Interpreting a zero norm of complex vector

**janie_t** · February 11th 2010, 05:07 AM

What can we conclude about the norm of a complex vector (dimension 2) when it's equal to 0? What does it mean? What can be said about their components x and y?

**dedust** · February 11th 2010, 05:48 AM

Originally Posted by janie_t

What can we conclude about the norm of a complex vector (dimension 2) when it's equal to 0? What does it mean? What can be said about their components x and y?

$z = x + iy$
$|z| = 0 \Longleftrightarrow z = 0$ , that is $x = 0$ and $y = 0$

**janie_t** · February 11th 2010, 05:50 AM

But what if x and y are not equal to 0 and the norm is 0. Basically x and y are complex vectors and their norm is 0 so what can I conclude from that?

**dedust** · February 11th 2010, 05:59 AM

Originally Posted by janie_t

But what if x and y are not equal to 0 and the norm is 0. Basically x and y are complex vectors and their norm is 0 so what can I conclude from that?

which one is it? x, y are the components or x,y are the complex vectors?

anyway, x,y should be 0

**janie_t** · February 11th 2010, 06:03 AM

Suppose you have the following vector x= (5+2i,-2+5i), then the norm ||x||=0.

What can you say about this vector?

**dedust** · February 11th 2010, 06:07 AM

Originally Posted by janie_t

Suppose you have the following vector x= (5+2i,-2+5i), then the norm ||x||=0.

What can you say about this vector?

what is your definition for norm here?

**janie_t** · February 11th 2010, 06:08 AM

Let x1 and x2 be the components of the vector x, then the norm of x is SQUARE ROOT [ (x1)^2 + (x2)^2 ]

**CaptainBlack** · February 11th 2010, 06:18 AM

Originally Posted by janie_t

Let x1 and x2 be the components of the vector x, then the norm of x is SQUARE ROOT [ (x1)^2 + (x2)^2 ]

That is not the norm of a 2-vector over $\mathbb{C}$ . Assuming $\bold{x}$ is a column vector the usual norm is:

$\|\bold{x}\|=\sqrt{\overline{\bold{x}}^T{\bold{x}} }=\sqrt{x_1\overline{x_1}+x_2\overline{x_2}} =\sqrt{|x_1|^2+|x_2|^2}$

which $0$ if and only if both $x_1$ and $x_2 =0$

CB

**dedust** · February 11th 2010, 06:22 AM

Originally Posted by janie_t

Let x1 and x2 be the components of the vector x, then the norm of x is SQUARE ROOT [ (x1)^2 + (x2)^2 ]

this is not define a "norm" because a norm must satisfy $||x|| = 0 \leftrightarrow x = 0$

in here, you have $||x|| = 0$ but $x \not = 0$

**janie_t** · February 11th 2010, 11:05 AM

I am talking about the Euclidean norm.

Suppose you have the following vector x= (5+2i,-2+5i), then how come I get ||x||=0. I don't understand why the Euclidean Norm of the vector is 0 when its components are not zero.

**Bruno J.** · February 11th 2010, 11:35 AM

Because it's not the Euclidean norm, as others have pointed out.

**janie_t** · February 11th 2010, 11:40 AM

x=(5+2i,-2+5i)

Well,
||x||^2 = (5+2i)^2 + (-2+5i)^2
||x||^2 = (25 + 10i + 4i^2) + ( 4 -10i + 25i^2)
||x||^2 = (25+10i-4) + (4-10i-25) if we let (i^2 = -1)
||x||^2 = 25 + 10i -4 + 4 - 10i - 25
||x||^2 = 0
||x|| = 0

? Is there any mistake in my calculations? The initial components of x were not 0, yet the norm is zero.

?

**Bruno J.** · February 11th 2010, 12:11 PM

I suggest you read this.

**janie_t** · February 11th 2010, 02:53 PM

Sorry for bothering you all! I found my mistake!

**HallsofIvy** · February 12th 2010, 04:37 AM

And it was exactly what everyone had been telling you all along?

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