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)



2 Answers 2


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 $1$.

  • $\begingroup$ When I do what you suggested I get $e^T M^{-1} e = 1$ $\endgroup$
    – Vagabond
    Oct 8, 2010 at 17:09
  • $\begingroup$ Correct! I wanted to write $\hat M$. I make the correction immediately. $\endgroup$ Oct 8, 2010 at 18:00
  • $\begingroup$ How do we proceed from here ? I would be eagerly waiting for the rest of the argument. $\endgroup$
    – Vagabond
    Oct 8, 2010 at 18:23
  • $\begingroup$ I make a correcion, because I forgot a factor $i$ in $N^{-1}e$. This does not change the rest. $\endgroup$ Oct 9, 2010 at 7:00
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    $\begingroup$ The $N$ you took if you eliminate the first row and column we get [-1 1; 1 -1] which is singular and we have the condition that none of the sub matrices are singular. I am not sure though if that affects the rest of the construction. $\endgroup$
    – Vagabond
    Oct 9, 2010 at 8:08

I don't know whether this satisfies all your conditions, but $$\pmatrix{1&1&1&1\cr1&-1&-1&-1\cr1&\alpha&\beta&\gamma\cr1&-\alpha&-\beta&-\gamma\cr}$$ is singular, and no two rows or columns are identical.

  • $\begingroup$ The requirement is none of the sub matrices are singular, I guess I have not been able to specify it clearly. $\endgroup$
    – Vagabond
    Oct 8, 2010 at 11:58
  • $\begingroup$ Edited the question, specifying the condition that none of the minors are singular. Thanks for pointing out. $\endgroup$
    – Vagabond
    Oct 8, 2010 at 12:17

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