Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am struggling with a Linear Algebra problem that involves finding the length of a vector

$w_1 = (i, 1, 0) \in W$ a vector space over $\mathbb{C}$

So, the way I did it is like this:

$$\|w_1\| = \sqrt{i^2 + 1^2 + 0^2} = \sqrt{-1 + 1 + 0} = 0$$

Is this correct?

share|improve this question
4  
That's a really short vector. You might want to check the definition of length in the complex case. –  André Nicolas Apr 19 '12 at 0:41
    
BTW, $W$ is a vector space over $\mathbb{C}$. You could say $W$ is $\mathbb{C}^3$, and $\mathbb{C}^3$ is a vector space over $\mathbb{C}$. –  Michael Hardy Apr 19 '12 at 0:46
1  
Thanks for that. What's the LaTEX for the complex symbol? ... Nevermind, I copied it from the answer below. –  adaptive Apr 19 '12 at 0:49
add comment

2 Answers

up vote 4 down vote accepted

The length of a vector $v=(z_1, z_2, \ldots, z_n)\in \mathbb{C}^n$ is $||v||=\sqrt{z_1\overline{z_1}+\ldots + z_n\overline{z_n}}$, where $\overline z$ denotes the complex conjugate of $z$.

share|improve this answer
    
Ooooh! I seeeee! Thanks. –  adaptive Apr 19 '12 at 0:47
add comment

No. You have to multiply each component not by itself, but by its complex conjugate. A component is real if and only if it is its own complex conjugate, so you multiply $0$ by $0$, getting $0^2$, and $1$ by $1$, getting $1^2$. But $i$ is not real. Its complex conjugate is $-i$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.