Let $V$ be a vector space over a field $K$ and let $B$ be a symmetric bilinear form on $V$. I want to show that the determinant of the matrix representing this bilinear form may depend on the choice of basis but the sign of the determinant does not.

Here is what I have so far:

Let $A$ be the matrix representing $B$ in some basis $\mathcal{B}$ and let $D$ be the matrix representing $B$ in the basis $\mathcal{D}$. Let $C$ denote the transition matrix from the basis $\mathcal{B}$ to the basis $\mathcal{D}. $ Then $$\det(A)=\det(C^tDC)=\det(C^t)\det(D)\det(C)=\det(C)^2\det(D)$$

I'm stuck at this point though. Also, I have no idea how to show that the sign won't change.

  • $\begingroup$ (At least) If the matrices are real, $\det(C)^2>0$ so $A$ and $D$ have the same sign. Since $\mathcal{B}$ and $\mathcal{D}$ are arbitrary, generically $\det(C)\neq1$, so I think there's nothing much to add. $\endgroup$
    – anderstood
    Nov 30, 2016 at 22:56
  • 1
    $\begingroup$ "Sign" has no meaning in a non-real field. $\endgroup$ Nov 30, 2016 at 23:02

1 Answer 1


You have shown that if $A,D$ are two matrices that represent the same bilinear form with respect to two different bases, then you have $\det(A) = x^2 \det(D)$ for some $0 \neq x \in K$.

When $K = \mathbb{R}$, this means precisely that $\det(A)$ and $\det(D)$ have the same sign (as $x^2 > 0$ is positive). For a general field, you don't have a notion of sign that divides the non-zero elements into "positive" and "negative" elements but your argument shows that the equivalence class of $\det(A)$ and $\det(D)$ in the quotient $K^{\times} / (K^{\times})^2$ is the same and this gives you an interesting invariant.

For example, when $K = \mathbb{C}$, then every non-zero complex number is a square and so you don't get anything interesting. However, when $K = \mathbb{Q}$, two rational numbers $a,b \in \mathbb{Q}^{\times}$ will define the same equivalence class if and only if $\frac{a}{b}$ is a square in $\mathbb{Q}$ which gives you much more information than only the sign. Thus, if you have two matrices $A,D$ over $\mathbb{R}$ with $\det(A) = 1$ and $\det(D) = 2$, you can't conclude that they cannot represent the same quadratic form over $\mathbb{R}$ but if the matrices are over $\mathbb{Q}$ then you know that they cannot represent the same quadratic form because you can't solve the equation $1 = \det(A) = \det(C)^2 \det(D) = x^2 \cdot 2$ with $x \in \mathbb{Q}$.

  • $\begingroup$ Interesting last paragraph. What is the $\times$ in $K^\times$? Non-zero? $\endgroup$
    – anderstood
    Dec 1, 2016 at 14:57
  • $\begingroup$ @anderstood: Yeah, $K^{\times}$ is the group of invertible (non-zero) elements of $K$. $\endgroup$
    – levap
    Dec 2, 2016 at 1:59
  • $\begingroup$ Hi, can you please explain how from $\det(A)=\det(C)^2\det(D)$ you concluded "the equivalence class of $\det(A)$ and $\det(D)$ in the quotient $K^{\times} / (K^{\times})^2$ is the same" $\endgroup$
    – Khal
    Sep 11, 2019 at 6:15

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