Normalizing a vector / Orthonormal?

1. Oct 10, 2015

RJLiberator

1. The problem statement, all variables and given/known data

Which of the following sets of vectors in C^3 is an orthogonal set? Orthonormal? Orthonormal basis?
2. Relevant equations
Orthogonal if v * w *= 0
Orthonormal if orthogonal and each vector is normalized

3. The attempt at a solution

We are giving in the first example:

v:= \begin{pmatrix}
1\\
1\\
1
\end{pmatrix}
w:= \begin{pmatrix}
1\\
-2\\
1
\end{pmatrix}
z:= \begin{pmatrix}
1\\
0\\
-1
\end{pmatrix}

1. It is pretty clear, by computation, to see that this is a set of orthogonal vectors. v*w=0, w*z=0 and v*z=0.
2. Orthonormal requirement 1 is now met, but is each vector normalized? Here is my question:
Is there a way to normalize vectors in general? I assume so from my notes.
If so, can this be normalized? From the looks of it, it doesn't seem so, but how would one go about this process?

With the way it is, IF it could be normalized, can you say this set is orthonormal? Or is it enough that the set is NOT currently normalized to conclude that it is NOT orthonormal?

2. Oct 10, 2015

pasmith

If $\|v\| \neq 0$ then $\frac{v}{\|v\|}$ has norm 1.

3. Oct 10, 2015

Krylov

The second sentence is correct. The set is not orthonormal, but we can obtain a new set (in the way suggested above) that is orthonormal.

4. Oct 10, 2015

RJLiberator

Okay, so what I am getting here is the following:

If ||v|| does NOT equal 0, then it can always be normalized to 1 using the equation posted in the first reply.
But since the set was NOT normalized to begin with, I can conclude that this is NOT orthonormal.

That makes sense to me ^^.

And since it is not orthonormal, it is automatically not an orthonormal basis.

5. Oct 10, 2015

Krylov

I agree.

Maybe just as an exercise, are the following statements true (give a proof) or false (give a counterexample)?
1. Every orthogonal set is linearly independent.
2. Every orthonormal set is linearly independent.
(Don't feel obliged, and do it only when you have already studied linear independence, which I suppose you have.)

6. Oct 10, 2015

RJLiberator

I'm going to jump at this one:
1. No, not every orthogonal set is L.I.
2. Yes! Every orthonormal set is linearly independent and this a linear basis.

Proof? Hm. That will take a bit more time today, but perhaps by the end of my assignment I will be able to easily prove it. I already had the intuition based off the definitions.

To drive the original post in this thread home:

Example 2:

{ 1/sqrt(6) * (1, -2, -i), 1/sqrt(3) (i, i, 1)}

In this example, we have to use complex conjugate by definition, correct?
In using the complex conjugate, we see these two vectors multiplied together do indeed equal 0 and we see orthogonality.

We also conclude that this is not orthonormal since it is not in normalized form.
Therefore it is not an orthonormal basis.

7. Oct 10, 2015

RJLiberator

Hm, it seems I am confused on one major aspect:

In this example: { 1/sqrt(6) * (1, -2, -i), 1/sqrt(3) (i, i, 1)}

If we apply the complex conjugate as I stated in my previous post, then YES it is orthogonal.

Error: NVM this part of my post, found my algebraic error.

This is NOT orthonormal as the first component multiplied by itself DOES equal 1, the second component multiplied by itself does NOT equal 1.

8. Oct 10, 2015

Staff: Mentor

Any orthogonal set that doesn't include the zero vector is linearly independent.

One thing you might not be clear on is what the "normal" part means. The only difference between an orthogonal set of nonzero vectors and an orthonormal set of vectors, is that the orthonormal vectors all have magnitude 1. It's very simple to normalize (i.e., make it length 1) -- just divide the vector by its magnitude. As long as the vector isn't the zero vector, you can always do this.

9. Oct 10, 2015

Ray Vickson

What is the justification of your answer to (1)? Can you actually find an orthogonal set whose members are linearly dependent? (I am betting not!
What is the justification of your answer to (1)? Can you actually give/describe/construct an orthogonal set whose members are not linearly independent?

What is the justification of your answer to (2)? Can you actually start with some orthonormal set and prove that it is a basis? (Waring: be very careful about the actual wording in (2).)

10. Oct 10, 2015

RJLiberator

Okay, I'm at a bit of a loss on this question and want to get back to the very basics of this question.

Let's take for example
(1/sqrt(6), -2/sqrt(6), -i/sqrt(6)) = vector 1 = v
(i/sqrt(3), i/sqrt(3), 1/sqrt(3)) = vector 2 = w

{ vector 1, vector 2 } is the set in question.

1. Is this set orthogonal?
YES. why?
|v> |w> = we must use the complex conjugate of w here, so we see that it does indeed equal 0 when we use the complex conjugate.
This concludes orthogonality.

2. Is it orthonormal?
Well, I'm not sure, but when I take:
|v>*|v> = 1 that is good.
|w>*|w> seems to equal -1/3.

So that would mean that this is NOT orthonormal.

3. This would mean that this cannot be an orthonormal basis.

Is my analysis here correct?

11. Oct 10, 2015

Staff: Mentor

Check your arithmetic. One definition of the norm for complex numbers is $|z| = \sqrt{z \cdot \bar{z}}$

12. Oct 10, 2015

RJLiberator

Beautiful. This is a great definition that I've been missing out here.

This could be the key to this problem!

But after trying to apply it here for the past 20 minutes, I can't seem to figure out how. Let's take our example:

(1/sqrt(6), -2/sqrt(6), -i/sqrt(6)) = vector 1 = v
(i/sqrt(3), i/sqrt(3), 1/sqrt(3)) = vector 2 = w

so If we use that definition of the norm

|v| = | (1/6+4/6-1/6) as we square each component, and namely (-i/sqrt(6))^2 = -1/6
The square root part of the equation turns out to be:

sqrt(<1/sqrt(6), -2/sqrt(6), -i/sqrt(6)> * <1/sqrt(6), -2/sqrt(6), i/sqrt(6)>)
and so we get
sqrt(1/6+4/6+1/6>
= 1

1 does not equal 4/6 so this is not orthogonal

But something tells me I am calculating this improperly.
My assignment has 4 different sets and I calculated them all to be non-orthonornmal, I highly doubt that all 4 of them are not orthonormal.
I must be doing something wrong.

13. Oct 10, 2015

Staff: Mentor

No, you're not doing this right.

Let's look at a simpler example: z = 1 + 2i, and calculate |z| two ways.
1. $|z| = \sqrt{1^2 + 2^2} = \sqrt{5}$ Note that you use the coefficient of i here, not i itself.
2. $|z| = \sqrt{z \cdot \bar{z}} = \sqrt{(1 + 2i) \cdot (1 - 2i)} = \sqrt{1 -4i^2} = \sqrt{5}$

14. Oct 10, 2015

RJLiberator

Well, I completely understand that example, but how do I translate it to this with 3 components?

Do I take it component by component?

That's what is giving me problems here.

15. Oct 11, 2015

Staff: Mentor

You could use the 2nd technique I listed before, i.e., $|z| = \sqrt{z \cdot \bar{z}}$.
What's the complex conjugate of $<\frac{1}{\sqrt{6}}, \frac{-2}{\sqrt{6}}, \frac{-i}{\sqrt{6}}>$?

16. Oct 11, 2015

RJLiberator

complex conjugate : $<\frac{1}{\sqrt{6}}, \frac{-2}{\sqrt{6}}, \frac{i}{\sqrt{6}}>$?

Isn't that what I did in post #12? Sorry for the lack of latex...

17. Oct 11, 2015

Staff: Mentor

Yes
In post #12 you did the same calculation twice, and got two different answers. The part quoted below is the one you got wrong.
The coefficient of i in the third element of your vector is $\frac{-1}{\sqrt{6}}$. When you square it, you get $\frac 1 6$ not $\frac{-1} 6$.

IMO, if you have to find the magnitude of a vector in $\mathbb{C}^2$ or higher dimensions, you're better off using the inner product definition -- $|z| = \sqrt{z \cdot \bar{z}}$. Less chance for making an error, I think.

18. Oct 11, 2015

Ray Vickson

I think you are still not "getting" it, despite having been told over and over again what to do. The examples above are all a bit simple, as each component is either real or pure imaginary. To really hammer home the concept, I suggest you do it for a "mixed" example, such as
$$\vec{v} = \langle 2+i, (1 - i)/2, -3i \rangle$$
What is $|\vec{v}|^2 = \langle \vec{v}, \vec{v} \rangle$ in this case?

19. Oct 11, 2015

RJLiberator

Okay, that makes sense to me -- regarding my calculation:
I am not seeing where I went wrong with my calculation, but I understand that that is how it is defined.

I think with a bit more work on my part soon, I will understand why I went wrong here.

With this being stated, let me try out Ray Vickson's suggestion:

$$\vec{v} = \langle 2+i, (1 - i)/2, -3i \rangle$$

$$\sqrt{\langle 2+i, (1 - i)/2, -3i \rangle * \langle 2-i, (1 + i)/2, +3i \rangle}$$

$$=\sqrt{(2+i)(2+i)+(1-i)/2*(1+i)/2+(-3i)(3i)}$$

$$=\sqrt{ 3+2i+1/2+9}$$

$$= \sqrt{12.5+2i}$$

20. Oct 11, 2015

Staff: Mentor

(Above) No.
The answer should be a real, nonnegative number.