# Symmetric bilinear form over $Mat_{2x2}(\mathbb{R})$

Let $$V = Mat_{2x2}(\mathbb{R})$$. Define $$\varphi(A,B) := det(A+B) - det(A) - det(B)$$.

I have to show that $$\varphi(A,B)$$ is a symmetric bilinear form. It is easy to see that $$\varphi$$ is symmetric. My problem is to show the bilinearity. I know i have to show that:

$$\varphi(\lambda(A)+\mu(C),B) = \lambda \varphi(A,B) + \mu \varphi(C,B)$$.

And the same for the second argument.

But I don't know how to get to the equality.

We can do the computaion in coordinate. Using the notation below $$$$X= \left[ \begin{matrix} x_1 & x_2\\ x_3 & x_4 \end{matrix} \right]$$$$ we have: $$\begin{multline} \varphi(A,B) = [(a_1+b_1)(a_4+b_4)-(a_2+b_2)(a_3+b_3)] - [a_1a_4 -a_2a_3] - [b_1b_4-b_2b_3] =\\ a_1b_4+b_1a_4-a_2b_3-b_2a_3 \end{multline}$$ At this step you have only to do the computation in order to find the bilinearity.
But I prefer another approach. Under the isomorphism of coordinates: $$$$f:Mat_{2\times 2}(\mathbb R)\longrightarrow \mathbb R^4, \quad f(X) = (x_1,x_2,x_3,x_4)^T$$$$ you can see that $$\varphi$$ corresponds to the function: $$\begin{gather} \tilde \varphi:\mathbb R^4 \times \mathbb R^4 \longrightarrow \mathbb R\\ (a,b) \longmapsto a^T M b \end{gather}$$ where $$a=(a_1,a_2,a_3,a_4)^T$$, $$b=(b_1,b_2,b_3,b_4)^T$$ and $$$$M= \left[ \begin{matrix} 0 & 0 & 0 &1\\ 0 & 0 & -1 & 0\\ 0 & -1 & 0 & 0\\ 1 & 0 & 0 &0 \end{matrix} \right]$$$$ and this is obviously a scalar product.
• So this M is the matrix associated to $\varphi$ with respect to the standard Basis of $V$, if I understand correctly? – MathematicalMoose Apr 2 '20 at 8:37