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.
Solving Linear Systems with Singular Matrices
1 Answers
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 $n < j$.
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0Yes, we use $0^0=1$ here. Why? Because it makes $\lambda^0$ a continuous function of $\lambda$ at $\lambda=0$. – 2012-06-25