Diagonalization differential equation I have a problem in solving the diagonalization of this differential equation :
$$\frac{d}{dt}\binom{x}{f} = \left(\begin{matrix} -\frac{1}{\tau} & 1 \\ 0 & -\frac{1}{\tau_{c}}\end{matrix}\right)\binom{x}{f}$$
Can anybody help me?
 A: Making the change of variables 
$$
p = (x,f)^{\dagger}\\
P = (X,F)^{\dagger}
$$
such that
$$
p = T\cdot P
$$
with $T$ invertible, we have
$$
\left(T\dot P\right) = A\cdot \left(T \cdot P\right)\Rightarrow \dot P = \left(T^{-1}\cdot A  \cdot T\right)\cdot P
$$
now choosing 
$$
T = \left(
\begin{array}{cc}
 \frac{\tau  \tau_c}{\tau_c-\tau} & 1 \\
 1 & 0 \\
\end{array}
\right)
$$
which is the eigenvectors matrix for $A$ we have finally
$$
\dot P = \left(
\begin{array}{cc}
 -\frac{1}{\tau_c} & 0 \\
 0 & -\frac{1}{\tau } \\
\end{array}
\right)\cdot P
$$
NOTE
Just in case you have not been introduced to the theory of eigenvalues, you can proceed with a generic invertible $T$ like 
$$
T = \left(
\begin{array}{cc}
 1 & t_1 \\
 0 & t_2 \\
\end{array}
\right)
$$
and then solve the matrix equation
$$
T^{-1}\cdot A\cdot T = \Lambda
$$
with 
$$
\Lambda = \left(
\begin{array}{cc}
 \lambda_1 & 0 \\
 0 & \lambda_2 \\
\end{array}
\right)
$$
arriving to the conditions
$$
\left(
\begin{array}{cc}
 -\frac{1}{\tau } & t_2 \left(\frac{t_1}{t_2 \tau_c}+1\right)-\frac{t_1}{\tau } \\
 0 & -\frac{1}{\tau_c} \\
\end{array}
\right) = \left(
\begin{array}{cc}
 \lambda_1 & 0 \\
 0 & \lambda_2 \\
\end{array}
\right)
$$
now choosing conveniently $t_1,t_2$ to calcell
$$
t_2 \left(\frac{t_1}{t_2 \tau_c}+1\right)-\frac{t_1}{\tau } 
$$
such as
$$
t_1 = \frac{\tau\tau_c t_2}{\tau_c-\tau}
$$
we solve the problem.
