Pauli matrices span the set of $2 \times 2$ matrices over $\mathbb{C}$? Define $B = \{\frac{1}{\sqrt{2}}I, \frac{1}{\sqrt{2}}X, \frac{1}{\sqrt{2}}Y, \frac{1}{\sqrt{2}}Z \}$, where:
$I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix}$,
$X = \begin{bmatrix} 0 & 1 \\ 1 & 0 \\\end{bmatrix}$,
$Y = \begin{bmatrix} 0 & i \\ -i & 0 \\ \end{bmatrix}$, and 
$Z = \begin{bmatrix} 1 & 0 \\ 0 & -1 \\ \end{bmatrix}$
I am trying to show that B (modified from the Pauli matrices) span the set of $2 \times 2$ matrices with entries in $\mathbb{C}$ (denote $\mathcal{M}_2(\mathbb{C})$).
If $A \in \mathcal{M}_2(\mathbb{C})$, then we can write: 
$A = \frac{1}{\sqrt{2}} \begin{bmatrix} a & b \\ c & d \\ \end{bmatrix}$ for some $a, b, c, d \in \mathbb{C}$.
I think that I want to show that $A = \frac{1}{\sqrt{2}}(\lambda_0I + \lambda_1X + \lambda_2Y + \lambda_3Z)$?
This means I would need to show that the system given by: $\{a = \lambda_0 + \lambda_3, \  b = \lambda_1 + \lambda_2i, \ c = \lambda_1 - \lambda_2i, \ d = \lambda_0 - \lambda_3\}$ is always solvable. How do I go about doing this? Or should I approach the question some other way?
 A: There is no need to fiddle around with proving everything is in the span. It should suffice to show that this set is a basis of the space by demonstrating it is a linearly independent set of maximal size.
Suppose there is a linear combination $\alpha I+\beta X+\gamma Y+\delta Z=0$.
This would entail that $\alpha +\delta=0$ and $\alpha-\delta=0$, and from that you would learn that $\alpha=\delta=0$.
Similarly, it also entails that $\beta+i\gamma=0$ and $\beta-i\gamma=0$, but that also implies $\beta=\gamma=0$.
Therefore the four items are linearly independent over $\mathbb C$. Since $M_4(\mathbb C)$ is four dimensional over $\mathbb C$, you are looking at a basis for $M_2(\mathbb C)$.

This should be an acceptable proof for any class beyond basic linear algebra.  The fact that $n$-linearly independent vectors in an $n$ dimensional space span the space is elementary.
A: You only need to show that you can construct the standard basis of $\mathcal{M}_2(\mathbb{C})$ as a linear combination of $I$, $X$, $Y$ and $Z$. Now this seems easy, e.g. $$\frac{1}{2}(I+Z) = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}.$$
Of course, you could also just solve your system of linear equations, e.g. by Gaussian elimination.
A: Once you set up your system, compile the matrix of coefficients and just check that this is nonzero.  This is all you need to guarantee unique solutions over $\mathbb{C}$.  I calculate a determinant of $4i$, so we're good.
