I have two complex column vectors with length $L$ given by $$\mathbb{x}=[x_1~x_2\cdots x_L]^T,~~~\mathbb{y}=[y_1~y_2\cdots y_L]^T.$$ Further, I have unit magnitude orthonormal vectors $\mathbb{c_1\cdots c_{L-1}}$ such that $\mathbb{c_i^Hc_j}=0~i\neq j,\text{and}~\mathbb{c_i^Hx}=0\text{ for all $i$}.$ In this case how to prove that following result is true $$\|\mathbb{x+y}\|^2=\left|\|\mathbb{x}\|+\frac{\mathbb{x^Hy}}{\|\mathbb{x}\|}\right|^2+\sum_{i=1}^{L-1}|\mathbb{c_i^Hy}|^2.~~~~\text{Eq. 1}$$ Any help in this regard will be much appreciated. Thanks in advance.
My attempt:
It is easy to see that l.h.s. of Eq. 1 is $\|\mathbb{x}\|^2+\|\mathbb{y}\|^2+2Re\{\mathbb{x^Hy}\}$. The r.h.s. becomes $$\|\mathbb{x}\|^2+\mathbb{x}^T\mathbb{y^*}+\mathbb{x^Hy}+\frac{\mathbb{x^Hy}\mathbb{x}^T\mathbb{y^*}}{\|\mathbb{x}\|^2}+\sum_{i=1}^{L-1}\mathbb{c_i^Hy}\mathbb{c_i}^T\mathbb{y^*}$$ and we can see that if this is to equal to the l.h.s. of Eq. 1 then we must have $$\frac{\mathbb{x^Hy}\mathbb{x}^T\mathbb{y^*}}{\|\mathbb{x}\|^2}+\sum_{i=1}^{L-1}\mathbb{c_i^Hy}\mathbb{c_i}^T\mathbb{y^*}=\|\mathbb{y}\|^2. ~~~~~~\text{Eq. 2}$$ So my question is how to prove that Eq. 2 is true? Any help in this regard will be much appreciated. Thanks in advance.
Apply a unitary transformation whose first columns are the $c_i$. Then w.l.o.g. $c_i=e_i$ with the canonical basis vectors. By the assumption, $x_i=0$ for $i=1,...,L-1$, that is, only $x_L\ne 0$. Then \begin{align} \|x+y\|^2&=|y_1|^2+\dots+|y_{L-1}|^2+|x_L+y_L|^2 \\ &=\sum_{i=1}^{L-1}|c_i^Hy|^2+\frac1{|x_L|^2}|x_L^*x_L+x_L^*y_L|^2 \\ &=\sum_{i=1}^{L-1}|c_i^Hy|^2+\frac1{\|x\|^2}\left|\|x\|^2+x^Hy\right|^2 \\ &=\sum_{i=1}^{L-1}|c_i^Hy|^2+\left|\|x\|+\frac{x^Hy}{\|x\|}\right|^2 \end{align} As the last two expressions do not make use of the properties of the canonical basis, they are true in general.