Is the linear dependence test also valid for matrices? I have the set of matrices
$
        \begin{pmatrix}
        1 & 0 \\
        0 & 0 \\
        \end{pmatrix}
$
$
        \begin{pmatrix}
        0 & 1 \\
        0 & 0 \\
        \end{pmatrix}
$
$
        \begin{pmatrix}
        0 & 0 \\
        1 & 0 \\
        \end{pmatrix}
$
$
        \begin{pmatrix}
        0 & 0 \\
        0 & 1 \\
        \end{pmatrix}
$
and I'm asked to check if said set is a basis of $
        \begin{pmatrix}
        a & b \\
        c & d \\
        \end{pmatrix}
$, that is, any 2x2 matrix. I know that in this case it is really simple to do  it like $$\pmatrix{a&b\cr c&d\cr}=\alpha\pmatrix{1&0\cr0&0\cr}+\beta\pmatrix{0&1\cr0&0\cr}+\gamma\pmatrix{0&0\cr1&0\cr}+\delta\pmatrix{0&0\cr0&1\cr}$$
But not all cases are that simple, so I was wondering if I could do it another way: When I'm asked to check if a set of vectors is a basis of a vector space $R^n$ I just see if said vectors are linearly independent, given that if I have n linearly independent vectors $\{(v1,...,vn),...,(u1,...un)\}$ they'll be a basis for $R^n$. Can I use the same test and say that if
$$\pmatrix{0&0\cr 0&0\cr}=\alpha\pmatrix{1&0\cr0&0\cr}+\beta\pmatrix{0&1\cr0&0\cr}+\gamma\pmatrix{0&0\cr1&0\cr}+\delta\pmatrix{0&0\cr0&1\cr}$$
Has only one solution $\alpha=\beta=\gamma=\delta=0$ then they are all linearly independent, and since it's all 2x2 it'll be a basis for all 2x2 matrices? Does that make any sense?

$\mathbb{R}^{N\times N}$ as a linear space (with addition between elements and multiplication by scalars) is no different than $\mathbb{R}^{2N}$ endowed with these same operations.  There is a bijective mapping between elements and operations in these two spaces, so any tricks you know about vectors is $\mathbb{R}^4$ are applicable to $\mathbb{R}^{2\times 2}$
