# Thread: Normalizing complex vector

1. ## Normalizing complex vector

Hi.
I'm trying to normalize the vector $\displaystyle v=(i,i,0)$, but I'm getting the wrong answer.

here's what I did:

$\displaystyle \hat{v}=\frac{v}{\left \| v \right \|}=\frac{(i,i,0)}{\sqrt{i^2+i^2+0}}=\frac{(i,i,0) }{\sqrt{-2}}=\frac{(i,i,0)}{2i}=(\frac{i}{2i},\frac{i}{2i}, 0)=(\frac{1}{2},\frac{1}{2},0)$

but $\displaystyle \hat{v}$ length is not 1, since: $\displaystyle \left \| \hat{v} \right \|=\sqrt{(\frac{1}{2})^2+(\frac{1}{2})^2+0}=\frac{ 1}{\sqrt{2}}\neq 1$

thanks in advanced!

2. ## Re: Normalizing complex vector

Originally Posted by Stormey
Hi.
I'm trying to normalize the vector $\displaystyle v=(i,i,0)$, but I'm getting the wrong answer.

here's what I did:

$\displaystyle \hat{v}=\frac{v}{\left \| v \right \|}=\frac{(i,i,0)}{\sqrt{i^2+i^2+0}}=\frac{(i,i,0) }{\sqrt{-2}}=\frac{(i,i,0)}{2i}=(\frac{i}{2i},\frac{i}{2i}, 0)=(\frac{1}{2},\frac{1}{2},0)$

but $\displaystyle \hat{v}$ length is not 1, since: $\displaystyle \left \| \hat{v} \right \|=\sqrt{(\frac{1}{2})^2+(\frac{1}{2})^2+0}=\frac{ 1}{\sqrt{2}}\neq 1$

thanks in advanced!
I'm not familiar with the norm you have. I'm pretty sure $\displaystyle \| v\| = \sqrt{v^T\overline v} = \sqrt{2}$.

Ignoring that for now, you went from $\displaystyle \sqrt{-2}$ to $\displaystyle 2i$ in the denominator instead of $\displaystyle i\sqrt{2}$.

3. ## Re: Normalizing complex vector

I'm not familiar with the norm you have. I'm pretty sure $\displaystyle \| v\| = \sqrt{v^T\overline v} = \sqrt{2}$.

Originally Posted by Gusbob
Ignoring that for now, you went from $\displaystyle \sqrt{-2}$ to $\displaystyle 2i$ in the denominator instead of $\displaystyle i\sqrt{2}$.

I'm always doing this stupid mistakes...

so:

$\displaystyle \hat{v}=\frac{v}{\left \| v \right \|}=\frac{(i,i,0)}{\sqrt{i^2+i^2+0}}=\frac{(i,i,0) }{\sqrt{-2}}=\frac{(i,i,0)}{\sqrt{2}i}=(\frac{i}{\sqrt{2}i} ,\frac{i}{\sqrt{2}i},0)=(\frac{1}{\sqrt{2}},\frac{ 1}{\sqrt{2}},0)$

and then:

$\displaystyle \left \| \hat{v} \right \|=\sqrt{(\frac{1}{\sqrt{2}})^2+(\frac{1}{\sqrt{2} })^2+0}=1$

Originally Posted by Gusbob
I'm not familiar with the norm you have. I'm pretty sure $\displaystyle \| v\| = \sqrt{v^T\overline v} = \sqrt{2}$.
What do you mean?
Isn't the norm is defined to be:
$\displaystyle \left \| x \right \|=\sqrt{|x_1|^2+|x_2|^2+...+|x_n|^2}$
?

4. ## Re: Normalizing complex vector

Originally Posted by Stormey
What do you mean?
Isn't the norm is defined to be:
$\displaystyle \left \| x \right \|=\sqrt{|x_1|^2+|x_2|^2+...+|x_n|^2}$
?
Yes, so in your case, $\displaystyle |x_1|=|i|=1,|x_2|=|i|=1,|x_3|=|0|=0$ and $\displaystyle \|x\| = \sqrt{1^2+1^2+0^2}=\sqrt{2}$

5. ## Re: Normalizing complex vector

Originally Posted by Gusbob
Yes, so in your case, $\displaystyle |x_1|=|i|=1,|x_2|=|i|=1,|x_3|=|0|=0$ and $\displaystyle \|x\| = \sqrt{1^2+1^2+0^2}=\sqrt{2}$
isn't $\displaystyle |x_1|,|x_2|,|x_3|$ supposed to be squared?

plus if $\displaystyle \left \| v \right \|=\sqrt{2}$, then:

$\displaystyle \hat{v}=\frac{(i,i,0)}{\sqrt{2}}=(\frac{i}{\sqrt{2 }},\frac{i}{\sqrt{2}},0)$

and then: $\displaystyle \left \| \hat{v} \right \|\neq 1$

which means that $\displaystyle \left \| v \right \|$ can't be equal to $\displaystyle \sqrt{2}$

6. ## Re: Normalizing complex vector

Yes, and they are squared! $\displaystyle 1^2= 1$ and $\displaystyle 0^2= 0$.

And the length of $\displaystyle \left(\frac{i}{\sqrt{2}},\frac{i}{\sqrt{2}}, 0\right)$ is 1:
$\displaystyle \sqrt{\left(\frac{i}{\sqrt{2}}\right)\left(\frac{-i}{\sqrt{2}}\right)+ \left(\frac{i}{\sqrt{2}}\right)\left(\frac{-i}{\sqrt{2}}\right)+ (0)(0)= \sqrt{\frac{1}{2}+ \frac{1}{2}+ 0}= \sqrt{1}= 1$

7. ## Re: Normalizing complex vector

Originally Posted by Stormey
. . . then:

$\displaystyle \hat{v}=\frac{(i,i,0)}{\sqrt{2}}=(\frac{i}{\sqrt{2 }},\frac{i}{\sqrt{2}},0)$
That looks right to me.

And $\displaystyle \left \| \hat{v} \right \|$ does equal 1.

And $\displaystyle \left \| v \right \|$ does equal $\displaystyle \sqrt{2}$.

8. ## Re: Normalizing complex vector

Oh... I forgot about the conjugate.

Thank you all for the help.

### normalizing complex vectors

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