global dimension Suppose we are given finite dimensional semisimple $k$-algebras $A_1,..., A_r$. now we consider the matrix algebra$A=\begin{pmatrix}
A_1 & M_{1,2} & \dots  & M_{1,r}      \\
0 & A_2  & \dots  & M_{2,r}    \\
\vdots & 0  & \ddots & \vdots \\
0  & \dots & 0  & A_r
\end{pmatrix}$
with finitely generated bimodules $M_{i,j}$. Now suppose $\mathrm{gldim}(A)$ is finite. Furthermoe suppose we are given a finite group $G$ acting on $A$ such that $\mathrm{char}(k)$ does not divide the order of $G$. Do we have $\mathrm{gldim}(A^G)<\infty$?
I guess that first of all one can consider 
$
\mathrm{gldim}\begin{pmatrix}
R & M \\
0 & S 
\end{pmatrix}=\mathrm{max}\{\mathrm{pdim}_R M+1, \mathrm{gldim}R\}
$, where $R$ and $S$ are finite dimensional semisimple $k$-algebras and $M$ a finitely generated $R-S$ bimodule and then proceed by induction...
 A: Yes, $A^G$ must have finite global dimension, as it has the same kind of triangular form as $A$.
The Jacobson radical of $A$ is
$$J(A)=\begin{pmatrix}
0 & M_{1,2} & \dots  & M_{1,r}      \\
0 & 0  & \dots  & M_{2,r}    \\
\vdots & 0  & \ddots & \vdots \\
0  & \dots & 0  & 0
\end{pmatrix},$$
and
$$A/J(A)=A_1\times\dots\times A_r,$$
which can be identified with a subalgebra of $A$ in the obvious way.
The primitive central idempotents $e_1,\dots,e_s$ of $A/J(A)$ can be ordered so that $i<j\Rightarrow e_jJ(A)e_i=0$ (list the central idempotents of $A_1$ first, then those of $A_2$, etc.). 
Since $G$ acts on $\{e_1,\dots,e_s\}$, preserving the property $e_jJ(A)e_i=0$, if $f_1,\dots,f_t$ are the $G$-orbit sums of the idempotents, then they can also be ordered so that $i<j\Rightarrow f_jJ(A)f_i=0$. Then taking $B_i=f_iAf_i$ and $N_{i,j}=f_iJ(A)f_j$, the triangular form of $A$ can be taken to be
$$A=\begin{pmatrix}
B_1 & N_{1,2} & \dots  & N_{1,t}      \\
0 & B_2  & \dots  & N_{2,t}    \\
\vdots & 0  & \ddots & \vdots \\
0  & \dots & 0  & B_t
\end{pmatrix},$$
which is now stable under the action of $G$, so
$$A^G=\begin{pmatrix}
B_1^G & N_{1,2}^G & \dots  & N_{1,t}^G      \\
0 & B_2^G  & \dots  & N_{2,t}^G    \\
\vdots & 0  & \ddots & \vdots \\
0  & \dots & 0  & B_t^G
\end{pmatrix}.$$
Since $\operatorname{char}(k)$ does not divide $|G|$, and $B_i$ is semisimple for each $i$, $B_i^G$ is also semisimple, and so $A^G$ has finite global dimension, as sketched in the last paragraph of the question.
