Inner product of complex vectors


by weetabixharry
Tags: complex, product, vectors
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#1
Dec18-12, 05:19 PM
P: 96
I have three (N x 1) complex vectors, a, b and c.

I know the following conditions:

(1) a and b are orthonormal (but length of c is unknown)
(2) c lies in the same 2D plane as a and b
(3) aHc = x (purely real, known)
(4) bHc = iy (purely imaginary, unknown)

where (.)H denotes Hermitian (conjugate) transpose, i is the imaginary unit and x,y are real numbers.

Given that I know x, can I deduce y?

My hunch is that (without the "purely real/imaginary" statements), these conditions would define y up to an arbitrary complex phase, but the "purely real/imaginary" conditions allow the phase to be known too. However, my reasoning relies on there being some sense of "angle" between a and c and between b and c... such that these angles sum to 90 for the orthonormality condition (1). I don't know if this is valid.
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tiny-tim
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#2
Dec18-12, 06:12 PM
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hi weetabixharry!
Quote Quote by weetabixharry View Post
(2) c lies in the same 2D plane as a and b
doesn't that mean that c must be a linear combination of a and b ?
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#3
Dec18-12, 06:54 PM
P: 96
Quote Quote by tiny-tim View Post
doesn't that mean that c must be a linear combination of a and b ?
Yes, where I guess the coefficients of the linear combination are complex scalars.

haruspex
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#4
Dec18-12, 10:44 PM
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Inner product of complex vectors


Quote Quote by weetabixharry View Post
Yes, where I guess the coefficients of the linear combination are complex scalars.
From which it follows that y=0?
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#5
Dec19-12, 12:03 AM
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Quote Quote by haruspex View Post
From which it follows that y=0?
Why?
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#6
Dec19-12, 12:12 AM
P: 96
Quote Quote by haruspex View Post
From which it follows that y=0?
c is a linear combination of a and b:

c = Aa + Bb

for A,B complex scalars.

Therefore, from (3) and (1), A = x
and, from (4) and (1), B = iy

I can't see why y=0

I guess, from this I have:

c = xa + iyb

which is 1 equation in 2 unknowns (y and c)... so I'm stumped.
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#7
Dec19-12, 12:51 AM
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You know facts about aHc, bHc. How can you combine that with with knowing c = Aa + Bb? Actually I was wrong to suggest y=0, but you can at least make progress this way.
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#8
Dec19-12, 01:21 AM
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Quote Quote by haruspex View Post
You know facts about aHc, bHc. How can you combine that with with knowing c = Aa + Bb? Actually I was wrong to suggest y=0, but you can at least make progress this way.
I combined these in my previous post to write A,B as functions of x,y.
Beyond that, I guess I can say:

|c|2 = x2 - y2
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#9
Dec19-12, 01:54 AM
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Suppose you found a y and c =xa+iyb which satisfied all the conditions. Wouldn't 2y and c' =xa+i2yb also satisfy them?
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#10
Dec19-12, 03:36 AM
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hi weetabixharry!

(just got up )
Quote Quote by weetabixharry View Post
c = xa + iyb
that's right!
which is 1 equation in 2 unknowns (y and c)...
so what's the answer to ?
Quote Quote by weetabixharry View Post
Given that I know x, can I deduce y?


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