Inner product of complex vectors

weetabixharry

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^Hc = x (purely real, known)
(4) b^Hc = 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.

tiny-tim

hi weetabixharry!
doesn't that mean that c must be a linear combination of a and b ?

weetabixharry

Yes, where I guess the coefficients of the linear combination are complex scalars.

haruspex

From which it follows that y=0?

weetabixharry

Why?

weetabixharry

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

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

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

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

hi weetabixharry!

(just got up )
that's right!
so what's the answer to … ?