= Solution
Writing $X=\begin{pmatrix}x&y\\z&w\end{pmatrix}$ and using the <Frobenius inner product>, the three <linear equations> are
$$
x+y+2z=0,\qquad x+y-2w=0,\qquad z+w=0.
$$
The first two differ by twice the third; equivalently $A-B=2C$. Thus there are only two independent constraints. Substituting $z=-w$ gives $y=2w-x$, so every element of the <vector subspace> has the unique expression
$$
X=x\begin{pmatrix}1&-1\\0&0\end{pmatrix}
+w\begin{pmatrix}0&2\\-1&1\end{pmatrix}.
$$
The two coefficient <matrices> satisfy all three constraints. Their <linear independence> follows by looking first at entry $(1,1)$ and then at entry $(2,2)$. Therefore a <basis> for this <orthogonal complement> is
$$
\boxed{\left\{\begin{pmatrix}1&-1\\0&0\end{pmatrix},
\begin{pmatrix}0&2\\-1&1\end{pmatrix}\right\}.}
$$
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