I have two complex column vectors $\mathbb{a}=[a_1~a_2~\cdots a_L]^T$ and $\mathbb{y}=[y_1~y_2\cdots y_L]^T$ with $\|\mathbb{a}\|=1$. I want to find the lower bound on the following $$\|\mathbb{y}\|^2-|\mathbb{y^Ha}|^2$$
In one of the research papers it is mentioned that if $\mathbb{y}=[y_1~y_2\cdots y_L]^T$ and $\mathbb{a}=[a_1~a_2\cdots a_L]^T$ are complex column vectors with length $L$ and $\|\mathbb{a}\|=1$ then $$\|\mathbb{y}\|^2-|\mathbb{y^Ha}|^2=\|\mathbb{u}_{(2,L)}\|^2~~~~~~~~~~~(\text{Eq. 1})$$ where $\mathbb{u}_{(2,L)}$ for a $L$ length complex vector $\mathbb{u}=[u_1~u_2~\cdots u_L]$ is defined as $\mathbb{u}_{(2,L)}=[u_2~u_3~\cdots u_L]$ and $\mathbb{u}$ is defined as $$\mathbb{u}=\mathbb{U^H(y-a)}$$ where $\mathbb{U^H}$ is a unitary matrix whose first row is $\mathbb{a^H}$ and superscript $\mathbb{H}$ denotes the Hermitian operator.
Actually, I was trying to verify Eq. 1 but when I evaluate l.h.s. and r.h.s. of Eq. 1 my results do not match (I have also shown my effort for this verification in What is wrong that I am doing in verifying a basic results about complex vectors?). So Now I am thinking maybe authors wanted to say that
$$\|\mathbb{y}\|^2-|\mathbb{y^Ha}|^2 \geq \|\mathbb{u}_{(2,L)}\|^2.~~~~~~\text{Eq. }2$$ Can somebody tell me whether Eq. 2 is right or wrong? I will be very thankful to you. Thanks in advance.
We have $u = U(y-a)$. Hence, $||u||^2 = u^H u = (y-a)^H U^H U (y-a) = (y-a)^H (y-a)$ (since $U$ is unitary). Simplifying we get: $$ ||u||^2 = y^H y + a^H a -a^H y - y^H a = y^H y + 1 -2 \mathrm{Real}(y^H a) \tag{1}. $$
On the other hand, we have $$||u_{(2, L)}||^2 = ||u||^2 - |a^H y - 1|^2 \tag{2}$$ because $$ u = U(y-a)= \begin{bmatrix} a^H (y-a)\\ u_2\\ \vdots\\ u_L \end{bmatrix}. $$ Also, $$ |a^H y - 1|^2 = |a^H y|^2 + 1 - 2 \mathrm{Real}(a^H y) = |y^H a|^2 + 1 - 2 \mathrm{Real}(y^H a) \tag{3}.$$
Replacing (1) and (3) in (2) we get $$ ||u_{(2, L)}||^2 = y^H y - |y^H a|^2 = ||y||^2 - |y^H a|^2 . $$
Note that when doing such derivations it is important to know that $a^H y \ne y^H a$. In fact, $a^H y = y^T a^* = (y^H a)^*$. This is why, $a^H y + y^H a = 2 \mathrm{Real}(y^H a)$.