dirk_mec1 said:
but shouldn't there be another summation sign for m?
No.
Suppose that instead of an infinite number of x and y, we only have two:
\aligned<br />
\dot x_1 &= a_{1,1} x_1 + b_{1,1} y_1 + a_{2,1} x_2 + b_{2,1} y_2 \\<br />
\dot y_1 &= c_{1,1} x_1 + d_{1,1} y_1 + c_{2,1} x_2 + d_{2,1} y_2 \\<br />
\dot x_2 &= a_{1,2} x_1 + b_{1,2} y_1 + a_{2,2} x_2 + b_{2,2} y_2 \\<br />
\dot y_2 &= c_{1,2} x_1 + d_{1,2} y_1 + c_{2,2} x_2 + d_{2,2} y_2<br />
\endaligned
Define the four-vector \vec u = [u_1, u_2, u_3, u_4]^T = [x_1, y_1, x_2, y_2]^T. The above becomes
\aligned<br />
\dot u_1 &= a_{1,1} u_1 + b_{1,1} u_2 + a_{2,1} u_3 + b_{2,1} u_4 \\<br />
\dot u_2 &= c_{1,1} u_1 + d_{1,1} u_2 + c_{2,1} u_3 + d_{2,1} u_4 \\<br />
\dot u_3 &= a_{1,2} u_1 + b_{1,2} u_2 + a_{2,2} u_3 + b_{2,2} u_4 \\<br />
\dot u_4 &= c_{1,2} u_1 + d_{1,2} u_2 + c_{2,2} u_3 + d_{2,2} u_4<br />
\endaligned
This is just a matrix-vector equation: \dot{\vec u} = \mathbf A \vec u if you treat u as a column vector (or \dot{\vec u} = vec u \mathbf A^T if you use row vectors). Each of those
a,
b,
c, and
d elements maps to exactly one of the elements of the state matrix
A.
This won't change when you go to infinite dimensional space.