Diagonalization to power of matrix












1












$begingroup$


$left(array{
2&3 \
5& 1
}right)^{20}$



$(lambda-1)(lambda-2)-15=0$



$lambda = (3pm sqrt{61})/2$



Is this problem should be solved in this method? Lambda is not tidy.






linear-algebra diagonalization







share|cite|improve this question











$endgroup$












  • $begingroup$
    Hint: en.wikipedia.org/wiki/Exponentiation_by_squaring
    $endgroup$
    – darij grinberg
    Jan 10 at 16:30










  • $begingroup$
    Random agglomerations of symbols are not useful explanations. What do you mean, and what are you doing?
    $endgroup$
    – user3482749
    Jan 10 at 16:30










  • $begingroup$
    @user3482749 OP is trying to find $$left(array{ 2&3 \ 5& 1 }right)^{20}$$
    $endgroup$
    – Shubham Johri
    Jan 10 at 16:35


















1












$begingroup$


$left(array{
2&3 \
5& 1
}right)^{20}$



$(lambda-1)(lambda-2)-15=0$



$lambda = (3pm sqrt{61})/2$



Is this problem should be solved in this method? Lambda is not tidy.






linear-algebra diagonalization







share|cite|improve this question











$endgroup$












  • $begingroup$
    Hint: en.wikipedia.org/wiki/Exponentiation_by_squaring
    $endgroup$
    – darij grinberg
    Jan 10 at 16:30










  • $begingroup$
    Random agglomerations of symbols are not useful explanations. What do you mean, and what are you doing?
    $endgroup$
    – user3482749
    Jan 10 at 16:30










  • $begingroup$
    @user3482749 OP is trying to find $$left(array{ 2&3 \ 5& 1 }right)^{20}$$
    $endgroup$
    – Shubham Johri
    Jan 10 at 16:35
















1












1








1





$begingroup$


$left(array{
2&3 \
5& 1
}right)^{20}$



$(lambda-1)(lambda-2)-15=0$



$lambda = (3pm sqrt{61})/2$



Is this problem should be solved in this method? Lambda is not tidy.






linear-algebra diagonalization







share|cite|improve this question











$endgroup$




$left(array{
2&3 \
5& 1
}right)^{20}$



$(lambda-1)(lambda-2)-15=0$



$lambda = (3pm sqrt{61})/2$



Is this problem should be solved in this method? Lambda is not tidy.






linear-algebra diagonalization




linear-algebra diagonalization






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 10 at 17:24









the_fox

2,90021537




2,90021537










asked Jan 10 at 16:27









4charwon4charwon

274




274












  • $begingroup$
    Hint: en.wikipedia.org/wiki/Exponentiation_by_squaring
    $endgroup$
    – darij grinberg
    Jan 10 at 16:30










  • $begingroup$
    Random agglomerations of symbols are not useful explanations. What do you mean, and what are you doing?
    $endgroup$
    – user3482749
    Jan 10 at 16:30










  • $begingroup$
    @user3482749 OP is trying to find $$left(array{ 2&3 \ 5& 1 }right)^{20}$$
    $endgroup$
    – Shubham Johri
    Jan 10 at 16:35




















  • $begingroup$
    Hint: en.wikipedia.org/wiki/Exponentiation_by_squaring
    $endgroup$
    – darij grinberg
    Jan 10 at 16:30










  • $begingroup$
    Random agglomerations of symbols are not useful explanations. What do you mean, and what are you doing?
    $endgroup$
    – user3482749
    Jan 10 at 16:30










  • $begingroup$
    @user3482749 OP is trying to find $$left(array{ 2&3 \ 5& 1 }right)^{20}$$
    $endgroup$
    – Shubham Johri
    Jan 10 at 16:35


















$begingroup$
Hint: en.wikipedia.org/wiki/Exponentiation_by_squaring
$endgroup$
– darij grinberg
Jan 10 at 16:30




$begingroup$
Hint: en.wikipedia.org/wiki/Exponentiation_by_squaring
$endgroup$
– darij grinberg
Jan 10 at 16:30












$begingroup$
Random agglomerations of symbols are not useful explanations. What do you mean, and what are you doing?
$endgroup$
– user3482749
Jan 10 at 16:30




$begingroup$
Random agglomerations of symbols are not useful explanations. What do you mean, and what are you doing?
$endgroup$
– user3482749
Jan 10 at 16:30












$begingroup$
@user3482749 OP is trying to find $$left(array{ 2&3 \ 5& 1 }right)^{20}$$
$endgroup$
– Shubham Johri
Jan 10 at 16:35






$begingroup$
@user3482749 OP is trying to find $$left(array{ 2&3 \ 5& 1 }right)^{20}$$
$endgroup$
– Shubham Johri
Jan 10 at 16:35












1 Answer
1






active

oldest

votes


















0












$begingroup$

I see two method two calculate the powers of the matrix in your example.



First method : you diagonalize your matrix. Then it is easy.



Second method : If I note $M$ your matrix, you want to calculate $M^{20}$. You know that $M^2 - M + 15 = 0$ (Cayley Hamilton theorem). So by putting $Q = X^2 - X + 15$, and $X^{20} = PQ + R $ where $deg(R) leq 1$ (so $R = a_{20} + b_{20} X), M^{20} = R(M).$ And it is possible to calculate $R$ by looking its value at your two lamdas. An another way is to make a recursion (but it is not necessary to calculate iterated calculus 20 times...).



If your question is : how I calculate $lambda^{20} = a_{20} times 3 pm b_{20} times sqrt{61}$, the answer is you do not have a better way to do that iterating calculus, so don't do it.



EDIT :



I've seen an another answer ; yes, it is a bit faster if you apply exponentiating by squaring, to try to calculate more explicitely $lambda^{20}$, but it is still dull. I advice you to keep $lambda^{20}$ in your result.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you for detail answer!!!
    $endgroup$
    – 4charwon
    Jan 10 at 16:57











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1 Answer
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1 Answer
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active

oldest

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active

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active

oldest

votes









0












$begingroup$

I see two method two calculate the powers of the matrix in your example.



First method : you diagonalize your matrix. Then it is easy.



Second method : If I note $M$ your matrix, you want to calculate $M^{20}$. You know that $M^2 - M + 15 = 0$ (Cayley Hamilton theorem). So by putting $Q = X^2 - X + 15$, and $X^{20} = PQ + R $ where $deg(R) leq 1$ (so $R = a_{20} + b_{20} X), M^{20} = R(M).$ And it is possible to calculate $R$ by looking its value at your two lamdas. An another way is to make a recursion (but it is not necessary to calculate iterated calculus 20 times...).



If your question is : how I calculate $lambda^{20} = a_{20} times 3 pm b_{20} times sqrt{61}$, the answer is you do not have a better way to do that iterating calculus, so don't do it.



EDIT :



I've seen an another answer ; yes, it is a bit faster if you apply exponentiating by squaring, to try to calculate more explicitely $lambda^{20}$, but it is still dull. I advice you to keep $lambda^{20}$ in your result.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you for detail answer!!!
    $endgroup$
    – 4charwon
    Jan 10 at 16:57
















0












$begingroup$

I see two method two calculate the powers of the matrix in your example.



First method : you diagonalize your matrix. Then it is easy.



Second method : If I note $M$ your matrix, you want to calculate $M^{20}$. You know that $M^2 - M + 15 = 0$ (Cayley Hamilton theorem). So by putting $Q = X^2 - X + 15$, and $X^{20} = PQ + R $ where $deg(R) leq 1$ (so $R = a_{20} + b_{20} X), M^{20} = R(M).$ And it is possible to calculate $R$ by looking its value at your two lamdas. An another way is to make a recursion (but it is not necessary to calculate iterated calculus 20 times...).



If your question is : how I calculate $lambda^{20} = a_{20} times 3 pm b_{20} times sqrt{61}$, the answer is you do not have a better way to do that iterating calculus, so don't do it.



EDIT :



I've seen an another answer ; yes, it is a bit faster if you apply exponentiating by squaring, to try to calculate more explicitely $lambda^{20}$, but it is still dull. I advice you to keep $lambda^{20}$ in your result.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you for detail answer!!!
    $endgroup$
    – 4charwon
    Jan 10 at 16:57














0












0








0





$begingroup$

I see two method two calculate the powers of the matrix in your example.



First method : you diagonalize your matrix. Then it is easy.



Second method : If I note $M$ your matrix, you want to calculate $M^{20}$. You know that $M^2 - M + 15 = 0$ (Cayley Hamilton theorem). So by putting $Q = X^2 - X + 15$, and $X^{20} = PQ + R $ where $deg(R) leq 1$ (so $R = a_{20} + b_{20} X), M^{20} = R(M).$ And it is possible to calculate $R$ by looking its value at your two lamdas. An another way is to make a recursion (but it is not necessary to calculate iterated calculus 20 times...).



If your question is : how I calculate $lambda^{20} = a_{20} times 3 pm b_{20} times sqrt{61}$, the answer is you do not have a better way to do that iterating calculus, so don't do it.



EDIT :



I've seen an another answer ; yes, it is a bit faster if you apply exponentiating by squaring, to try to calculate more explicitely $lambda^{20}$, but it is still dull. I advice you to keep $lambda^{20}$ in your result.






share|cite|improve this answer











$endgroup$



I see two method two calculate the powers of the matrix in your example.



First method : you diagonalize your matrix. Then it is easy.



Second method : If I note $M$ your matrix, you want to calculate $M^{20}$. You know that $M^2 - M + 15 = 0$ (Cayley Hamilton theorem). So by putting $Q = X^2 - X + 15$, and $X^{20} = PQ + R $ where $deg(R) leq 1$ (so $R = a_{20} + b_{20} X), M^{20} = R(M).$ And it is possible to calculate $R$ by looking its value at your two lamdas. An another way is to make a recursion (but it is not necessary to calculate iterated calculus 20 times...).



If your question is : how I calculate $lambda^{20} = a_{20} times 3 pm b_{20} times sqrt{61}$, the answer is you do not have a better way to do that iterating calculus, so don't do it.



EDIT :



I've seen an another answer ; yes, it is a bit faster if you apply exponentiating by squaring, to try to calculate more explicitely $lambda^{20}$, but it is still dull. I advice you to keep $lambda^{20}$ in your result.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 10 at 16:45

























answered Jan 10 at 16:39









DLeMeurDLeMeur

3148




3148












  • $begingroup$
    Thank you for detail answer!!!
    $endgroup$
    – 4charwon
    Jan 10 at 16:57


















  • $begingroup$
    Thank you for detail answer!!!
    $endgroup$
    – 4charwon
    Jan 10 at 16:57
















$begingroup$
Thank you for detail answer!!!
$endgroup$
– 4charwon
Jan 10 at 16:57




$begingroup$
Thank you for detail answer!!!
$endgroup$
– 4charwon
Jan 10 at 16:57


















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