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,