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.
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.