# Determinant of symmetric $2 \times 2$ block matrix

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.

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

## 1 Answer

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