Why will the result of two different vector multiple the same matrix be the same?$||\mathbf H_{AB} f_{0.1}||^2 \approx ||\mathbf H_{AB} f_{0.2}||^2$

Question

First,i use matlab to create three rayleigh fading channel

$\mathbf H_A = \frac{1}{\sqrt2}\times [randn(5,6) + j \ randn(5,6)]$,$\mathbf H_A$ is a $5$ by $6$ matrix which elements are complex

$\mathbf H_{AB} = \frac{1}{\sqrt2}\times [randn(2,6) + j \ randn(2,6)]$,$\mathbf H_{AB}$ is a $2$ by $6$ matrix which elements are complex

$\mathbf H_{AC} = \frac{1}{\sqrt2}\times [randn(2,6) + j \ randn(2,6)]$,$\mathbf H_{AC}$ is a $2$ by $6$ matrix which elements are complex

then calculate their null space ,$\vec n_A$,to let $\mathbf H_A \vec n_A=\vec 0$,and find $\vec v_{AB}$ and $\vec v_{AC}$,which is from the SVD result.Now because some reason,i have to let $\vec v_{AB}$ and $\vec v_{AC}$ projected onto the $\vec n_A$ to become a new vector ,$\vec f_{AB}$ and $\vec f_{AC}$.That is,

$\vec f_{AB}=\vec n_A (\vec n_A^H \vec n_A)^{-1}\vec n_A^H \vec v_{AB}$ , H is conjugate transpose,$\vec f_{AB}$ is $6$ by $1$ vector

$\vec f_{AC}=\vec n_A (\vec n_A^H \vec n_A)^{-1}\vec n_A^H \vec v_{AC}$ , H is conjugate transpose,$\vec f_{AC}$ is $6$ by $1$ vector

Now i assume a new vector is the sum of part of $\vec f_{AB}$ and part of $\vec f_{AC}$,and normalize the new vector.that is

$f_{new}=\frac{\alpha \vec f_{AB}+(1-\alpha)\vec f_{AC}}{\alpha^2 ||\vec f_{AB}||^2+(1-\alpha)^2 ||\vec f_{AC}||^2}$,and $||f_{new}||^2=1$

So now we can know we will have different $f_new$ vector with the different $\alpha$.

Now ,i call $f_{new}=f_{0.1} $ when $\alpha=0.1;$ and $f_{new}=f_{0.2} $ when $\alpha=0.2;$.As we know,$f_{0.1} \neq f_{0.2}$,but why is $||\mathbf H_{AB} f_{0.1}||^2 \approx ||\mathbf H_{AB} f_{0.2}||^2...\approx ||\mathbf H_{AB} f_{0.9}||^2$ ?$||\mathbf H_{AC} f_{0.1}||^2 \approx ||\mathbf H_{AC} f_{0.2}||^2...\approx ||\mathbf H_{AC} f_{0.9}||^2$their difference unit is $ 10^{-15}$

Answer 1

A random $5\times 6$ matrix is going almost sure to have a one dimensional kernel, i.e. the construction is equivalent to taking a random vector $\vec n_A$. Now you project two vectors on $\vec n_A$ and get two parallel vectors $\vec f_{AB}=b\vec n_A$, $\vec f_{AC}=c\vec n_A$. Normalization gives $$ \vec f_{new}=\frac{\alpha b\vec n_A+(1-\alpha)c\vec n_A}{\|\alpha b\vec n_A+(1-\alpha)c\vec n_A\|}=\frac{d\vec n_A}{\|d\vec n_A\|}=\pm\frac{\vec n_A}{\|\vec n_A\|}. $$ It does not depend on $\alpha$.

P.S. It is something wrong with your normalization formula, it does not make $\|f_{new}\|=1$.