Why will $(\vec u_1)^H H^H_{AB} H_{AB}\vec u_1=(\vec u_2)^HH^H_{AB}H_{AB}\vec u_2 $,when both $\vec u_1$ and $\vec u_2$ are unit vector

Question

$\mathbf H_A$ is a $5$ by $6$ matrix,and it just has only one null space vector.

Now,i create two $6$ by $1$ random vectors,$\vec v_1$ and $\vec v_2$

and i use their vector ,$\vec v^⊥_1$ and $\vec v ^⊥_2$, which are both orthogonal to the $\mathbf H_A$ matrix

$\vec v^⊥_1=\vec v_1-\vec v^{||}_1$,and $\vec u_1=\frac{\vec v^⊥_1}{||\vec v^⊥_1||_2}$

$\vec v^⊥_2=\vec v_2-\vec v^{||}_2$,and $\vec u_2=\frac{\vec v^⊥_2}{||\vec v^⊥_2||_2}$

$\vec v^{||}_1$ and $\vec v^{||}_2$ are $\vec v_1$ and $\vec v_2$ that project onto the $\mathbf H_A$ ,that is,

$\vec v^{||}_1=\mathbf H_A'(\mathbf H_A\mathbf H_A')^{-1}\mathbf H_A\vec v_1$

$\vec v^{||}_2=\mathbf H_A'(\mathbf H_A\mathbf H_A')^{-1}\mathbf H_A\vec v_2$

Now,i create a $2$ by $6$ random matrix, $\mathbf H_{AB}$,and found $(\vec u_1)^H\mathbf H^H_{AB}\mathbf H_{AB}\vec u_1=(\vec u_2)^H\mathbf H^H_{AB}\mathbf H_{AB}\vec u_2=$scalar value,not a matrix

Why?$\vec u_1 \neq \vec u_ㄉ$,why does the result two different vector multiply the same matrix become the same?

when i change the size of $\mathbf H_A$ to $5$ by $7$ and change $\vec v_1$ and $\vec v_2$ to $7$ by $1$,to let the $\mathbf H_A$ has two null space vector,$(\vec u_1)^H\mathbf H^H_{AB}\mathbf H_{AB}\vec u_1\neq (\vec u_2)^H\mathbf H^H_{AB}\mathbf H_{AB}\vec u_2$

It seems that the reason will have a relation with the number of null space vector,but i don't know how to explain it,and not very sure about that

Answer 1

First ,you said that the orthogonal vector of $v_1$ and $v_2$ are orthogonal to $H_A$,and there is just a single null space vector in the $H_A$.

That is ,your $u_1$ and $u_2$ are both actually projected onto that single null space vector ,so you can imagine that null space vector is a line,and the direction of $u_1$ and $u_2$ can be the same or opposite as this null space vector,because no matter which direction of the $u_1$ and $u_2$ are,they both orthogonal to $H_A$

So you can find the your $u_1=u_2$(same direction) or $u_1=-u_2$(opposite),so that is why $||u_1 H_A||^2=||u_2 H_A||^2$

And when the number of null space vector is larger than $1$,take $2$ for example,you can image that if a vector can be as a line,so two vectors can become a plane,so in this situation,your $u_1$ won't be the same as $u_2$ or $-u_2$ in most of cases,that is , $||u_1 H_A||^2 \neq ||u_2 H_A||^2$ in most of cases