Exhibit a basis for the given space and prove that it is a basis.
The space of 2 x 2 matrices.
I don't understand how to this at all, can someone take me through this example step by step and kind of explain why/concepts involved.
Thanks so much.
I can't understand what Roam said at all since the words "independent" and "spanning" apply to sets of vectors, not individual vectors or matrices.
millerst, you probably know that a basis for, say, $\displaystyle R^4$, the set of quadruples, (a, b, c, d), is {(1, 0, 0, 0), (0, 1, 0, 0), (0, 0, 1, 0), (0, 0, 0, 1)} since we can write (a, b, c, d)= (a, 0, 0, 0)+ (0, b, 0, 0)+ (0, 0, c, 0)+ (0, 0, 0, d)= a(1, 0, 0, 0)+ b(0, 1, 0, 0)+ c(0, 0, 1, 0)+ d(0, 0, 0, 1).
Okay, any matrix in your space can be written as
$\displaystyle \begin{bmatrix}a & b \\ c & d\end{bmatrix}$
You want matrices M1, M2, M3, M4 so that
$\displaystyle \begin{bmatrix}a & b \\ c & d\end{bmatrix}= aM1+ bM2+ cM3+ dM4$
there are obvious matrices that do that.
The standard basis for $\displaystyle M_{22}$ consists of the following four matrices:
$\displaystyle \left\{ \begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix}, \begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix}, \begin{bmatrix}0 & 0 \\ 1 & 0\end{bmatrix}, \begin{bmatrix}0 & 0 \\ 0 & 1\end{bmatrix} \right\}$
Because the space of 2x2 matrices has dimension 4, so can't be spanned by just two vectors.
Sorry for my carelessness...