How to show that matrices are linearly independent as vectors in $ M_{2×2}(\mathbb C) $ over $ \mathbb C $? Let
$$
    A=\begin{pmatrix}
    1 & 2 \\
    1 & 4  \\
    \end{pmatrix}, B=\begin{pmatrix}
    2 & 3 \\
    3 & 5  \\
    \end{pmatrix}, C=\begin{pmatrix}
    -1 & 1 \\
    -2 & -3  \\
    \end{pmatrix} ∈ M_{2×2}(\mathbb C).
$$
Show that A, B, C are linearly independent as vectors in $ M_{2×2}(\mathbb C) $ over $ \mathbb C $.
How do I show that matrices are linearly independent? And does it matter that it is over $ \mathbb C $?
I had an idea that it has something to do with determinants but I'm not sure from there.
 A: With linear independence for matrices, it is exactly the same as with linear independence for vectors. Hence, the following two statements are equivalent even for matrix spaces: 
1) $
(A,B,C) \ \text{is a linearly independent set over} \ \mathbb{C} 
$
2) $\forall \lambda_i \in \mathbb{C}:\lambda_1 A + \lambda_2B+\lambda_3C=0 \Longrightarrow \lambda_1=\lambda_2=\lambda_3=0$.
To show this, one can also proceed very similarly like for vectors. For your specific problem, the first equation of 2) would yield a linear, homogenous system of 4 equations with 3 variables. If this system has a non-trivial solution, your matrices are linearly dependent; if not, they aren't.
Edit:
Also, in the general case it does matter whether you seek for linear independence over $\mathbb{R}$, $\mathbb{C}$ or some other field. The scalars $\lambda_i$ are namely then element of different sets, and for $\lambda_i \in \mathbb{C}$ it might be possible to find such a non-trivial solutions, whereas it might be impossible for $\lambda_i \in \mathbb{R}$.
An example for this happening is the following:
$$
A=\pmatrix{1 & 1 \\ 1 & 0} \text{and} \ B=\pmatrix{i & i \\ i & 0} \in M_{2 \times 2}(\mathbb{C})
$$
are linearly independent over $\mathbb{R}$, but aren't over $\mathbb{C}$.
Edit 2:
As mentioned in the comments, it is also possible to write them as vectors of $\mathbb{C}^4$. This is because the mapping
$$
\phi:  \left\{
\begin{array}{ll}
M_{2 \times 2}(\mathbb{C}) \to \mathbb{C}^4 \\ 
\pmatrix{a & b \\ c & d} \mapsto \pmatrix{a \\ b \\ c \\ d} \\
\end{array} \right.
$$
is an isomorphism. Linearity is quite easy to check and bijectivity follows from the fact that 
$$
\ker(\phi)=\left\{\pmatrix{0 & 0 \\ 0 & 0}\right\}
$$
and every element of $\mathbb{C}^4$ has a unique inverse image.
In your specific question, this wouldn't help a lot nevertheless, because you wouldn't get a square matrix by writing $\phi(A)$, $\phi(B)$ and $\phi(C)$ together into a matrix. Therefore, you can't apply the determinant and can just check the linear independence by the first method I proposed.
