Can we recover a vector from one element of resulted vector after multiplication?

Question

I have a matrix $X = \begin{bmatrix} 0.5000 + 0.5000i & 0.5000 - 0.5000i\\ 0.5000 - 0.5000i & 0.5000 + 0.5000i \end{bmatrix}$ multiplied with a column containing a complex number and its conjugate, as below:

$y = \begin{bmatrix} y_1\\ y_2 \end{bmatrix} = \begin{bmatrix} 0.5000 + 0.5000i & 0.5000 - 0.5000i\\ 0.5000 - 0.5000i & 0.5000 + 0.5000i \end{bmatrix} \times \begin{bmatrix} s\\ s' \end{bmatrix}$

I am wondering if we can recover $s$ from only $y_1$ or $y_2$. I mean, as long as the vector $s$ is only containing a complex number with its conjugate, so we can estimate $s$ from only $y_1$. But, I don't know how can we estimate it.

Answer 1

I think the answer is there is no way to recover $s$. Here I will be using a superscript $*$ to indicate the complex conjugate.

First lets expand the matrix multiplication: $$y_1 = \frac{1}{2}\left[(1+i)s + (1-i)s^*\right]$$ $$y_2 = \frac{1}{2}\left[(1-i)s + (1+i)s^*\right]$$

Let $a = 1+i:$

$$y_1 = \frac{1}{2}\left[as + a^*s^*\right]$$ $$y_2 = \frac{1}{2}\left[a^*s + as^*\right]$$

We can immediately see here that the term in the square bracket, for either case, is the sum of a number and its complex conjugate and therefore must be real. This means there is no hope of recovering the value of $s$ is only $y_1$ or $y_2$ is provided unless $s$ is real.

To see what's going on lets solve for $y_1$ $$y_1 = \frac{1}{2}\left[(1+i)s + (1-i)s^*\right]$$

Let $s = s_R + is_I$ $$y_1 = \frac{1}{2}\left[(1+i)(s_R + is_I) + (1-i)(s_R + is_I)^*\right]$$ $$y_1 = \frac{1}{2}\left[(1+i)(s_R + is_I) + (1-i)(s_R - is_I)\right]$$ $$y_1 = \frac{1}{2}\left[s_R + is_I + is_R - s_I + s_R - is_I - is_R -s_I\right]$$ $$y_1 = \frac{1}{2}[2s_R - 2s_I] = s_R - s_I$$

Similarly the expression for $y_1$ can be determined to be $y_2 = s_R + s_I$. So neither expression by itself will let you solve for a complex $s$, but both together will.