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.

  • $\begingroup$ What do you mean by "the equivalent question"? Are the two questions logically equivalent? $\endgroup$ – Omnomnomnom Jun 12 '15 at 1:06
  • 1
    $\begingroup$ @PeterWoolfitt : A quick python script shows that there are only $5040$ distinct equations. :) $\endgroup$ – Alexey Burdin Jun 12 '15 at 1:08
  • 3
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

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.

  • $\begingroup$ 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

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..

  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ Is the permutation matrix generate all the permutations? It shouldn't be. $\endgroup$ – user99914 Jun 12 '15 at 1:35
  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.