# What this symbol mean

1. Nov 22, 2008

### transgalactic

x=a1,b1,c1
y=a2,b2,c2
<x,y>+<x,y>= (a1*a2 +b1*b2 +c1*c2) + (a1*a2 +b1*b2 +c1*c2)=2*(a1*a2 +b1*b2 +c1*c2)

i have <x,y> + <x,y> it written that it equals

2Re<x,y>

what 2Re<x,y> means ??

2. Nov 22, 2008

### cristo

Staff Emeritus
Re means "the real part of..."

3. Nov 22, 2008

### transgalactic

buts its not a complex number its a vector operation

why to involve therms from complex numbers in that formula when developed it other wise

x=a1,b1,c1
y=a2,b2,c2
<x,y>+<x,y>= (a1*a2 +b1*b2 +c1*c2) + (a1*a2 +b1*b2 +c1*c2)=2*(a1*a2 +b1*b2 +c1*c2)

why am i mistaken?

http://img179.imageshack.us/img179/7205/97924861fq7.gif [Broken]

and what the meaning of |< >| marked in blue in the link

Last edited by a moderator: May 3, 2017
4. Nov 22, 2008

### winter85

if you look carefully, it says <x,y> + <y,x>, not <x,y> + <x,y>.

if the inner product is defined on a complex vector space, which means that a1,b1,c1, a2,b2,c2 are complex numbers, then <x,y> is a complex number, and <y,x> is the complex conjugate of <x,y>
in which case | < > | means the complex norm of the complex number given by the inner product.

5. Nov 22, 2008

### jambaugh

I believe that the vectors are supposed to be complex and the asterisks should represent complex conjugates. Note then too that the order in the inner product matters. Your formula is then a bit off and should read:

$$\langle x,y \rangle + \langle y , x \rangle = \cdots = 2\Re(\langle x , y \rangle )$$

Reversing the inner product effects a complex conjugate of the value thus adding both you cancel out the imaginary part and get twice the real part.

6. Nov 22, 2008

### transgalactic

when i wrote astrix i ment multiplication

7. Nov 22, 2008

### transgalactic

whats the full formula for <x,y>
(including the complex part)

??

8. Nov 22, 2008

### transgalactic

9. Nov 22, 2008

### jambaugh

I know you ment it as multiplication but...
1.) If you are working with complex vectors (implied by the $$\Re$$), and
2.) If you are working with a Hermitian inner product, then
the complex conjugate is required:

$$\langle x , y \rangle = x_1^* \cdot y_1 + x_2^*\cdot y_2 + \ldots$$

(where I am using * for complex conjugate and $$\cdot$$ for multiplication.)

If you are not working with a Hermitian inner product then the formula is just plain wrong.
If you are not working with possibly complex vectors then the formula is trivial:
$$\langle x,y\rangle + \langle x , y \rangle = 2\langle x , y \rangle$$

See what I mean?

You didn't state from where you got your formula or for what reason you are writing it. You asked about the $$\Re$$ = "Re" symbol and I can only assume it came from some assignment or text. I am also extrapolating that the text must be talking about complex vectors with a Hermitian inner product.

Go back to your source of information and recheck it without the assumptions that the vectors are real or that $$\langle x , y \rangle = \langle y , x \rangle$$

10. Nov 22, 2008

### transgalactic

11. Nov 22, 2008

### transgalactic

why are they
opening the 2ab part
not as <x,y> + <x,y>

but as
<x,y> + <y,x>

??

12. Nov 22, 2008

### cristo

Staff Emeritus
For the reasons stated above. Note that that section of the wikipedia article proves the inequality for an arbitrary inner product space, that is, a vector space endowed with an inner product. There is nothing that says the vectors have to be real

If I may, I'd suggest you didn't try and learn things from wikipedia, but instead invest in a good book.

13. Nov 22, 2008

### jambaugh

It's about vectors real or complex. In the real case $$Re( \langle x, y \rangle) = \langle x,y \rangle$$ hence the "or equals" in the first inequality relation.

In the complex case they are in effect invoking the triangle inequality on the complex norm:

Let $$Re(\langle x,y \rangle) = a+bi$$
Then $$|a| \le |a|+|b| \le |a+bi| = \sqrt{a^2 + b^2}$$
where a, b are real numbers and the second inequality is the trangle inequality on the complex norm.

Note that we can write the complex squared norm norm as:
$$a^2+b^2= (a+bi)^*\cdot (a+bi) = c^* c$$
where $$c = a+bi$$ our complex number. Recall that $$i^2 = -1$$ and $$(a+bi)^* = a-bi$$ (where a,b are real).

Thence the magnitude of a complex number is $$|c| = \sqrt{c^* c} = \sqrt{a^2 + b^2}$$

Also $$Re(c) \equiv \Re(c) = a$$ and $$Im(c) = \Im(c) = b$$ is just a way to indicate the real and imaginary parts of a complex number.

That is all the details implicit in the wiki article. Sit down and work them out on paper and you should be able to see what's goin' on.

14. Nov 22, 2008

### transgalactic

whats a complex conjugate??

15. Nov 22, 2008

### Pere Callahan

Maybe you should start learning the basics.... and then go on to more advanced topics....

16. Nov 22, 2008

### transgalactic

ok now i know whats complex conjugate

know i will read it all over again

17. Nov 22, 2008

### transgalactic

what the values of x1 ,y1 ,x2 ,y2

$$\langle x , y \rangle = x_1^* \cdot y_1 + x_2^*\cdot y_2 + \ldots$$

Last edited: Nov 22, 2008
18. Nov 24, 2008

### jambaugh

I was using subscripted x's to indicate the x vector's coordinates. Your a,b,c my x_1 x_2 x_3.
That way we can talk about a vector with more than 3 or even 26 coordinates/dimensions.

BTW One way to think about complex conjugation is to remember we invented i to be one of the two square roots of -1 The other of course is -i
$$i^2=-1, i = \pm \sqrt{-1}$$.
Complex conjugation switches this choice, i.e. where ever an i occurs replace it with -i.