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Let $M$ be a $3\times 3$ real matrix with at least $3$ distinct elements and the property that any permutation of it's elements gives a matrix with determinant $0$.

Must $M$ contain exactly seven $0$s?

This question is a special case of my previous question: Exactly $n-1$ nonzero elements if $\det(A)=0$ for every arrangement

Thanks to user Holonomia, we know a counterexample to the analogous question for $2\times 2$ matrices: Any $2\times 2$ matrix with two $1$s and two $-1$s has determinant $0$.

However, I have not been able to make much progress on the $3\times 3$ case. Theoretically the property gives us a system of $9!$ equations (with some symmetries and repeats), but I haven't found a good way to compute with this idea.

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  • $\begingroup$ What do you mean by "the equivalent question"? Are the two questions logically equivalent? $\endgroup$ – Omnomnomnom Jun 12 '15 at 1:06
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    $\begingroup$ @PeterWoolfitt : A quick python script shows that there are only $5040$ distinct equations. :) $\endgroup$ – Alexey Burdin Jun 12 '15 at 1:08
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    $\begingroup$ Partial answer: If $M$ has at least $3$ zero entries but less than $7$, then its entries can be permuted to make an upper triangular matrix with non-zero-entries on the diagonal. Such a matrix must have a non-zero determinant. We therefore deduce that if a counterexample exists, it has at most $2$ zero-entries. $\endgroup$ – Omnomnomnom Jun 12 '15 at 1:37
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    $\begingroup$ Partial partial answer: It can't have exactly two zero too: If not, set the first row to be $(c, 0, 0)$ Then the $(1, 1)$ minor would have zero determinant. Then the entries of this minor must be $a, a, -a, a$ (or $a, a, a, a)$ for some $a\neq 0$. By permutating $(c, 0, 0)$ to $(0,c, 0)$, then all the other six entries are $\pm a$, for some $a\neq 0$. But then (1) if they are all $a$, $a$ has to be $c$ and that violate your assumption. If not, then there're $\ge 3$ $a$'s and 1 $-a$. But that is not possible as you can make the minor to be $\begin{bmatrix} a & a \\ -a & a \end{bmatrix}$. $\endgroup$ – user99914 Jun 12 '15 at 1:56
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    $\begingroup$ I don't know how helpful this is (if at all), but if you treat the equations defining your determinant condition as a linear system in variables of the form $M_{ij}M_{kl}M_{mn}$, you can iteratively eliminate variables and end up with only 75 distinct equations. $\endgroup$ – David Zhang Jun 12 '15 at 3:09
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Let $a, b, c$ be three distinct elements in the entries of the matrix. We will be using this lemma a lot:

Woolfitt's lemma: Let the matrix be given by $$(*)\ \ \ \ \begin{bmatrix} a & b& c \\ \cdot & \cdot & \cdot \\ \cdot& \cdot& \cdot \end{bmatrix},$$

then $A=B=C$, where $A, B, C$ are the minors with respect to $a, b, c$ respectively.

(See the comment for the proof)

Now we split into two cases.

  • First case: there is yet another entry $d$ not equals to $a, b, c$. Then consider the matrix

\begin{bmatrix} d & b& c \\ a & \cdot & \cdot \\ \cdot & \cdot & \cdot \end{bmatrix}

and use the lemma. Then one also get $D' = B' =C'$ (Three new minors). But actually $A = D'$. Thus they are all the same. Then there will be two zeroes in the matrix (the $(3, 2)$ and $(3, 3)$ entries). By my comment in the question, there has to be seven zeroes and we are done.

  • Second case: All the other entries are $a, b, c$: By symmetry, assume that we have at least three $a$'s. By another comment below the question, we can assume $a\neq 0$ (or we are done).

Claim: There are three $b$'s (or $c$): If not, there has to be five $a$'s, but if we expand the first row of

$$\begin{bmatrix} a & b& c \\ a & a& e \\ a& a& f \end{bmatrix},$$

then by the lemma, $$0 = C = A = \det \begin{bmatrix} a & e \\ a& f\end{bmatrix}\Rightarrow e = f$$

One can check that this $e$ cannot be $a$ (or there will be seven $a$'s). Then the claim is shown.

Now we are almost done: the matrix can be written as

$$\begin{bmatrix} a & b& c \\ a & b& e \\ a& b& f \end{bmatrix},$$

Again using the lemma, we have $e = f$. By the claim, $e=f = c$. (It can't be $a$, $b$). Thus the matrix is

\begin{bmatrix} a & b& c \\ a & b& c \\ a& b& c \end{bmatrix}

But then

$$\det \begin{bmatrix} a & a& a \\ b& b& c \\ c & b& c\end{bmatrix} = -a(bc-c^2 ) + a(b^2 - bc)= a(c-b)^2\neq 0$$

This last contradiction concludes the proof.

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  • $\begingroup$ Unless I'm missing something, $$\left[\begin{matrix}a&b&c\\b&c&a\\c&a&b\end{matrix}\right]$$ doesn't have to be nonsingular. Indeed if we let $(a,b,c)=(3,-1,-2)$, then we get a matrix with determinant $0$: The determinant is $$3abc-a^3-b^3-c^3$$ and for the values I gave this is $$18-27+1+8=0$$ $\endgroup$ – Peter Woolfitt Jun 12 '15 at 5:44
  • $\begingroup$ By taking the difference in the equations $$0=aA+bB+cC$$ and $$0=bA+aB+cC$$ we can get $A=B$, and so by symmetry we have $$A=B=C$$ but I don't see the argument that these are also equal to $0$. $\endgroup$ – Peter Woolfitt Jun 12 '15 at 5:47
  • $\begingroup$ @PeterWoolfitt : You are right. When I was writing I was thinking $a, b, c\ge 0$ (hence the mistake). I guess I could fix that as I haven't made full use of the symmetry. Thanks for spotting it out! $\endgroup$ – user99914 Jun 12 '15 at 6:36
  • $\begingroup$ @PeterWoolfitt : The answer is updated. I hope I am correct this time...... $\endgroup$ – user99914 Jun 12 '15 at 8:36
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For $3 \times 3$ we can use three $+1$, three $e^{2\pi \mathbf{i}/3}$ and three $e^{4\pi\mathbf{i}/3}$ and the determinant is $0$.

$$ A = \det \left( \begin{array}{ccc} 1 & e^{2\pi \mathbf{i}/3} & e^{4\pi \mathbf{i}/3}\\ e^{2\pi \mathbf{i}/3} & e^{4\pi \mathbf{i}/3} & 1\\ e^{4\pi \mathbf{i}/3} & 1 & e^{2\pi \mathbf{i}/3} \end{array} \right) = 0 $$

So we don't need zeros..

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    $\begingroup$ This is not a true counterexample. I believe that {{1,e^(2i pi/3),e^(4i pi/3)},{e^(2i pi/3),1,1},{e^(4i pi/3),e^(2i pi/3),e^(4i pi/3)}} has a non-zero determinant. $\endgroup$ – TokenToucan Jun 12 '15 at 1:34
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    $\begingroup$ Is the permutation matrix generate all the permutations? It shouldn't be. $\endgroup$ – user99914 Jun 12 '15 at 1:35
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    $\begingroup$ The sigma to which you refer is a row permutation, whereas the question asks about any reordering of the elements in the matrix. $\endgroup$ – TokenToucan Jun 12 '15 at 1:35
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    $\begingroup$ Oh, that's right, the $3\times 3$ permutation matrices don't generate all the matrices I am interested in, since I want any reordering of the elements $\endgroup$ – Peter Woolfitt Jun 12 '15 at 1:35

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