
#1
Dec1812, 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) a^{H}c = x (purely real, known) (4) b^{H}c = 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. 



#2
Dec1812, 06:12 PM

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hi weetabixharry!




#3
Dec1812, 06:54 PM

P: 96





#4
Dec1812, 10:44 PM

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Inner product of complex vectors 



#5
Dec1912, 12:03 AM

P: 96





#6
Dec1912, 12:12 AM

P: 96

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. 



#7
Dec1912, 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.




#8
Dec1912, 01:21 AM

P: 96

Beyond that, I guess I can say: c^{2} = x^{2}  y^{2} 



#9
Dec1912, 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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