# Linear Combination Proof of Orthonormal basis

1. Oct 11, 2015

### RJLiberator

1. The problem statement, all variables and given/known data
Assume that $(|v_1>, |v_2>, |v_3>)$ is an orthonormal basis for V. Show that any vector in V which is orthogonal to $|v_3>$ can be expressed as a linear combination of $|v_1>$ and $|v_2>$.

2. Relevant equations
Orthonormality conditions:
|v_i>*|v_j> = 0 if i≠j OR 1 if i=j.

3. The attempt at a solution

I dont know how to mathematically prove this obvious one.
I understand the definitions here.

Orthonormal basis implies that the set of three vectors lives in dimension three and are all orthogonal to one another. This means the |v_1> *|v_2> = 0.

If some vector is orthogonal to |v_3> that means |x> |v_3> = 0.

I am pretty sure we have to use sum notation for an inner product and orthonormality conditions to prove this statement.

Hints on starting out properly?
Here's the only lead I have:

|x> = c*|v_1> + t*|v_2>
as it is a linear combination of v_1 and v_2. But we don't want to assume the proof.

2. Oct 11, 2015

### Staff: Mentor

It helps to think of the geometry here. Since $v_1, v_2$ and $v_3$ form a basis for V, then they span V. If $v \in V$ is orthogonal to $v_3$, then it must lie in the two-dimensional subspace of V spanned by $v_1$ and $v_2$; i.e., the plane that is spanned by these two vectors. At this point, I would draw a sketch, showing the plane of $v_1$ and $v_2$, with $v_3$ sticking straight out of the plane. Recognizing that $v_1$ and $v_2$ lie in this plane, and are a basis for that two-D subspace of V, then v must be a linear combination of $v_1$ and $v_2$. When you have a handle on the geometry, it should be easier to write a proof that fleshes out the details.

3. Oct 11, 2015

### HallsofIvy

Staff Emeritus
Since v1, v2, and v3 are a basis, any vector, v, can be written as v= av1+ bv2+ cv3. Now, take the inner product of both sides with v.

4. Oct 11, 2015

### RJLiberator

Thanks guys, what you have stated I understand.

Taking the starting point that hallsofivy has gifted me, it's clear to see that c = 0, and a and b can be any numbers. This essentially is the proof.
HallsofIvy suggests taking the inner product of both sides with v.

LHS: <v|v> = |v|^2
RHS: <v|av_1> + <v|bv_2> + <v|cv_3>
a*<v|v_1>+b*<v|v_1>+c*<v|v_3> = |v|^2

I'm not really sure what this proves, is this not assuming the proof?

5. Oct 11, 2015

### Staff: Mentor

What else are you given in this problem? There is information given that you aren't using.

Your last line should be an equation showing that v is a linear combination of $v_1$ and $v_2$.

6. Oct 11, 2015

### RJLiberator

Is it really as simple as stating that c must = 0 by the definition of orthogonal, and so low and behold, v must be a linear combination of $v_1$ and $v_2$ ?

Why do we need the inner product of both sides?

Can we not just jump immediately to that step? C = 0, so linear combination is seen.

7. Oct 12, 2015

### Staff: Mentor

No, there's nothing that says that c = 0.

It's convenient and helpful for reaching the conclusion.
No.

8. Oct 12, 2015

### RJLiberator

Is the piece of information that you are alluding to the fact that the set is an orthonormal basis?

9. Oct 12, 2015

### RJLiberator

Okay, so after taking the inner product, it is clear to see that the inner product of <v|v_3> = 0 because of the definition of orthogonality.

And now we have
<v|v> = a<v|v_1>+b<v|v_2>

Which shows that it is a linear combination of the other two vectors.

10. Oct 12, 2015

### HallsofIvy

Staff Emeritus
Actually, I miswrote. I meant to say "take the dot product of each side with v1, v2, and v3 in turn".

11. Oct 12, 2015

### RJLiberator

Ah, now this is the grove I needed.

Taking the dot product in turn shows what |v_i> * |v> is clearly due to the orthogonal and orthonormal rules.

It can be seen that |v_3>|v> must be 0 due to definition of orthogonality and so c=0.
|v_2>|v> = b
|v_1>|v> = a

and so the answer seems clear. :D

With c = 0, all you need to do is go back to basis, plug it in and we have a linear combination representation of v.

12. Oct 12, 2015

### Staff: Mentor

More to the point, $v_3 \cdot v = 0$, because this is given.
A simpler proof:
$v = av_1 + bv_2 + cv_3$, since $\{v_1, v_2, v_3\}$ is a basis for V
$\Rightarrow cv_3 = v - av_1 - bv_2$
$\Rightarrow cv_3 \cdot v_3 = v \cdot v_3 - av_1 \cdot v_3 - bv_2 \cdot v_3$
$\Rightarrow c = 0 - 0 - 0 = 0$ Can you give a justification for each of the four dot products?
Hence $v = av_1 + bv_2$, as required.