Consider the application $\text{det}: M_2(K) \to K$
- What is the isotropic cone of det in $M_2(K)$?
- Consider the vector subspace $D$ of the diagonal matrices of $M_2(K)$. Is it true that $M_2(K)=D \bigoplus D^{\perp}$?
- Show, if possible without calculation, that every hyperplane of $M_2(K)$ contains an invertible matrix.
Let $B = (E_1, E_2, E_3, E_4)$ the canonical basis of $M_2(K)$, then a matrix $A \in M_2(K)$ can be written as $A = aE_1 + bE_2 + cE_3 + dE_4$. Thus $\text{det}(A) = ad - cb$ is a quadratic form. To find the isotropic cone, we search for the matrices in $M_2(K)$ that satisfy \begin{aligned} \det{A} = 0 &\Leftrightarrow ad - cb = 0 \end{aligned} We conclude that the isotropic cone consists of matrices with rows that are not linearly independent or matrices with a first row that's zero.
In order to sensibly talk about a bilinear form, I assume $\text{char }\mathbb K\neq 2$. With $V:=M_2\big(\mathbb K\big)$, your quadratic form (given by the determinant) implies the symmetric bilinear form $V\times V\longrightarrow \mathbb K$ given by $\langle x ,y\rangle = \frac{1}{2}\cdot\Big(\det\big(x+y\big) - \det\big(x\big) - \det\big(y\big)\Big)$, for $x,y \in V$
I suggest considering the basis
$\mathbf B :=\bigg[\begin{array}{c|c|c|c|c}E_1 & E_4 & E_2& E_3 \end{array}\bigg]=\bigg[\begin{array}{c|c|c|c}b_1 & b_2& b_{3} &b_4 \end{array}\bigg]$
Define matrix $G$ where $g_{i,j}:=\langle b_i, b_j\rangle$, then
$G=\frac{1}{2}\cdot\left[\begin{matrix}0 & 1 & 0 & 0\\1 & 0 & 0 & 0\\0 & 0 & 0 & -1\\0 & 0 & -1 & 0\end{matrix}\right]$
Since $(2G)^2=I$ we conclude the bilinear form is non-degenerate (non-singular). (Note: In the special case of $\mathbb K:=\mathbb R$ then $2G$ being an involution, and $\text{trace}\big(G\big)=0$ tells us the bilinear form has signature $(2,2)$.)
For the first bullet point:
The isotropic cone consists of the set of all vectors that are self-orthogonal under the bilinear form. What you've said is right, the isotropic cone consists of singular matrices, i.e. $V-GL_2\big(\mathbb K\big)$.
For the second bullet point:
The bilinear form with respect to vector subspace $D$ is characterized by the leading $2\times 2$ principal submatrix of $G$. Up to re-scaling this is a permutation matrix so the form is non-degenerate on $D$, which means $D\cap D^\perp=\big\{0\big\}$.
Since the bilinear form is non-degenerate on $V$:
It is always the case that $\dim D+\dim D^\perp = \dim V$ (ref e.g. 'alternative proof' in below link) thus
$\dim \big(D\oplus D^\perp\big) = \dim \big(D+D^\perp\big) =\dim D+\dim D^\perp - \dim \big(D\cap D^\perp\big) = \dim V$
$\implies V= D\oplus D^\perp$
since $\big(D\oplus D^\perp\big)\subseteq V$ and their dimensions are the same.
For the third bullet point:
A hyperplane is a subspace $W$ such that $\dim W=\dim V -1 =3$. But for a non-degenerate bilinear form, a totally isotropic subspace has dimension $\leq \frac{\dim V}{2}=2$, hence $W$ is not totally isotropic. See Problem on a Bilinear pairing
That is, there is some $w,w'\in W$ such that $\langle w,w'\rangle \neq 0$. This implies (at least) one of the vectors in $\Big\{w,w', (w+w')\Big\}$ isn't self orthogonal under the bilinear form, i.e. consider $\langle w+w', w+w'\rangle = \langle w,w\rangle+\langle w',w'\rangle+2\cdot\langle w,w'\rangle $
And checking the first bullet point, this means that (at least) one of those vectors is $\in GL_2\big(\mathbb K\big)$.