How Do Imaginary Components Affect Vectors in a 2D Complex Vector Space?

[weetabixharry]

I'm having difficulty understanding the nature of a plane in complex space.

Specifically, I have two complex (N \times 1) vectors, \underline{u}_1 and \underline{u}_2, which are orthonormal:
\underline{u}_1^H\underline{u}_2 = 0 \\<br /> \Vert \underline{u}_k \Vert = 1 \ \ \ (k=1,2)
So, \underline{u}_1 and \underline{u}_2 span a 2D subspace (plane) and any vector in that plane can be written in the form:

\underline{v} = a_1\underline{u}_1 + a_2\underline{u}_2
(for some complex scalars a_1 and a_2). What I want to know is whether certain properties of a_1 and a_2 can tell us something about the nature of \underline{v} (in the context of \underline{u}_1 and \underline{u}_2). Specifically, what if a_1, a_2 are purely imaginary (i.e. \Re e\{a_1\}=\Re e\{a_2\}=0)? I feel that such a construction must have specific (visualisable/intuitive) properties, but I can't see what they might be. Can anyone shed any light on this?

 

[chiro]

Hey weetabixharry.

The only thing I would have to offer merely as a suggestion is to interpret the multiplications themselves of complex numbers visually, which can be interpreted as rotations and scalings corresponding to the polar magnitudes and the angle descriptions.

In terms of linear combinations though, they are just sums of complex numbers which have the standard vector space, arrow-like behaviour but in terms of the multiplication parts, the geometric interpretation in the polar form is probably a good way to answer your question.

 

[weetabixharry]

interpret the multiplications themselves of complex numbers visually, which can be interpreted as rotations and scalings corresponding to the polar magnitudes and the angle descriptions.

Thanks for the response. I've done some more thinking (and reading) but am not convinced that I've really got to the bottom of the issue.

From what I read, one interesting perspective is to interpret the magnitude of the inner product of two (unit-length) complex vectors as being the cosine of the "angle", \theta, between them: cos(\theta) \triangleq \frac{\vert \underline{w}_1^H\underline{w}_2 \vert}{\Vert \underline{w}_1 \Vert \Vert\underline{w}_2 \Vert}
Then, the argument/phase of the inner product represents the "phase", \varphi, between the two vectors:\varphi \triangleq arg \{\underline{w}_1^H\underline{w}_2\}
It seems that this interpretation has the useful property that scaling either/both vector by a complex number (scalar) does not change the "angle" between them.

Returning to my original post, I suppose my problem is that I can't see the role of this "phase" between the two vectors. Specifically, if the phase is zero, then what does that actually tell us? Does this allow us to, in some sense, study the geometry as if it were in real space (since the differences in the two vectors' directions are fully characterised by the real angle \theta)?

 

[chiro]

If the phase is 0 or pi, then the vectors are linearly dependent (or non-zero. scalar multiples of one another).

Phase is the first way to establish orientation in geometry. With complex numbers, the phase gives the rotation and establishes orientation since the multiplication of vectors in the complex space adds the phases and multiplies the lengths.

The orientation part is critical and has a few interpretations. One interpretation is the lie group interpretation where you can form a group with a unit length set of complex vectors.

You can't do this normally with real numbers because of division by zero and also because of the situations involved with dealing with large numbers, so if we need to form a group, we need to remove zero. The unitary complex vectors with the group operation as multiplication provides a way of relating complex numbers, multiplication and geometry.

The other thing is that it fits nicely with existing geometric theory that is based around two important things: distance and angle. The angle corresponds to an inner product and the distance corresponds to either a norm or a metric. This is why complex numbers are kind of a basic relation to geometry in the form of having both distance and angle (real numbers have no orientation and orientation is through angle).

So in a sense multiplying two unit complex numbers tells you the geometric relative orientation between those vectors. Think of it like for example having a way of saying how 'similar' two vectors are. The angle does this and you will always get cos(angle) between -1 and 1 inclusive where 1 means 'directly similar' and -1 means 'completely opposite'.

This one attribute obtained from the inner product gives a relation not only with respect to the relative orientation, but for the relative length as well. So in this sense we get a kind of 'combined' indicator for relativity that is a 'length-phase' measure in same way that we use a formulation of 'space-time' to describe a distance in a 4-dimensional entangled geometry that entangles space with time.

Think about for example two vectors that are dis-similar. When the inner-product is less than 0 we know that they are pointing in somewhat different directions. We could for example keep the value the same by changing the orientation, but then changing the length. Similar case for the positive inner product value. When the inner-product is 0 however, we know that either a vector is zero or the phase is pi/2 and this means no matter how you scale the length of a vector, it doesn't change it's inner product which is useful as well.

 

[weetabixharry]

If the phase is 0 or pi, then the vectors are linearly dependent (or non-zero scalar multiples of one another).

If I understand correctly, this is the "angle" which I defined as \theta, above. Therefore, where you talk about "orientation", I assume this refers to the thing I'm having difficulty understanding (I defined it as \varphi).

I agree with your discussion of inner products (similar things can be said about real-valued vectors). However, I suppose my question is about orientation. Specifically, when the orientations of two complex vectors are the same

arg \{\underline{w}_1^H\underline{w}_2\} = 0

and whether this simplifies analysis in some way.

 

[chiro]

If I understand correctly, this is the "angle" which I defined as \theta, above. Therefore, where you talk about "orientation", I assume this refers to the thing I'm having difficulty understanding (I defined it as \varphi).

I agree with your discussion of inner products (similar things can be said about real-valued vectors). However, I suppose my question is about orientation. Specifically, when the orientations of two complex vectors are the same

arg \{\underline{w}_1^H\underline{w}_2\} = 0

and whether this simplifies analysis in some way.

When you use w^h is this vector a conjugate?

 

[weetabixharry]

When you use w^h is this vector a conjugate?

Yes, sorry, a superscript (.)^H denotes the complex conjugate transpose (and all my vectors are column vectors).

 

[chiro]

Yes, sorry, a superscript (.)^H denotes the complex conjugate transpose (and all my vectors are column vectors).

Well if this is the case then it means both vectors have the same phase since a conjugate inverts the phase with respect to 2pi (i.e. conjugate basically calculates a vector with phase 2pi - normal-phase) but it keeps the length the same.

Does this help?

 

[weetabixharry]

both vectors have the same phase

Yes - I think that is what I was trying to say with this equation:
arg \{\underline{w}_1^H\underline{w}_2\} = 0When this is the case, does it help/simplify the analysis in any way? I feel as though it means that the two complex vectors somehow lie in the same real 2D subspace (?).

 

[Tinyboss]

I'm having difficulty understanding the nature of a plane in complex space.

Specifically, I have two complex (N \times 1) vectors, \underline{u}_1 and \underline{u}_2, which are orthonormal:
\underline{u}_1^H\underline{u}_2 = 0 \\<br /> \Vert \underline{u}_k \Vert = 1 \ \ \ (k=1,2)
So, \underline{u}_1 and \underline{u}_2 span a 2D subspace (plane) and any vector in that plane can be written in the form:

\underline{v} = a_1\underline{u}_1 + a_2\underline{u}_2
(for some complex scalars a_1 and a_2). What I want to know is whether certain properties of a_1 and a_2 can tell us something about the nature of \underline{v} (in the context of \underline{u}_1 and \underline{u}_2). Specifically, what if a_1, a_2 are purely imaginary (i.e. \Re e\{a_1\}=\Re e\{a_2\}=0)? I feel that such a construction must have specific (visualisable/intuitive) properties, but I can't see what they might be. Can anyone shed any light on this?

Pardon me if I'm recounting the obvious, but we're not working in the complex plane here--there are no two orthonormal complex numbers in the plane, just as there are no two nonzero real numbers that aren't scalar multiples of one another. The span of two linearly independent vectors in a complex vector space, sometimes called "a complex plane", has four real dimensions. "The complex plane" is used to refer to the set of complex numbers, which (like every field) is a 1-dimensional vector space over itself. But one complex dimension is two real dimensions, and since we think in real dimensions, we traditionally call that a plane, even though "complex line" is more appropriate, and is often used when talking about a one (complex) dimensional subspace of a complex vector space (which has two real dimensions and so "looks" like a plane).

So, to get back to your specific question: the complex span of u_1,u_2 is a four-dimensional real vector space. By complex span I mean \{a_1u_1+a_2u_2\mid a_1,a_2\in\mathbb{C}\}. Now if you constrain a_1,a_2 to be pure imaginary, you have that ia_1,ia_2 are real, and so the span can be written as \{x_1(iu_1)+x_2(iu_2)\mid x_1,x_2\in\mathbb{R}\}, and so the geometric interpretation is that your subspace has two real dimensions. Or, you could write the span as \{x(iu_1)+(yi)u_2\mid x+yi\in\mathbb{C}\} to make it clear that it's also a one-complex-dimensional subspace.

 

[weetabixharry]

we're not working in the complex plane here

Ah yes, sorry - it seems that I got the terminology wrong. You are correct that I'm not talking about the complex plane (as in a real axis and an imaginary one). I'm talking about a two-dimensional complex vector space.

The rest of what you have written seems to have arrived at what I guessed at in my last post. I'll have to take some time to think about the details, but I think I'm getting there!

Thanks very much for taking the time to respond!

 

[Tinyboss]

Ah yes, sorry - it seems that I got the terminology wrong. You are correct that I'm not talking about the complex plane (as in a real axis and an imaginary one). I'm talking about a two-dimensional complex vector space.

The rest of what you have written seems to have arrived at what I guessed at in my last post. I'll have to take some time to think about the details, but I think I'm getting there!

Thanks very much for taking the time to respond!

You didn't get it wrong--"complex plane" is certainly correct terminology for what you're working with! It's just that it's also used in the other sense, which can be confusing. I thought that probably you had it right, but I wasn't 100% sure, and anyway it's good to have it clarified for others who will come across this thread later. Good luck with your investigation.

 

[zeraoulia]

if the function f(z*) is analytic if the function f(z) is analytic

Hi, there
I ask if the function f(z*) is analytic if the function f(z) is analytic. Here z* is the complex conjugate of z and f: A---C the complex plane.

 

[chiro]

Hi, there
I ask if the function f(z*) is analytic if the function f(z) is analytic. Here z* is the complex conjugate of z and f: A---C the complex plane.

Being analytic means that you have to consider some neighbourhood and only in trivial cases can this be generalized for both cases without any other specific information.

Do you know where f(z) is analytic?

 

[Tinyboss]

Hi, there
I ask if the function f(z*) is analytic if the function f(z) is analytic. Here z* is the complex conjugate of z and f: A---C the complex plane.

Hint: what if f is the identity map?

 

[zeraoulia]

Do you know where f(z) is analytic?

Yes, f(z) is analytic in the domain: 0<Re(z)<1. Re(z) is the real part of z.

 

[chiro]

Do you know where f(z) is analytic?

Yes, f(z) is analytic in the domain: 0<Re(z)<1. Re(z) is the real part of z.

It should also be analytic there for the conjugate as well since the conjugate doesn't change the real part.

 

[zeraoulia]

Thank you very much Chiro

 

[zeraoulia]

Can one tel me about this problem: Let us consider the series (sum an/n*s) if the coffecient an is depend on s, then it is true to consider this serie as a Dirichlet serie or not.

 

[Bacle2]

I'm having difficulty understanding the nature of a plane in complex space.

Specifically, I have two complex (N \times 1) vectors, \underline{u}_1 and \underline{u}_2, which are orthonormal:
\underline{u}_1^H\underline{u}_2 = 0 \\<br /> \Vert \underline{u}_k \Vert = 1 \ \ \ (k=1,2)
So, \underline{u}_1 and \underline{u}_2 span a 2D subspace (plane) and any vector in that plane can be written in the form:

\underline{v} = a_1\underline{u}_1 + a_2\underline{u}_2
(for some complex scalars a_1 and a_2). What I want to know is whether certain properties of a_1 and a_2 can tell us something about the nature of \underline{v} (in the context of \underline{u}_1 and \underline{u}_2). Specifically, what if a_1, a_2 are purely imaginary (i.e. \Re e\{a_1\}=\Re e\{a_2\}=0)? I feel that such a construction must have specific (visualisable/intuitive) properties, but I can't see what they might be. Can anyone shed any light on this?

It may help you to remember that a plane is one-dimensional as a complex subspace,

but two-dimensional over the reals. Not a brilliant insight, but it helped me unconfuse

things a bit.

 

[zeraoulia]

If f(z) is analytic in the domain: 0<Re(z)<1

If f(z) is analytic in the domain: 0<Re(z)<1. Re(z) is the real part of z, then show that
(1) f(z-bar) is also analytic in the domain: 0<Re(z)<1. z-bar is the conjugate of z
(2) f(1-z) is also analytic in the domain: 0<Re(z)<1.
I am not sure about an erlier answer. Can one give to me a reference for this topic.

 

[Bacle2]

I think Chiro correctly answered your question:

Another answer is just using Cauchy-Riemann to double-check. If

f(z)=u+iv , what is f(z-bar)?

For (2), it is just a matter of translation. 1-z just shifts/translates every term by 1.

See if that shift lands you in the domain where f is analytic.

 

[zeraoulia]

f(z-bar)=f(z*) the image of the conjugate z* of z under f

 

[Bacle2]

Right, but if

z=a+ib , then z^* =a-ib

Similarly,

f(z)=U+iV , then f(z^*)=...

 

[zeraoulia]

Yes, I remark that if s is in D=0<re(s)<1, then s* and 1-s are alos in D.

 

[christoff]

I'm talking about a two-dimensional complex vector space.

Ohhhhhhh...