True of False: Every 3-dimensional subspace of $ \Bbb R^{2 \times 2}$ contains at least one invertible matrix. The true or false question states:
"True of False: Every 3-dimensional subspace of $ \Bbb R^{2 \times 2}$ contains at least one invertible matrix."
Here the $ \Bbb R^{2 \times 2}$ represents the space of all two by two matrices. It seems like this is true, but I am not sure how to prove or disprove the statement. (If such is true, then it's easy to see that every 3-dimensional subspace of $ \Bbb R^{2 \times 2}$ contains infinitely many invertible matrices)

Here's a nice solution using the fact that $\Bbb R^{2 \times 2}$ has a "dot-product" given by
$$
\DeclareMathOperator{\tr}{Tr}
\langle A,B \rangle = \tr(AB^T)
$$
With that, we can describe any dimesnion $3$ subspace by
$$
S = \{A : \tr(AM) = 0\}
$$
for a fixed non-zero matrix $M$.  If $M$ is invertible, then we can note that
$$
A = \pmatrix{1&0\\0&-1}M^{-1}
$$
is an element of $S$.  If $M$ is not invertible, then $M = uv^T$  for column vectors $u$ and $v$.  It suffices to select an invertible $A$ such that $Au$ is perpendicular to $v$.
