Register to reply

Inner product of complex vectors

by weetabixharry
Tags: complex, product, vectors
Share this thread:
weetabixharry
#1
Dec18-12, 05:19 PM
P: 108
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.
Phys.Org News Partner Science news on Phys.org
Bees able to spot which flowers offer best rewards before landing
Classic Lewis Carroll character inspires new ecological model
When cooperation counts: Researchers find sperm benefit from grouping together in mice
tiny-tim
#2
Dec18-12, 06:12 PM
Sci Advisor
HW Helper
Thanks
tiny-tim's Avatar
P: 26,157
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 ?
weetabixharry
#3
Dec18-12, 06:54 PM
P: 108
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
#4
Dec18-12, 10:44 PM
Homework
Sci Advisor
HW Helper
Thanks
P: 9,656
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?
weetabixharry
#5
Dec19-12, 12:03 AM
P: 108
Quote Quote by haruspex View Post
From which it follows that y=0?
Why?
weetabixharry
#6
Dec19-12, 12:12 AM
P: 108
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.
haruspex
#7
Dec19-12, 12:51 AM
Homework
Sci Advisor
HW Helper
Thanks
P: 9,656
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.
weetabixharry
#8
Dec19-12, 01:21 AM
P: 108
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
haruspex
#9
Dec19-12, 01:54 AM
Homework
Sci Advisor
HW Helper
Thanks
P: 9,656
Suppose you found a y and c =xa+iyb which satisfied all the conditions. Wouldn't 2y and c' =xa+i2yb also satisfy them?
tiny-tim
#10
Dec19-12, 03:36 AM
Sci Advisor
HW Helper
Thanks
tiny-tim's Avatar
P: 26,157
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?


Register to reply

Related Discussions
Inner Product for Vectors in Complex Space Linear & Abstract Algebra 10
Cross product for complex vectors Linear & Abstract Algebra 4
Angle between 2 vectors using 1) Dot product and 2) cross product gives diff. answer? Calculus & Beyond Homework 8
Cross product of complex vectors Linear & Abstract Algebra 2
Hermitian inner product btw 2 complex vectors & angle btw them Linear & Abstract Algebra 1