Showing a certain endomorphism algebra is basic Let $A$ be a finite dimensional algebra over $\mathbb{C}$. Let $P_1, \dots P_n$ be the projective indecomposable $A$-modules, pairwise non-isomorphic, and define $P=P_1\oplus\dots\oplus P_n$. 
Let $B=\text{End}_A(P)^{op}$.
I want to show that $B/\text{rad}(B)\cong \mathbb{C}^k$ for some nonnegative integer $k$, where $\text{rad}(B)$ is the Jacobson radical of $B$.
What I've tried: We can say that $\text{End}_A(P)^{op}=(\bigoplus_{i,j}\text{Hom}_A(P_i,P_j))^{op}$. Maps between indecomposables are nilpotent or isomorphisms. If $i \neq j$, then by assumption $P_i\not\cong P_j$. Therefore the elements of  $\text{Hom}_A(P_i,P_j)$ are nilpotent. Does the radical contain these nilpotent elements? I'm not really sure how to deal with the other summands or the $^{op}$. 
Auslander, Reiten & Smalo state, but do not prove, a proposition on page 36 of their book, which would imply what I want to show.
 A: While looking at $\text{End}_A(P)^{op}=(\bigoplus_{i,j}\text{Hom}_A(P_i,P_j))^{op}$ you might want to think of elements of $\text{End}_A(P)^{op}$ as matrices $(\varphi_{i,j})_{i,j}$ where $\varphi_{i,j} \in \text{Hom}_A(P_i,P_j)$ where multiplication is given by 
$$ ((\psi_{i,j})(\varphi_{i,j}))_{l,m} = \sum_{k} \varphi_{k,m} \circ \psi_{l,k}$$
You might want to recall that the radical of $P_i$ is the unique maximal submodule of $P_i$. In particular, for $i \neq j$ and $\varphi_{i,j} \in \text{Hom}(P_i,P_j)$ we have that $\text{im}(\varphi_{j,i})$ is contained in the radical of $P_i$.
Thus, for an endomorphism $(\varphi_{i,j})$ to be an isomorphism all the $\varphi_{i,i}$ have to be isomorphisms otherwise $(\varphi_{i,j})$ would not be surjective.
Now, to calculate the Jacobson radical, you need to find those $(\varphi_{i,j})$ which  satisfy that $\text{id}_P + \psi_1(\varphi_{i,j})\psi_2 $ is an isomorphism for all endomorphisms $\psi_1$ and $\psi_2$. If one of the $\varphi_{i,i}$ were an isomorphism, then we could find $\psi_1$ and $\psi_2$ (for example take $\psi_1$ to be diagonal with entries $\text{id}_{P_j}$ and $-\varphi_{i,i}^{-1}$ and $\psi_2 = \text{id}_P$) so that $(\psi_1(\varphi_{i,j})\psi_2)_{i,i} = - \text{id}_{P_i}$ and then $\text{id}_P + \psi_1(\varphi_{i,j})\psi_2$ is not an isomorphism. So for $(\varphi_{i,i})$ to be contained in the Jacobson radical, we have to have $\varphi_{i,i}$ not to be an isomorphism for all $i$ and this turns out to be sufficient, too.
This description of the Jacobson radical immediately gives the desired result, however, you might want to argue a little bit more abstractly.
For any finite dimensional $\mathbb{C}$-algebra $B$ you have $B \cong \bigoplus_{ i = 1}^n Q_i^{m_i}$ as left $B$-module, where the $Q_1,\dots, Q_n$ are representatives for the isomorphism classes of projective indecomposable $B$-modules. The modules $Q_i/J(B)Q_i$ are representatives for the isomorphism classes of simple $B$-modules.
We have $$B/J(B) \cong \bigoplus_{ i = 1}^n (Q_i/J(B) Q_i )^{m_i}$$ and now the Artin-Wedderburn Theorem tells us that $m_i = \dim_{\mathbb{C}}(Q_i/J(B) Q_i)$
Idempotents of the endomorphism ring correspond to direct sum decompositions of $P$ as well as to projective submodules of the endomorphism ring. This gives a correspondence between the indecomposable direct summands of $P$ and the indecomposable projective modules of $\text{End}_A(P)^{op}$ (a direct summand $X$ of $P$ corresponds to the projective module $\text{Hom}_A(P,X)$).
Thus $B =\text{End}_{A}(P)^{op} = \bigoplus_i Q_i$ (with $m_i = 1$ for all $i$ by the definition of $P$) and comparing with the general result above, we find that $Q_i/J(B) Q_i$ is one-dimensional for all $i$ and so $B/J(B) \cong \mathbb{C}^n$.
