Find a basis of $M_2(F)$ so that every member of the basis is idempotent Let $V=M_{2\times 2}(F)$ (the space of 2x2 matrices with coefficients in a field $F$). Find a basis $\{A_1,A_2,A_3,A_4\}$ of $V$ so that $A_j^2=A_j$ for all $j$.
My attempt. Let $A_j$ be $$\begin{pmatrix} a_1 & a_2\\ a_3 & a_4\\ \end{pmatrix}. $$
We want to have $A_j^2=A_j$, so $$\begin{pmatrix} a_1^2+a_2a_3 & a_1a_2+a_2a_4\\ a_1a_3+a_3a_4 & a_3a_2+a_4^2\\ \end{pmatrix} = \begin{pmatrix} a_1 & a_2\\ a_3 & a_4\\ \end{pmatrix}. $$ 
If we let $a_1=1$ and $a_2=a_3=a_4=0$ then the matrix $$\begin{pmatrix} 1 & 0\\ 0 & 0\\ \end{pmatrix} $$ meets the property. Similarly if we let $a_4=1$ and $a_1=a_2=a_3=0$ the matrix $$\begin{pmatrix} 0 & 0\\ 0 & 1\\ \end{pmatrix} $$ also meets the property, but I´m having trouble finding the other two matrices, can you help me please? I would really appreciate it :)
 A: Hint.  If $a_2=0$ and $a_3=0$ then the matrix is
$$\pmatrix{a_1&0\cr0&a_4\cr}\ ,$$
which is no good because it is a linear combination of the two matrices you have already.  So you must have $a_2$ or $a_3$ or both non-zero.  To keep things as simple as possible, try starting with $a_3=0$ and $a_2\ne0$. . . 
Good luck!
A: An idempotent basis for V is: 
$\{A_1, A_2, A_3, A_4\} =\{[1, 0; 0, 0], [1, 0; 1, 0], [0, 1; 0, 1], [0, 0; 0, 1]\}$
To see this, first of all notice that A1^2 = A1, A2^2 = A2, A3^2 = A3, and A4^2 = A4.
Next, to show that $\mathrm{ span}\{A_1, A_2, A_3, A_4\} = V$, notice that A2- A1 = [0, 0; 1, 0] and A3 - A4 = [0, 1; 0, 0]. Thus we have that each matrix v = [a1,a2;a3,a4] in V can be written as a linear combination of A1, A2, A3, A4 as follows. v = a1(A1)+a2(A3-A4)+a3(A2-A1)+a4(A4).
Moreover, 0 = [0 0; 0 0] = a1(A1)+a2(A2)+a3(A3)+a4(A4) = [a1+a2, a3, a2, a3+a4] => a1 = a2 = a3 = a4 = 0. Therefore, A1, A2, A3, and A4 are also linearly independent. 
So we can conclude that A1, A2, A3, and A4 are a basis for V. 
