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I had a question where they ask me if a matrix is convergent. Matrix is as below:

A = {{-1.7, -12.6, -12.6}, {-1.2, -5.6, -6.6}, {1.5, 7.8, 8.8}}

I put it in matrix form:

MatrixForm[A]

And used the command:

MatrixPower[A, n]

With increasingly big n until I got (at n=100 000, with I believe is enough to see convergence)

{{-2.84217*10^-14, -1.13687*10^-13, -1.42109*10^-13}, {-1., -4., \
-5.}, {1., 4., 5.}}

So I wrote that the matrix was not convergent for its first row, and convergent to {-1, -4, -5} for row 2, and convergent towards {1, 4, 5} for row 3. And it is incorrect.

I even tried: MatrixPower[A, n] Limit[n->Infinity] but it did not work either.

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    $\begingroup$ Your MatrixForm does nothing for your calculation. It merely displays A as a matrix. It does not alter A. $\endgroup$
    – John Doty
    May 9, 2018 at 13:12
  • $\begingroup$ Yes, sure, I just wrote down all the stuff I was doing with the problem :) $\endgroup$
    – Dovendyr
    May 9, 2018 at 13:44
  • $\begingroup$ Possible duplicate of Why does MatrixForm affect calculations? $\endgroup$
    – Yves Klett
    May 9, 2018 at 15:27

1 Answer 1

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This the correct way to write your final command

MatrixPower[A, n] // Limit[#, n -> Infinity] &  

And the result is

{{-3.20081*10^-14, -1.28033*10^-13, -1.60041*10^-13}, 
 {-1., -4., -5.}, 
 {1., 4., 5.}}

Hope this helps.

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  • $\begingroup$ Thank you @Konstantinos. But you agree that it converges to -1,-4,-5 and 1,4,5? $\endgroup$
    – Dovendyr
    May 9, 2018 at 12:55
  • $\begingroup$ It seems so. I don't understand why you said that the first row is not converging though. It seems that it does I think to some small numbers. $\endgroup$
    – user49048
    May 9, 2018 at 12:57
  • $\begingroup$ It's my fault... I thought that the first row was not convergent since it looked so weird compared to row 2 and 3! Thank you so much for your help! $\endgroup$
    – Dovendyr
    May 9, 2018 at 13:01
  • $\begingroup$ Don't mention it. Cheers!!! $\endgroup$
    – user49048
    May 9, 2018 at 13:19

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