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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 $P=P_1oplusdotsoplus P_n$.
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 $P_inotcong P_j$. 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.

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 $B/J(B) cong mathbb{C}^n$.