Solving Linear Systems with Singular Matrices

Question

Good morning! For (say, homogenous) linear systems of the form $$x_{n+1} = A x_n,$$ where $A$ is a nonsingular matrix, each initial value problem can be solved by the method of finding a general solution by means of eigenvalues of $A$. However, for singular matrices, this method need not to be successful for all initial value problems (because of zero eigenvalues) and I was unable to find references for such case. So my question is, is there any general method of solving such systems for singular matrices? Thank you in advance.

Answer 1

The trouble with your method is not when $A$ is singular, it's when $A$ is not diagonalizable. The solution of the initial value problem $x_{n+1} = A x_n$, $x_0$ given, is $x_n = A^n x_0$. Now we can write $A = S^{-1} J S$ where $S$ is invertible and $J$ is in Jordan canonical form, and so $x_n = S^{-1} J^n S x_0$. For a $d \times d$ Jordan block $$ J = \pmatrix{\lambda & 1 & 0 & \ldots & 0 & 0\cr 0 & \lambda & 1 & \ldots & 0 & 0\cr \ldots & \ldots & \ldots & \ldots & \ldots & \ldots \cr 0 & 0 & 0 & \ldots & \lambda & 1\cr 0 & 0 & 0 & \ldots & 0 & \lambda\cr}$$ $$ J^n = \pmatrix{ \lambda^n & {n \choose 1} \lambda^{n-1} & {n \choose 2} \lambda^{n-2} & \ldots & {n \choose {d-2}} \lambda^{n-d+2} & {n \choose {d-1}} \lambda^{n-d+1}\cr 0 & \lambda^n & {n \choose 1} \lambda^{n-1} & \ldots & {n \choose {d-3}} \lambda^{n-d+3} & {n \choose {d-2}} \lambda^{n-d+2}\cr & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots \cr 0 & 0 & 0 & \ldots & \lambda^n & {n \choose 1} \lambda^{n-1}\cr 0 & 0 & 0 & \ldots & 0 & \lambda^n\cr}$$ where ${n \choose j} \lambda^{n-j} = 0$ when $n < j$.