values. But thank you so much for all your help, I truly understand this now!