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Find $A^{100}$ where

$$A = \begin{bmatrix} 2 & 2 & 1\\ 1 & 3 & 1\\ 1 & 2 & 2 \end{bmatrix}$$

Please give me the complete solution.

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Diwakar yadav
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    Diagonalize it. – David P Apr 28 '19 at 06:19
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    @David Peterson it is from function of square matrix say A raise to n,tan,cot,e,sin – Diwakar yadav Apr 28 '19 at 06:21
  • If my eigen values is repeating 5,1,1 then I have to take derivatives of equation – Diwakar yadav Apr 28 '19 at 06:25
  • https://www.wolframalpha.com/input/?i=diagonalize+%7B%7B2,2,1%7D,%7B1,3,1%7D,%7B1,2,2%7D%7D – David P Apr 28 '19 at 06:26
  • Every time someone posts this type of question, there is someone recommending to diagonalize the matrix. Face palm! I wonder which bad textbook is to blame for popularizing that wasteful strategy among students. – user647486 Apr 28 '19 at 06:29
  • David Peterson I don,t want to diagonalize it I want to find the value of A^100 – Diwakar yadav Apr 28 '19 at 06:33
  • Why don't you want to diagonalize? If it helps you to get what you want.. – QFi Apr 28 '19 at 06:37
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    So, @user647486, are you going to post an answer to explain a better way? – Gerry Myerson Apr 28 '19 at 06:39
  • QFi I want to solve by function of square matrices if say in examination if they ask to solve by function of square matrices then .........? – Diwakar yadav Apr 28 '19 at 06:43
  • @GerryMyerson I have answered this question many times. The best way is to compute the characteristic polynomial, then compute the remainder of the division of $x^{100}$ by that polynomial. This can be done by indeterminate coefficients evaluating at the roots of the characteristic polynomial. The result is the evaluation of the remainder at $A$. This is not only more efficient, but also don't make students catastrophically fail when they are given matrices that are not diagonalizable. – user647486 Apr 28 '19 at 06:46
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    @Diwakaryadav Then... mention this in your question! – QFi Apr 28 '19 at 06:47
  • @user647486 The method you are suggesting involves finding the roots of the characteristic polynomial, among other things, which in this case is the same as finding the eigenvalues. Overall it's much more complicated than simply diagonalizing the matrix, which works here. – Zarrax Apr 28 '19 at 06:55
  • @Zarrax To diagonalize you need to compute not only the eigenvalues, but also a basis of eigenvectors, compute inverse of the matrix of eigenvectors, and multiply the three matrices back after you raise the diagonal to the relevant power. Compare to compute the eigenvalues, find the remainder (which is only one system of equations instead of three), and evaluating the remainder in the matrix. You simply don't know what you are talking about. What might be is longer to explain for the instructors, which explains why the less suitable method has been popularized. – user647486 Apr 28 '19 at 07:03
  • @user647486, isn't there a bit of a complication when the characteristic polynomial has repeated roots? – Gerry Myerson Apr 28 '19 at 08:29
  • @GerryMyerson Which is solved, like always, by evaluating the derivative of the equation. – user647486 Apr 28 '19 at 08:36
  • @user647486 Solving your system of undetermined coefficients will be no easier than inverting the diagonalization matrix. Your method effectively requires finding the eigenvalues and inverting a matrix, after which the matrix has to be plugged into a second order polynomial. Not to mention, your method is conceptually more difficult and less relevant for a student. The concepts of changing basis and eigenvalues are important and can be used to solve various problems, such as this one. – Zarrax Apr 28 '19 at 14:18
  • See the related questions https://math.stackexchange.com/questions/1206528/find-matrix-a50?rq=1 and https://math.stackexchange.com/questions/2861926/xa100-where-a-beginbmatrix-1-2-3-4-endbmatrix?rq=1 from the handy list at right for many ways to do this. – amd Apr 28 '19 at 17:39

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For this matrix there is a simple method to find any power. Start by defining

$$ B := A - I_3 = \begin{bmatrix} 1& 2& 1\\ 1& 2& 1\\ 1& 2& 1 \end{bmatrix}. \tag{1} $$

Check that $\ B^2 = 4B,\ $ and that $\ A^2 = (I_3 + B)^2 = I_3 + 6B.\ $ In general, $\ A^n = I_3 + a_n\,B\ $ where $\ a_n \!=\!(5^n \!-\! 1)/4 = 5a_{n-1}\!+\!1\ $ is OEIS sequence A003463.

There is no need to diagonalize or compute eigenvalues in this case, just use simple observation and algebra.

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