Consider the following fragment from Paulsen's book "Completely bounded maps and operator algebras". Here $M_n= M_n(\mathbb{C})$ are the complex $n \times n$ matrices.
How can we prove this? I tried the following: the subspaces are $$\mathcal{K}=\bigoplus_{i=1}^n\pi(E_{i,i})\mathcal{K}.$$ However, are the spaces $\pi(E_{i,i})\mathcal{K}$ $n$-dimensional? I guess not, because otherwise $\mathcal{K}$ would be finite-dimensional and we can represent $M_n$ on an infinite-dimensional space. Hence, we need to look for other subspaces.
The subspaces $\pi(E_{jj})K$, $j=1,\ldots,n$, are pairwise isomorphic. This is because $\pi(E_{kj})$ is a linear isomorphism between $\pi(E_{jj})K$ and $\pi(E_{kk})K$. Let $\{f_r\}$ be an orthonormal basis of $\pi(E_{11})K$. Now form, for each $r$, $$K_r=\operatorname{span}\{\pi(E_{j1})f_r:\ j=1,\ldots,n\}.$$ Then $\dim K_r=n$, and $K=\bigoplus_rK_r$ (if we assume, as is usually done implicitly, that $\pi$ is non-degenerate). For any $A\in M_n(\mathbb C)$, \begin{align} \pi(A)\bigg(\sum_jx_j\pi(E_{j1})f_r\bigg) &=\pi\bigg(\sum_{a,b}A_{ab}E_{ab}\bigg)\bigg(\sum_jx_j\pi(E_{j1})f_r\bigg)\\[0.3cm] &=\sum_{a,b}A_{ab}x_b\pi(E_{a1})f_r\\[0.3cm] &=\sum_a\bigg(\sum_bA_{ab}x_b\bigg)\,\pi(E_{a1})f_r, \end{align} so $\pi(A)$ acts as the identity on $K_r$ (with the orthonormal basis chosen above).