General form of an element of the othogonal basis of $q$ Let $$q \begin{pmatrix}a & b \\ c & d\end{pmatrix}= (a-b)^2+(b-c)^2+(c-d)^2$$ quadratic form on $M_2(\mathbb{R})$. How can I prove that every orthogonal basis $B$ of $M_2(\mathbb{R})$  has also a vector of the form $\begin{pmatrix}\alpha & \alpha \\ \alpha & \alpha\end{pmatrix}$ (with $\alpha\neq 0$)?
 A: Consider a map $f:M_2(\mathbb{R}) \to \mathbb{R}^3$
$$
\left(\begin{matrix} a &b \\ c& d\end{matrix}\right) \mapsto (a-b,b-c,c-d).
$$
Then $q(m) = ||f(m)||^2$ for any $m\in M_2(\mathbb{R})$, where $||\cdot||$ is the standard Euclidean norm on $\mathbb{R}^3$. So if $m_1,m_2,m_3,m_4$ are pairwise orthogonal elements of $M_2(\mathbb{R})$, then $f(m_i)$ are paiwise orthogonal in $\mathbb{R}^3$. Since there can be at most three pairwise orthogonal non-zero vectors in $\mathbb{R}^3$, at least one of $f(m_i)$ must be zero, and the kernel of $f$ cosists of matrices of the form $\left(\begin{matrix} a &a \\ a& a\end{matrix}\right)$.
A: The associated bilinear form reads
$$\begin{align}
B(x,y) &=\tfrac12 (q(x+y)-q(x)-q(y)) \\
       &=(a_x-b_x)(a_y-b_y)+(b_x-c_x)(b_y-c_y)+(c_x-d_x)(c_y-d_y)
\end{align}$$
Thus a vector $v$ of the form $\begin{pmatrix}\alpha & \alpha \\ \alpha & \alpha\end{pmatrix}$ is orthogonal to every other vector.
Now suppose a basis $\{b_i\}$ didn't contain such a vector $v$. Then $v$ would be linear combination of the basis vectors. But nonzero pairwise orthogonal vectors are always linearly independent. So let $v=\sum_{i=1}^4 \lambda_i b_i$. Then for some $b_i$ with $\lambda_i \neq 0$ we have $B(b_i,v)=0=\lambda_i\, B(b_i,b_i)$. But the only nonzero self-orthogonal vector is $v$, since $q(x)=0 \implies x=0 \vee x=v$.
