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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.

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Please state your question more carefully if you expect us to answer carefully. Where does your condition 2. end? And if the answer is "just before 'what'", then the condition is still not clear; can you give an example? Which geometric series are you talking about? –  Marc van Leeuwen Jun 14 '12 at 7:49
    
What does it mean for every element contained in the same row to share the same positive value? Also, $|\lambda_i|<1$ for each eigenvalue is both necessary and sufficient. –  anon Jun 14 '12 at 7:54
    
thanks, I mean every entry is the same except the diagonal one. –  Johnny Jun 14 '12 at 8:18
    
So matrices of the form $$\begin{pmatrix}r_1\\r_2\\ \vdots\\r_n\end{pmatrix}(1~~1~~\cdots~~1)-\begin{pmatrix}r_1 & 0 & \cdots & 0 & 0 \\ 0 & r_2 & \cdots & 0 & 0 \\ 0 & 0 & \cdots & r_{n-1} & 0 \\ 0 & 0 & \cdots & 0 & r_n\end{pmatrix},$$ if I understand you correctly. –  anon Jun 14 '12 at 8:25
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1 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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$\,\,\,$I think you meant in the Necessary condition that "...this only converges in the complex..." BTW, +1 for the succint and nice exposition. –  DonAntonio Jun 14 '12 at 15:26
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