The difficulty of generate complex Hadamard matrix

Question

A complex $n\times n$ matrix $A=[a_{ij}]$ is called a Hadamard matrix if $A^{+}A=nI$ and $|a_{ij}|=1$ holds for all $i,j$, where $A^{+}$ denotes the conjugate transposed matrix of $A$, and a vector $x=(x_1,x_2.....x_n)\in \mathbb{C}^n$ is called unit if $|x_i|=1$ holds for all $i$.

For $n>5$, are there $d\times n$ matrix $F=[f_{ij}]$ which satisfies the following properties:

1. $FF^{+}=dI$,

2. $|f_{ij}|=1$ , holds for all $i,j$,

3. there is no unit vector $x\in \mathbb{C}^n$ satisfies $F x^{+}=0$.

Which can be regarded as the difficulty of generate complex Hadamard matrix, actually, it is wanted to show the following method to find a complex Hadamard matrix does not always work:

Let $S$ be an empty set, choose a unit vector $x$ which is orthogonal to all the elements in $S$, and put $x$ in $S$, until there is no unit vector which orthogonal to all elements in $S$. Now what we ask can be regarded as is it possible that the algorithm stops but $|S|< n$.

Answer 1

The circulant matrix $G$ with $3$ rows, $8$ columns and with first row

\begin{equation*} G= \begin{bmatrix} -1 & -1 & 1 & -1 & 1 & 1 & 1 & 1\\ \end{bmatrix} \end{equation*} satisfies

(1) $$ GG^{T} = 8I, $$

(2)

All entries in $G$ are $-1$ or $1$,

(3)

The only solution to: $XG=0$ is $X=0.$

Thus, there is no unit vector with $XG=0.$