[Math] sufficient and necessary conditions for convergence of geometric series of matrices

convergence-divergencematrices

A is a square matrix with the following properties:
1. the diagonal elements are zero.
2. every element in the same row shares the same positive value.

What is the sufficient and necessary conditions for the convergence of the geometric series?
the conditions, such as, the absolute value of eigenvalues is less than one, is not necessary.

Thanks.

Best Answer

That all eigenvalues $\lambda_i$ of a matrix $A$ are strictly $<1$ in abs. value is both necessary and sufficient.

  • Necessary: If $R=I+A+A^2+\cdots$ is to converge and $\lambda_i,v_i$ is an eigenpair of $A$, then $Rv_i$ must be well-defined, and additionally $S_nv_i\to Rv_i$ where $S_n$ are the partial sums, because matrix multiplication is continuous, but $S_nv_i$ can be written as $1+\lambda_i+\lambda_i^2+\cdots+\lambda_i^{n-1}$, and this only converges in the reals if $|\lambda_i|<1$ (it is the usual geometric series).
  • Sufficient: Suppose $|\lambda_i|<1$ for each $i$. Then $\det(I-A)\ne0$ since $\det(\lambda I-A)$ does not have any root $\lambda=1$, hence $(I-A)^{-1}$ is well defined and $S_n=(I-A)^{-1}(I+A^n)$. All eigenvalues of the powers $A^n$ tend to zero as $r\to\infty$ hence $A^n\to0$ and thus $S_n\to (I-A)^{-1}$.

A couple of items require some further justification using the underlying topology (defined e.g. by the metric induced by the Frobenius norm): continuity of matrix multiplication (in the example this should be clear because matrix multiplication is a linear map..), and the fact that eigenvalues approaching zero implies a sequence of matrices converges to zero (perhaps we could consider the action of $S_n$ on the subspaces in the Jordan aka generalized eigenvector decomposition?).

(There are probably some standard arguments for these things, but I may not have hit them all because I'm pulling them out of my hat instead of going on experience here.)

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