If $v$ is an $n\times 1$ nonzero vector with constant complex entries (i.e., $v\in \mathbb{C}^{n\times 1}$), is there any way to find an $n\times 1$ matrix $A$ so that $$Av^{*}=I_{n}$$
I tried to use $A=(vv^{*})^{-1}\;v$, but we cannot guarantee that $vv^{*}$ to be invertible!
I appreciate any help, thanks!
Let $$v= \begin {pmatrix} v_1\\v_2\\...\\v_n\end {pmatrix}$$ and $$A= \begin {pmatrix} a_1\\a_2\\...\\a_n\end {pmatrix}$$ For $$ Av^{*}=I_{n}$$ we need to have $a_1v_1=1$ and $a_1v_i=0$ for $i=2,3,...,n$
That is $v_2=v_3=...=v_n=0$
On the other hand you need $a_2v_2=1$ which is not possible due to $v_2=0$
Thus the answer is no, we can not find such an $A$
There is a known Rank of Matrix multiplication inequality
$$ \operatorname{rank}(AB) \leq \operatorname{min} \{\operatorname{rank}(A), \operatorname{rank}(B)\} $$
We know $\operatorname{rank}(I_n) = n$ and the rank of an $n \times 1$ vector is at most $1$.
So, unless $n =1$ ...
