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let $V$ be a vector space of all $2 \times 2$ hermitian matrices with entries from $\mathbb C$, over the field $\mathbb R$. prove that $q(v)=\det(v)$ is a quadratic form.

I tried to prove that $f(u,v)=$$1\over4$$(q(u+v)-q(u-v))$ is a bilinear form but i'm stuck with that... can anyone point me in the right direction?

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What's your definition of quadratic form? One solution might be to observe that an explicit formula for $\det(v)$ is a homogeneous quadratic polynomial in the entries of $v$. – Ted Shifrin May 9 '13 at 20:46
my definition: $q(v)$ is a quadratic form if there exists a symmetric bilinear form $f:V \times V \to F$ such that $q(v) = f(v,v)$. I didn't quite understand your suggestion though... – izikgo May 9 '13 at 21:01
up vote 2 down vote accepted

$$\det\pmatrix{x&z+wi\\ z-wi&y}=xy - z^2 - w^2= \pmatrix{x&y&z&w}\pmatrix{0&\tfrac12&0&0\\ \tfrac12&0&0&0\\ 0&0&-1&0\\ 0&0&0&-1}\pmatrix{x\\ y\\ z\\ w}.$$

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didn't think about looking at the matrix defining $f$ at all! thanks! – izikgo May 10 '13 at 12:18
though I think there is a little mistake in your answer. $V$ is over the field $\mathbb R$, therefore $x$ and $y$ both need to be represented by 2 element from the basis (unlike the other 2 entries which go together, both of which need are represented by 2 element from the basis). So $dim(V) = 6$, there for $f$ needs to be a $6 \times 6$ matrix, and the vectors need to be of size 6 too. But thanks for giving me the direction! – izikgo May 11 '13 at 17:53
@izikgo The matrices are Hermitian, so the diagonal entries must be real. You may of course embed the 4-dimensional $\mathbb{R}$-linear space of Hermitian matrices into the 6-dimensional space of complex matrices, but this is not necessary. In case you need the embedding. The matrix representation of the quadratic is permutation-similar to the direct sum of the present one with a $2\times2$ zero block. – user1551 May 12 '13 at 1:17
yeah i realized that after I commented that hermitian matrix enforce the diagonal to be real... thanks! – izikgo May 12 '13 at 7:14
@izikgo No. $q(x)=\|x\|_2^2+1$ is not a quadratic form, but $\frac14\left(q(u+v)-q(u-v)\right)$ is symmetric in $u$ and $v$. – user1551 May 13 '13 at 12:14

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