Quadratic form on the space of Hermitian matrices Let $V$ be the real vector space of all hermitian matrices in $\mathbb{C}^{2\times 2}$ and let $q:V\to \mathbb{R}$ be the transformation defined as $$q(A)=2.\det A$$ where $A\in V.$ Find the dimention of $V$ and prove $q$ is a quadratic form and find its symmetric bilinear form.  Find the signature of $q$.


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*Dimension of $V$ is 4. Easy to show by finding a basis 

*proving $q$ is a quadratic form. Showed that $q(\lambda A)= \lambda^2 q(v)$. Then I wrote the matrices as column vectors and found the matrix representing $B(v,v)=q(v)$
$$\begin{pmatrix}a&b&b^*&d\end{pmatrix}\begin{pmatrix}0&0&0&1\\ 0&0&-1&0\\ 0&-1&0&0\\ 1&0&0&0\end{pmatrix}\begin{pmatrix}a\\ b\\ b^*\\ d\end{pmatrix}=\begin{pmatrix}2ad-2bb^*\end{pmatrix}=2\det \begin{pmatrix} a&b\\b^*&c\end{pmatrix}=q(A)$$
Since the matrix of the form is symmetric then the bilinear form admits a quadratic form. Is this sufficient?
Signature is $(2,-2,0)$ which can be read directly from the matrix since its already anti upper triangular and has its eigenvalues on the anti diagonal.


$\bf{Questions}$
1. Am I allowed to write the matrices as vectors to find the the matrix of the bilinear form?
2. is the way I showed it's a quadratic form correct?
3. Do I still need to test bilinearity or ist it implied by the matrix?
 A: That's not quite right. When you write $\pmatrix{a&b&b^\ast&d}$, are these coordinates with respect to some basis of $V$?
If you pick the ordered basis $\{E_{11},\ E_{12}+E_{21},\ iE_{12}-iE_{21},\ E_{22}\}$, since
$$
2\det\pmatrix{a&b+ic\\ b-ic&d}=2ad-2(b^2+c^2),
$$
the matrix representation of the symmetric bilinear form would be
$$
\pmatrix{0&0&0&1\\ 0&-2&0&0\\ 0&0&-2&0\\ 1&0&0&0}
$$
and clearly, its signature is different from the one you claimed.
And for your other questions:


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*Yes, but writing a matrix as a "vector" doesn't always mean to reshape a 2D array as a 1D array. What you need, instead, is to first pick an ordered basis of the matrix space, and then write the matrix as a coordinate vector with respect to this basis. In your answer, your $b$ is the coordinate of the "vector" $\pmatrix{0&1\\ 0&0}$, but this vector is not a basis vector in the real linear space of Hermitian matrices. (It isn't even Hermitian in the first place!)

*To show that $q$ is a quadratic form, it suffices to pick a basis of $V$ and show that with respect to this basis, $q$ can be expressed as a homogeneous polynomial of degree $2$. Note that this is a stronger condition than that $q$ is a homogeneous function of degree $2$. That is, merely proving that $q(\lambda A)=\lambda^2q(A)$ is not enough, because homogeneous functions are not necessarily polynomials.

*As long as $q(A)=(a,b,c,d)M(a,b,c,d)^T$ for some matrix $M$ (where $(a,b,c,d)^T$ is the coordinate vector for $A$ with respect to some ordered basis of $V$), bilinearity of the mapping $(x,y)\mapsto x^TMy$ follows automatically and you don't need to prove that explicitly. That said, when the questions asks you to "find its symmetric bilinear form", I am not sure whether it expects you to give a coordinate-free answer or not. Just writing down $M$ might be fine, but the question may expect you to express $b(A,B)=\frac12(q(A+B)-q(A)-q(B))$ in terms of determinants as well. Anyway, it is handy to find the signature of the symmetric bilinear form $b$ by looking at the matrix representation $M$.

