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Prove that the determinant in $M_2(\Bbb R)$ is a quadratic form of signature $(2,2)$.

I found the first part: the symmetric bilinear form

$$B(M,N)=\frac{1}{2} ( \det(M+N) - \det (M) -\det (N) )$$

shows that $\det$ is a quadratic form. But I can't find the signature. I don't understand this quantity and I don't find a clear definition.

Thank you for your help.

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2 Answers 2

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First of all, $\det \begin{pmatrix}a & b \\ c & d\end{pmatrix} = ad-bc$, which is a homogeneous 2nd-order polynomial, which means it is a quadratic form.

To compute the signature, note that $ad=\left({a+d\over 2}\right)^2-\left({a-d\over 2}\right)^2$, and similarly for $bc$. Thus,

$$ad-bc = \left({a+d\over 2}\right)^2-\left({a-d\over 2}\right)^2-\left({b+c\over 2}\right)^2+\left({b-c\over 2}\right)^2$$

The signature of a quadratic form is the number of positive and negative coefficients when the form is expressed as a sum of squares in some basis, like $x_1^2+x_2^2-x_3^2-\dots$. We have 2 positive and 2 negative coefficients in the expression for $\det$ above, thus the signature is $(2,2)$.

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The signature of a quadratic form is the number of positive and negative coefficients in a Gauss' reduction as a linear combination of squares of linear forms. This signature does not depend on the decomposition by Sylvester's law if inertia.

Hint:

If $M=\begin{pmatrix} a&b \\c&d \end{pmatrix}$, $\;\det M=ad-bc$. You can use the identity $$xy=\tfrac12\bigl((x+y)^2-(x-y)^2\bigr).$$

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