calculating signature and showing group homomorphism I got this question in my group theory book. I think I understand the theory behind it but can't seem to use it to get a solution im happy with.
Let $V = M_2(\mathbb F)$. For $x,y \in V$ define $B(x,y) = det(x+y)-det(x)-det(y)$.


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*Show that $B$ is a symetric nondegenerate form on V, and calculate the signature of $B$ when $\mathbb F$=$\mathbb R$.

*Let $G=SL(2,\mathbb F)\times SL(2,\mathbb F)$, and $\phi\colon G\rightarrow$ $GL(V)$ by $\phi(a,b)v=avb^t$ for $a,b \in SL(2,\mathbb F)$ and $v \in V$. show that $\phi$ is a group homomorphism and $\phi(G) \subset SO(V,B)$. determine $Ker(\phi)$ (I think you can determine this with jordan canonical form)
Thanks in advance.
 A: (1) Do you know what you're trying to show? You want to show that $B(x,y)=B(y,x)$ (that's the symmetric part), that $B(u+v,w)=B(u,w)+B(v,w)$ and $B(\lambda x,y)=\lambda B(x,y)$ (this is linearity in the first argument; linearity in the second follows from symmetry), and that for all $x\in M_2(\Bbb F)$ there exists a $y\in M_2(\Bbb F)$ such that $B(x,y)\ne0$ (with symmetry, this means nondegeneracy).
In order to do these things, you're going to have to actually write out what $B(x,y)$ looks like explicitly when $x=(\begin{smallmatrix}a&b\\c&d\end{smallmatrix})$, $y=(\begin{smallmatrix}\bar{a}&\bar{b}\\\bar{c}&\bar{d}\end{smallmatrix})$. In order to compute the signature of $B$, you need to compute the underlying matrix (this is easy), and then count the number of zeros which are positive, negative or zero. You can do this by computing the characteristic polynomial, which happens to be easy in this case, and then finding its roots.
(2) Specifying a group homomorphism $G\to{\rm GL}(V)$ means having a linear group action of $G$ on the space $V$. Check that $(a,b)v:=avb^t$ defines a group action of ${\rm SL}_2(\Bbb F)\times{\rm SL}_2(\Bbb F)$ on $V=M_2(\Bbb F)$, and then check that the action is linear, i.e. $(a,b)(v+w)=(a,b)v+(a,b)w$. This is just basic properties of matrix/vector multiplication.
The meaning of $\phi(G)\subset SO(V,B)$ is that (i) as a linear map on $V=M_2(\Bbb F)$, each $(a,b)\in G$ acts by a matrix with determinant $1$ (for which it suffices to check that each $(a,{\rm Id}_2)$ and $({\rm Id}_2,b)$ act by matrices of determinant $1$ - do you know why?), and (ii) $B((a,b)v,(a,b)w)=B(v,w)$ for every choice of $(a,b)\in G={\rm SL}_2(\Bbb F)\times{\rm SL}_2(\Bbb F)$ and $v,w\in M_2(\Bbb F)$. To do (i), notice that (say) the left action of ${\rm SL}_2(\Bbb F)$ on $M_2(\Bbb F)$ corresponds to two copies of ${\rm SL}_2(\Bbb F)$ acting on $\Bbb F^2$ (since it acts on the left and right columns of matrices in $M_2(\Bbb F)$ independently). To do (ii), use the product formula for determinants (i.e. $\det(uv)=\det(u)\det(v)$).
Finally, to compute $\ker\phi$, plug $x={\rm Id}_2$ into $axb^t=x$, simplify to $ax=xa^t$. Write this out in full, multiply out on both sides, equate both sides. Since $x$'s entries are unknowns, we can equate coefficients on both sides and solve for the entries of the matrix $a$. It will narrow down fairly quickly, and then you add in the fact $\det a=1$ and you'll be done.
