Let $\mathbb K$ be a field and $M_n(\mathbb K)$ be the ring of the $n\times n$ matrices with entries in $\mathbb K$.
Let $C_j\subset M_n(\mathbb K)$ be the subspace of all matrices which have all the entries zero except possibly those in the $j$-th column.
How can I show that if $I$ is a left ideal of $M_n(\mathbb K)$ then $I$ is direct sum of the column spaces $C_j$?
A small correction first, $I$ is isomorphic to a direct sum of $C_j$'s (and all of the $C_i$ are isomorphic between themselves) - not necessarily equal.
The reason is that $M_n(K)$ is a direct sum of isomorphic (minimal) left ideals
$$M_n(K) = \oplus_{j=1}^n C_j$$
and that should finish it with some theory of semi simple modules.
Or, you can reason as follows: every left ideal $I$ of $M_n(K)$ is of the form
$$I = I_B = \{ A \mid A \cdot B = 0 \}$$
for some matrix $B$; equivalently, $I = I_W$ consists of all the matrices which are $0$ on a given subspace $W$ of $\mathbb{K}^n$. For $W$ = $K e_{l+1} + \cdots + K e_{n}$ we get $I_W = C_1 \oplus \cdots \oplus C_l$.
How can I show that if $I$ is a left ideal of $M_n(K)$ then $I$ is direct sum of the column spaces $C_j$?
This idea is mistaken. If this were true, then $M_n(K)$ would have only $2^n$ left ideals, one for every possible distinct direct sum of $n$ column spaces; however, it's known that for an infinite field $K$, there are infinitely many left ideals (actually there are already infinitely many maximal left ideals...).
Another way to see it is to take the subset of matrices which are zero off the first two columns and which also have their first two columns matching. It's easy to see that this is another left ideal, and it's easy to see it isn't a sum of the column spaces.
The main feature about column spaces is that they're simple modules, and I know sometimes beginners mistakenly think the simple modules that show up in a decomposition of $R$ are all the simple modules that exist, but of course there are many more.
What is true is that there is only one isotype of minimal left ideal $L$, and that any other left ideal of $R$ is isomorphic to $L^i$ for some $i$ with $0\leq i\leq n$ ($L^0=\{0\}$.)
