Are three matrices linearly independent and form a basis of $M_2(\mathbb R)$? I know how to prove whether or not vectors are linearly independent, but can't apply the same thing to matrices it seems. Given three 2x2 matrices, for example:
$$A = \begin {bmatrix} -1&1 \\\\ -1&1 \ \end{bmatrix}$$
 $$B = \begin {bmatrix} 1&1 \\\\ -1&-1 \ \end{bmatrix}$$
 $$C = \begin {bmatrix} -1&1 \\\\ 1&-1 \ \end{bmatrix}$$
I want to test whether or not these are linearly dependent. So with vectors I would do something like:
$$ c_1A + c_2B + c_3C = 0$$
Where the $c_i$'s are some scalar constants, and prove that the only solution of that is when $$c_1 = c_2 = c_3 = 0$$  
So how do I go about solving this:
$$ c_1 \begin {bmatrix} -1&1 \\\\ -1&1 \ \end{bmatrix} + c_2 \begin {bmatrix} 1&1 \\\\ -1&-1 \ \end{bmatrix} + c_3 \begin {bmatrix} -1&1 \\\\ 1&-1 \ \end{bmatrix} = 0$$
Or I am going about this completely the wrong way?
Any help would be hugely appreciated.
 A: It's perfectly fine (except that, if you want to prove independence, you need to show $c_1=c_2=c_3=0$, not that their sum is 0). Next, do the matrix arithmetic on the left hand side:
$$ c_1 \begin {bmatrix} -1&1 \\\\ -1&1 \ \end{bmatrix} + c_2 \begin {bmatrix} 1&1 \\\\ -1&-1 \ \end{bmatrix} + c_3 \begin {bmatrix} -1&1 \\\\ 1&-1 \ \end{bmatrix} = 
 \begin {bmatrix} -c_1+c_2-c_3&c_1+c_2+c_3\\\\  -c_1-c_2+c_3 & c_1-c_2-c_3 \ \end{bmatrix}
={\bf 0}.
$$
Since a matrix is the zero matrix if and only if each of its components is 0, you get the system of equations
$$\eqalign{
 -c_1+c_2-c_3&=0\cr c_1+c_2+c_3&=0  \cr -c_1-c_2+c_3&=0 \cr c_1-c_2-c_3&=0 }
$$
A: You're going about it exactly the right way. EDIT: As David Mitra points out, you have to prove that $c_1=c_2=c_3=0$, not just that $c_1+c_2+c_3=0$. 
In fact, you can just think of the matrices as being vectors of length 4:
$$\begin{pmatrix}a & b \\ c& d\end{pmatrix}\mapsto (a,b,c,d)$$
and use your knowledge about the linear independence of vectors. 
