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Let $M$ be a symmetric $4 \times 4$ matrix and write it as a $2 \times 2$ block matrix $ M = \begin{bmatrix} A & B \\ B^T & C \end{bmatrix} $. Suppose that $\det(C) = 0$.

What can we say about the determinant of $M$? Ideally I would like $\det(M)$ to be a square. When does that occur?

EDIT: The entries in $M$ are linear forms with coefficients in $\mathbb{C}$. When I say that determinant should be a square, I mean as a polynomial, not as a number.

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  • $\begingroup$ If $C=0, \det B\ne 0$ then $\det M$ cannot be a square. $\endgroup$
    – Minz
    Apr 21, 2017 at 11:15
  • $\begingroup$ @Minz, then $\det(M) = -\det(B)^2$, which is a square over $\mathbb{C}$. $\endgroup$
    – user44413
    Apr 21, 2017 at 11:26
  • $\begingroup$ all numbers in $\mathbb C$ are squares $\endgroup$
    – Minz
    Apr 21, 2017 at 11:33
  • $\begingroup$ @Minz, I have updated the question now. I am interested in square polynomials. $\endgroup$
    – user44413
    Apr 21, 2017 at 11:37
  • $\begingroup$ then I believe the devil only knows. $\endgroup$
    – Minz
    Apr 21, 2017 at 11:41

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I may be misunderstanding the problem, but assume the entries of $M$ to be distinct indeterminates. Then if $x$ is an indeterminate, we have \begin{equation} \det \begin{bmatrix} -1 & 0 & 1 & 0\\ 0 & 1 & 0 & 1\\ 1 & 0 & x & 0\\ 0 & 1 & 0 & 0\\ \end{bmatrix} = x, \end{equation} so in general $\det(M)$ does not appear to be a square.

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