Diagonalization to power of matrix
$begingroup$
$left(array{
2&3 \
5& 1
}right)^{20}$
$(lambda-1)(lambda-2)-15=0$
$lambda = (3pm sqrt{61})/2$
Is this problem should be solved in this method? Lambda is not tidy.
linear-algebra diagonalization
edited Jan 10 at 17:24
the_fox
2,90021537
asked Jan 10 at 16:27
4charwon4charwon
274
$endgroup$
add a comment |
$begingroup$
$left(array{
2&3 \
5& 1
}right)^{20}$
$(lambda-1)(lambda-2)-15=0$
$lambda = (3pm sqrt{61})/2$
Is this problem should be solved in this method? Lambda is not tidy.
linear-algebra diagonalization
edited Jan 10 at 17:24
the_fox
2,90021537
asked Jan 10 at 16:27
4charwon4charwon
274
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
$left(array{
2&3 \
5& 1
}right)^{20}$
$(lambda-1)(lambda-2)-15=0$
$lambda = (3pm sqrt{61})/2$
Is this problem should be solved in this method? Lambda is not tidy.
linear-algebra diagonalization
edited Jan 10 at 17:24
the_fox
2,90021537
asked Jan 10 at 16:27
4charwon4charwon
274
$endgroup$
$left(array{
2&3 \
5& 1
}right)^{20}$
$(lambda-1)(lambda-2)-15=0$
$lambda = (3pm sqrt{61})/2$
Is this problem should be solved in this method? Lambda is not tidy.
linear-algebra diagonalization
linear-algebra diagonalization
edited Jan 10 at 17:24
the_fox
2,90021537
asked Jan 10 at 16:27
4charwon4charwon
274
edited Jan 10 at 17:24
the_fox
2,90021537
asked Jan 10 at 16:27
4charwon4charwon
274
edited Jan 10 at 17:24
the_fox
2,90021537
edited Jan 10 at 17:24
the_fox
2,90021537
edited Jan 10 at 17:24
the_fox
2,90021537
2,90021537
asked Jan 10 at 16:27
4charwon4charwon
274
asked Jan 10 at 16:27
4charwon4charwon
274
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered Jan 10 at 16:39
DLeMeurDLeMeur
3148
$endgroup$
$begingroup$
Thank you for detail answer!!!
$endgroup$
– 4charwon
Jan 10 at 16:57
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
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.
answered Jan 10 at 16:39
DLeMeurDLeMeur
3148
$endgroup$
$begingroup$
Thank you for detail answer!!!
$endgroup$
– 4charwon
Jan 10 at 16:57
add a comment |
$begingroup$
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.
answered Jan 10 at 16:39
DLeMeurDLeMeur
3148
$endgroup$
$begingroup$
Thank you for detail answer!!!
$endgroup$
– 4charwon
Jan 10 at 16:57
add a comment |
$begingroup$
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.
answered Jan 10 at 16:39
DLeMeurDLeMeur
3148
$endgroup$
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.
answered Jan 10 at 16:39
DLeMeurDLeMeur
3148
edited Jan 10 at 16:45
answered Jan 10 at 16:39
DLeMeurDLeMeur
3148
answered Jan 10 at 16:39
DLeMeurDLeMeur
3148
answered Jan 10 at 16:39
DLeMeurDLeMeur
3148
3148
$begingroup$
Thank you for detail answer!!!
$endgroup$
– 4charwon
Jan 10 at 16:57
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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