<s>[INST] Find a basis for the subspace $\left\{\begin{bmatrix}x & y \\ z & t\end{bmatrix}, x-y-z = 0\right\}$ The exercise gives me the subspace
$$\left\{\begin{bmatrix}x & y \\ z & t\end{bmatrix}, x-y-z = 0\right\}$$
and ask me to show that these two sets are basis for this subspace:
$$B = \left\{\begin{bmatrix}1 & 1 \\ 0 & 0\end{bmatrix}, \begin{bmatrix}1 & 0 \\ 1 & 0\end{bmatrix}, \begin{bmatrix}0 & 0 \\ 0 & 1\end{bmatrix}\right\}$$
$$C = \left\{\begin{bmatrix}1 & 0 \\ 1 & 0\end{bmatrix}, \begin{bmatrix}0 & -1 \\ 1 & 0\end{bmatrix}, \begin{bmatrix}0 & 0 \\ 0 & 1\end{bmatrix}\right\}$$
for the basis $B$ it's easy to see that if we solve for $x = y + z$ then a matrix 
$$\begin{bmatrix}x & y \\ z & t\end{bmatrix}$$
can be written as
$$\begin{bmatrix}x & y \\ z & t\end{bmatrix} = y\begin{bmatrix}1 & 1 \\ 0 & 0\end{bmatrix} + z \begin{bmatrix}1 & 0 \\ 1 & 0\end{bmatrix} + t\begin{bmatrix}0 & 0 \\ 0 & 1\end{bmatrix}$$
but what can I do to verify that basis $C$ generates the subspace? (I know I also must verify that these sets are linearly independent)
 [/INST] This is similar to what you did with B:
$$
\begin{bmatrix}x & y \\ z & t\end{bmatrix} = x\begin{bmatrix}1 & 0 \\ 1 & 0\end{bmatrix} + (-y) \begin{bmatrix}0 & -1 \\ 1 & 0\end{bmatrix} + t\begin{bmatrix}0 & 0 \\ 0 & 1\end{bmatrix}
$$
</s>