Linear Combination Proof of Orthonormal basis

[RJLiberator]

Homework Statement

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 $|v_2>$.

Homework Equations

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

The Attempt at a Solution

[/B]
I don't 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.

 

[Mark44]

Homework Statement

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 $|v_2>$.

Homework Equations

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

The Attempt at a Solution

[/B]
I don't 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.

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.

 

[HallsofIvy]

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.

 

[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?

 

[Mark44]

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?

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 $v_2$.

 

[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.

 

[Mark44]

Is it really as simple as stating that c must = 0 by the definition of orthogonal

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

I asked this before, but you didn't answer.

What else are you given in this problem?

, 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?

It's convenient and helpful for reaching the conclusion.

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

No.

 

[RJLiberator]

What else are you given in this problem?

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

 

[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.

 

[HallsofIvy]

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

 

[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.

 

[Mark44]

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

More to the point, $v_3 \cdot v = 0$, because this is given.

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.

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.