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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)

For the problem I am interested in I have additional information about the matrix $A$. For example I know it is a generalized vandermonde matrix.

Then the results of Schlickewei, Hans Peter; Viola, Carlo Generalized Vandermonde determinants. Acta Arith. 95 (2000), no. 2, 123--137. seems to imply that if the ratio of the entry's are not roots of unity then the choice of the exponent set is rather limited its always finite and except for a few exceptional cases actually very very small.

Here is a brief summary,

My requirement is for the $n=4$, the entries of the matrix have modulus 1, the matrix is a generalised vandermonde matrix in the sense of Schlickwel and Viola, yes in light of the $n=3$ case I do know no proper subdeterminant vanishes. Well I also know the entries are algebraic integers, and we can assume the ratios are not roots of unity.

My question is

Do we know that the solution set is small i.e., in the non exceptional case as mentioned in the above mentioned paper

.

Thankyou,

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,

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)

In fact, I have additional information about the matrix $A$ it is a generalized vandermonde matrix.

Then the results of Schlickewei, Hans Peter; Viola, Carlo Generalized Vandermonde determinants. Acta Arith. 95 (2000), no. 2, 123--137. seems to imply that if the ratio of the entry's are not roots of unity then the choice of the exponent set is rather limited its always finite and except for a few exceptional cases actually very very small.

Here is a brief summary,

My requirement is for the $n=4$, the entries of the matrix have modulus 1, the matrix is a generalised vandermonde matrix in the sense of Schlickwel and Viola, yes in light of the $n=3$ case I do know no proper subdeterminant vanishes. Well I also know the entries are algebraic integers, and we can assume the ratios are not roots of unity.

My question is

Do we know that the solution set is small i.e., in the non exceptional case as mentioned in the above mentioned paper

.

Thankyou,

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)

For the problem I am interested in I have additional information about the matrix $A$. For example I know it is a generalized vandermonde matrix.

Then the results of Schlickewei, Hans Peter; Viola, Carlo Generalized Vandermonde determinants. Acta Arith. 95 (2000), no. 2, 123--137. seems to imply that if the ratio of the entry's are not roots of unity then the choice of the exponent set is rather limited its always finite and except for a few exceptional cases actually very very small.

Here is a brief summary,

My requirement is for the $n=4$, the entries of the matrix have modulus 1, the matrix is a generalised vandermonde matrix in the sense of Schlickwel and Viola, yes in light of the $n=3$ case I do know no proper subdeterminant vanishes. Well I also know the entries are algebraic integers, and we can assume the ratios are not roots of unity.

My question is

Do we know that the solution set is small i.e., in the non exceptional case as mentioned in the above mentioned paper

.

Thankyou,

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 $n=4$ ?

In fact, I have additional information about the matrix $A$ it is a generalized vandermonde matrix.

Then the results of Schlickewei, Hans Peter; Viola, Carlo Generalized Vandermonde determinants. Acta Arith. 95 (2000), no. 2, 123--137. seems to imply that if the ratio of the entry's are not roots of unity then the choice of the exponent set is rather limited its always finite and except for a few exceptional cases actually very very small.

Here is a brief summary,

My requirement is for the $n=4$, the entries of the matrix have modulus 1, the matrix is a generalised vandermonde matrix in the sense of Schlickwel and Viola, yes in light of the $n=3$ case I do know no proper subdeterminant vanishes. Well I also know the entries are algebraic integers, and we can assume the ratios are not roots of unity.

My question is

Do we know that the solution set is small i.e., in the non exceptional case as mentioned in the above mentioned paper

.

Thankyou,

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)

In fact, I have additional information about the matrix $A$ it is a generalized vandermonde matrix.

Then the results of Schlickewei, Hans Peter; Viola, Carlo Generalized Vandermonde determinants. Acta Arith. 95 (2000), no. 2, 123--137. seems to imply that if the ratio of the entry's are not roots of unity then the choice of the exponent set is rather limited its always finite and except for a few exceptional cases actually very very small.

Here is a brief summary,

My requirement is for the $n=4$, the entries of the matrix have modulus 1, the matrix is a generalised vandermonde matrix in the sense of Schlickwel and Viola, yes in light of the $n=3$ case I do know no proper subdeterminant vanishes. Well I also know the entries are algebraic integers, and we can assume the ratios are not roots of unity.

My question is

Do we know that the solution set is small i.e., in the non exceptional case as mentioned in the above mentioned paper

.

Thankyou,