Originally Posted by

**lo2** I guess:

$\displaystyle \begin{pmatrix}1 & 0 \\ 0 & 0 \end{pmatrix} , \begin{pmatrix}0 & 1 \\ 1 & 0 \end{pmatrix} , \begin{pmatrix}0 &0 \\ 0 & 1 \end{pmatrix}$

That's it. Now to *show* that it is a basis you have to say that the three matrices are linearly independent and you have to show that

$\displaystyle \left\{\begin{pmatrix}1 & 0 \\ 0 & 0 \end{pmatrix} , \begin{pmatrix}0 & 1 \\ 1 & 0 \end{pmatrix} , \begin{pmatrix}0 &0 \\ 0 & 1 \end{pmatrix}\right\}$

is a spanning set of $\displaystyle U$ (that's more or less what I did in my previous post).

I agree.