Let $A$ be a $n \times n$ matrix all of whose entries has modulus 1. 


Suppose the matrix $A$ is singular. 


We will assume without loss of generality that all the entries in the first row and the first column of the matrix are 1. 


Observe when $n=2$ the matrix $A$ can be then singular if and only if $a\_{2,2}=1$ as well.


A slightly less trivial observation is that the same thing happens when $n=3$, 
that is the matrix $A$ is singular if and only if two of the rows or columns 
are identical.


\begin{equation}
\left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & \alpha\_{2,2} & \alpha\_{2,3} \\
1 & \alpha\_{3,2} & \alpha\_{3,3} \\
\end{array}\right| = 0
\end{equation}


So the matrix $A$ is singular iff $(\alpha\_{2,2}-1)(\alpha\_{3,3}-1)=(\alpha\_{2,3}-1)(\alpha\_{3,2}-1)$.


Let us assume without loss of generality that $\alpha\_{2,2} \neq 1$ and $\alpha\_{3,2} \neq 1$.


Consider the circle $C\_1(t)= (\alpha\_{2,2}-1) (e^{2 \pi i t}-1) $ and $C\_2(t)=(\alpha\_{2,3}-1) (e^{2 \pi i t}-1), t\in [0,1]$.


Since, the two circles either are identical and in that case $\alpha\_{i,2}=\alpha\_{i,3}$ that is the second and third columns are identical, or else as two distinct circles can intersect in at most two points we get similarly two of the rows or columns are identical.


Now, probably it is too much to expect the same result for all $n$.



> 
> But my requirement is only for $n=4$,
>  is it true that a similar result holds
>  for $n=4$ ? 
> 
> 
> Edit: I forgot to mention that I am interested in the case when the matrix is singular > > and none of its sub matrices are singular. (thanks @ Gerry Myerson for pointing it out) 
> 
> 
> 


Thankyou,



structure of singular matrices whose entries have modulus one
Your first calculation for $3\times3$ matrices applies in full generality: If your matrix writes blockwise $[1 \quad e^T ; e \quad M]$, with $e^T=(1,\ldots,1)$, and if $M$ is non-singular (implied by your assumption), then the property that $\det A=0$ is equivalent to $e^T\hat M e=\det M$, that is $e^TM^{-1}e=1$, or to $\det(J-M)=0$, with $J=ee^T$ the matrix with $1$s everywhere. Just use the Sherman-Morrison formula $\det(B+xy^T)=(\det B)(1+y^tB^{-1}x)$.


**Edit** after 3 hours. Take the $3\times3$ matrix $N:=[1 \quad i \quad -1;i \quad -1 \quad1;-1 \quad1 \quad-1]$. We have $N^{-1}e=(-i-1,-2i,-i)^T$. Let $D$ be a diagonal matrix with unit entries on the diagonal, so that $M:=ND^{-1}$ is still an admissible matrix. Then
$$e^TM^{-1}e=e^TDN^{-1}e=(-i-1)z\_1-2iz\_2-iz\_3.$$
**Claim**: There exist unit numbers $z\_j$ such that the right-hand side equals $1$. There exists actually a lot of them Consequence: the matrix $A$ is singular. Yet it does not have two equal rows or columns. *Proof* of the claim: we may search for unit numbers $y\_j$ such that $\sqrt2 y\_1+2y\_2+y\_3=1$. Taking $y\_3=y\_1$, we just have $(1+\sqrt2)y+2y'$, which covers a corona $(\sqrt2-1)\le |z|\le 3+\sqrt2$. In particular the equation $(1+\sqrt2)y+2y'=1$ has a solution. If instead we choose $y\_3=e^{i\epsilon}y\_1$ with a small enough $\epsilon$, the number $\sqrt2 y\_1+y\_3$ covers a circle of raidus $\rho$ close to $\sqrt2-1$, and the corona obtained by adding $2y\_2$ still contain $1$.


