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)
 A: This is similar to what you did with B:
$$
\begin{bmatrix}x & y \\ z & t\end{bmatrix} = x\begin{bmatrix}1 & 0 \\ 1 & 0\end{bmatrix} + (-y) \begin{bmatrix}0 & -1 \\ 1 & 0\end{bmatrix} + t\begin{bmatrix}0 & 0 \\ 0 & 1\end{bmatrix}
$$
A: Once you know that $B$ is a basis, you also know that the subspace has dimension $3$.
Thus, if you prove that (a) $C$ is linearly independent, and (b) every member of $C$ is in the subspace, you know you have a linear independent set from the subspace whose size is the dimension of the subspace. You should know that such a set must be a basis.
(This is true when the dimension is finite).
A: To show that a basis $(v_1, \ldots, v_k)$ spans a particular vector space $V$, that is, that $$\text{span}(v_1, \ldots, v_k) = V,$$ we can (as usual) show that both
$$\text{span}(v_1, \ldots, v_k) \subseteq V \qquad\text{and}\qquad \text{span}(v_1, \ldots, v_k) \supseteq V.$$
If $V$ has finite dimension $k$, then $V$ and the span have the same dimension, and in particular, it's enough to show either containment.

Since your subspace has dimension $3$ (it is the kernel of a nonzero map from a four-dimensional space to a one-dimensional space, so this follows from the Rank-Nullity Theorem) and the linearly independent set $C$ has three elements, it's enough to check that each element of $C$ is in your subspace, but this is just arithmetic.

