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 quick proof which uses special properties of the field $\mathbb{R}$.  Consider the set of matrices of the form $\begin{pmatrix}a & -b \\ b & a\end{pmatrix}$.  Note that every nonzero matrix in this set is invertible, since such a matrix has determinant $a^2+b^2$ which is nonzero unless $a=b=0$ (here is where we use the fact that our field is $\mathbb{R}$).  But these matrices form a $2$-dimensional subspace of $\mathbb{R}^{2\times 2}$, which must have nontrivial intersection with any $3$-dimensional subspace.  So any $3$-dimensional subspace contains a nonzero matrix of this form, which is invertible.

OK, now here's a more complicated proof that works over any field.  Let $V\subseteq\mathbb{R}^{2\times 2}$ be $3$-dimensional and let $\{e_1,e_2\}$ be a basis for $\mathbb{R}^2$.  Let $W$ be the $2$-dimensional subspace of $\mathbb{R}^{2\times 2}$ consisting of all $A$ such that $A(e_1)=0$.  Note that $\dim V=\dim V\cap W+\dim V/(V\cap W)$ and $V\cap W$ and $V/(V\cap W)$ are each at most $2$-dimensional.  So one has dimension $1$, and the other has dimension $2$.
Suppose $\dim V\cap W=1$ so $\dim V/(V\cap W)=2$.  Let $A\in V\cap W$ be nonzero, so $A(e_1)=0$ and $A(e_2)\neq 0$.  Note that $\dim V/(V\cap W)=2$ means that every element of $\mathbb{R}^{2\times 2}/W$ has a representative in $V$.  That is, for any matrix $B$, there is $C\in V$ such that $B-C\in W$, which means $B(e_1)=C(e_1)$.  In particular, choosing $B$ such that $B(e_1)$ is linearly independent from $A(e_2)$, there is some $C\in V$ such that $C(e_1)$ is linearly independent from $A(e_2)$.  If $C$ is invertible, we're done.  Otherwise, $C(e_2)$ is a multiple of $C(e_1)$, and so $C(e_2)+A(e_2)$ is not a multiple of $C(e_1)$.  Taking $D=C+A$, we then have that $D(e_1)=C(e_1)$ and $D(e_2)=C(e_2)+A(e_2)$ are linearly independent.  Thus $D$ is an invertible element of $V$.
The case that $\dim V\cap W=2$ and $\dim V/(V\cap W)=1$ is similar.  Let $A\in V\setminus (V\cap W)$, so $A(e_1)\neq 0$.  If $A$ is invertible, we're done; otherwise $A(e_2)$ is a multiple of $A(e_1)$. Since $\dim V\cap W=2$, we have $W\subset V$.  In particular, let $B$ be a matrix such that $B(e_1)=0$ and $B(e_2)$ is not a multiple of $A(e_1)$.  Then $A(e_2)+B(e_2)$ is not a multiple of $A(e_1)$, and $B\in W\subset V$.  So $C=A+B\in V$ is invertible since $C(e_1)=A(e_1)$ and $C(e_2)=A(e_2)+B(e_2)$ are linearly independent.
(In fact, with a little work you can prove you can always choose $e_1$ so that you're in the first case, so the second case is unnecessary.)
