Mixing
$A^100 $ where $A = beginbmatrix 1 &2 \ 3& 4 endbmatrix$ [duplicate]
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This question already has an answer here:
Finding a 2x2 Matrix raised to the power of 1000
5 answers
Compute $A^100 $ where $A = beginbmatrix 1 &2 \ 3& 4 endbmatrix$.
I can calculate $A^100$ using a calculator, but my question is that is there any short formula/method or is their any trick to find the $A^100$?
matrices
edited Jul 25 at 2:16
Math Lover
12.3k21232
asked Jul 25 at 1:08
stupid
54018
marked as duplicate by Claude Leibovici, Delta-u, Arnaud D., Mostafa Ayaz, Adrian Keister Jul 25 at 13:14
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
 |Â
show 6 more comments
up vote
4
down vote
favorite
This question already has an answer here:
Finding a 2x2 Matrix raised to the power of 1000
5 answers
Compute $A^100 $ where $A = beginbmatrix 1 &2 \ 3& 4 endbmatrix$.
I can calculate $A^100$ using a calculator, but my question is that is there any short formula/method or is their any trick to find the $A^100$?
matrices
edited Jul 25 at 2:16
Math Lover
12.3k21232
asked Jul 25 at 1:08
stupid
54018
marked as duplicate by Claude Leibovici, Delta-u, Arnaud D., Mostafa Ayaz, Adrian Keister Jul 25 at 13:14
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
2
Hint: Diagonalization (you have unique eigenvalues) and similar approaches are methods. See: en.wikipedia.org/wiki/Diagonalizable_matrix#Diagonalization.
– Moo
Jul 25 at 1:09
1
To elaborate, supposing that $A$ can be diagonalized in the form $A=SDS^-1$ where $D$ is a diagonal matrix, one would have $A^n = (SDS^-1)^n$ which by expanding and induction you would see simplifies dramatically to be $A^n = SD^n S^-1$, remembering also that the power of a diagonal matrix is extremely easy to compute.
– JMoravitz
Jul 25 at 1:12
2
No. don't use diagonalization. That is a bad advice. You waste time computing eigenvalues, eigenvectors, operations that potentially take you out of the ring of coefficients. The ideal (no pun intended) solution for this problem is the following: Compute the characteristic polynomial $p(x)$, which will have degree only $2$. By Hamilton-Cayley $p(A)=0$. Divide $x^100=p(x)q(x)+ax+b$. Then $A^100=p(A)q(A)+aA+b=aA+b$.
– user578878
Jul 25 at 1:17
Does it have to be the matrix you've written down? It isn't the best example to start with in attempting to understand this process. You may want to consider something simpler, like $$beginpmatrix1&2\2&1endpmatrix$$, which has a nice diagonal form.
– Chickenmancer
Jul 25 at 1:20
2
To complete my comment above. Since $x^100=p(x)q(x)+ax+b$ and we only need $a,b$, then we can evaluate at the roots $r1,r2=frac5pmsqrt332$ of $p(x)$. We get $r_1^100=ar_1+b$ and $r_2^100=ar_2+b$. From where $a=fracr_1^100-r_2^100r_1-r_2$ and $b=fracr_1r_2^100-r_2r_1^100r_1-r_2$. Therefore, $A^100=fracr_1^100-r_2^100r_1-r_2A+fracr_1r_2^100-r_2r_1^100r_1-r_2$.
– user578878
Jul 25 at 1:34
 |Â
show 6 more comments
up vote
4
down vote
favorite
up vote
4
down vote
favorite
This question already has an answer here:
Finding a 2x2 Matrix raised to the power of 1000
5 answers
Compute $A^100 $ where $A = beginbmatrix 1 &2 \ 3& 4 endbmatrix$.
I can calculate $A^100$ using a calculator, but my question is that is there any short formula/method or is their any trick to find the $A^100$?
matrices
edited Jul 25 at 2:16
Math Lover
12.3k21232
asked Jul 25 at 1:08
stupid
54018
This question already has an answer here:
Finding a 2x2 Matrix raised to the power of 1000
5 answers
Compute $A^100 $ where $A = beginbmatrix 1 &2 \ 3& 4 endbmatrix$.
I can calculate $A^100$ using a calculator, but my question is that is there any short formula/method or is their any trick to find the $A^100$?
This question already has an answer here:
Finding a 2x2 Matrix raised to the power of 1000
5 answers
matrices
edited Jul 25 at 2:16
Math Lover
12.3k21232
asked Jul 25 at 1:08
stupid
54018
edited Jul 25 at 2:16
Math Lover
12.3k21232
edited Jul 25 at 2:16
Math Lover
12.3k21232
edited Jul 25 at 2:16
Math Lover
12.3k21232
12.3k21232
asked Jul 25 at 1:08
stupid
54018
asked Jul 25 at 1:08
stupid
54018
asked Jul 25 at 1:08
stupid
54018
54018
marked as duplicate by Claude Leibovici, Delta-u, Arnaud D., Mostafa Ayaz, Adrian Keister Jul 25 at 13:14
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Claude Leibovici, Delta-u, Arnaud D., Mostafa Ayaz, Adrian Keister Jul 25 at 13:14
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
2
Hint: Diagonalization (you have unique eigenvalues) and similar approaches are methods. See: en.wikipedia.org/wiki/Diagonalizable_matrix#Diagonalization.
– Moo
Jul 25 at 1:09
1
To elaborate, supposing that $A$ can be diagonalized in the form $A=SDS^-1$ where $D$ is a diagonal matrix, one would have $A^n = (SDS^-1)^n$ which by expanding and induction you would see simplifies dramatically to be $A^n = SD^n S^-1$, remembering also that the power of a diagonal matrix is extremely easy to compute.
– JMoravitz
Jul 25 at 1:12
2
No. don't use diagonalization. That is a bad advice. You waste time computing eigenvalues, eigenvectors, operations that potentially take you out of the ring of coefficients. The ideal (no pun intended) solution for this problem is the following: Compute the characteristic polynomial $p(x)$, which will have degree only $2$. By Hamilton-Cayley $p(A)=0$. Divide $x^100=p(x)q(x)+ax+b$. Then $A^100=p(A)q(A)+aA+b=aA+b$.
– user578878
Jul 25 at 1:17
Does it have to be the matrix you've written down? It isn't the best example to start with in attempting to understand this process. You may want to consider something simpler, like $$beginpmatrix1&2\2&1endpmatrix$$, which has a nice diagonal form.
– Chickenmancer
Jul 25 at 1:20
2
To complete my comment above. Since $x^100=p(x)q(x)+ax+b$ and we only need $a,b$, then we can evaluate at the roots $r1,r2=frac5pmsqrt332$ of $p(x)$. We get $r_1^100=ar_1+b$ and $r_2^100=ar_2+b$. From where $a=fracr_1^100-r_2^100r_1-r_2$ and $b=fracr_1r_2^100-r_2r_1^100r_1-r_2$. Therefore, $A^100=fracr_1^100-r_2^100r_1-r_2A+fracr_1r_2^100-r_2r_1^100r_1-r_2$.
– user578878
Jul 25 at 1:34
 |Â
show 6 more comments
2
Hint: Diagonalization (you have unique eigenvalues) and similar approaches are methods. See: en.wikipedia.org/wiki/Diagonalizable_matrix#Diagonalization.
– Moo
Jul 25 at 1:09
1
To elaborate, supposing that $A$ can be diagonalized in the form $A=SDS^-1$ where $D$ is a diagonal matrix, one would have $A^n = (SDS^-1)^n$ which by expanding and induction you would see simplifies dramatically to be $A^n = SD^n S^-1$, remembering also that the power of a diagonal matrix is extremely easy to compute.
– JMoravitz
Jul 25 at 1:12
2
No. don't use diagonalization. That is a bad advice. You waste time computing eigenvalues, eigenvectors, operations that potentially take you out of the ring of coefficients. The ideal (no pun intended) solution for this problem is the following: Compute the characteristic polynomial $p(x)$, which will have degree only $2$. By Hamilton-Cayley $p(A)=0$. Divide $x^100=p(x)q(x)+ax+b$. Then $A^100=p(A)q(A)+aA+b=aA+b$.
– user578878
Jul 25 at 1:17
Does it have to be the matrix you've written down? It isn't the best example to start with in attempting to understand this process. You may want to consider something simpler, like $$beginpmatrix1&2\2&1endpmatrix$$, which has a nice diagonal form.
– Chickenmancer
Jul 25 at 1:20
2
To complete my comment above. Since $x^100=p(x)q(x)+ax+b$ and we only need $a,b$, then we can evaluate at the roots $r1,r2=frac5pmsqrt332$ of $p(x)$. We get $r_1^100=ar_1+b$ and $r_2^100=ar_2+b$. From where $a=fracr_1^100-r_2^100r_1-r_2$ and $b=fracr_1r_2^100-r_2r_1^100r_1-r_2$. Therefore, $A^100=fracr_1^100-r_2^100r_1-r_2A+fracr_1r_2^100-r_2r_1^100r_1-r_2$.
– user578878
Jul 25 at 1:34
2
2
Hint: Diagonalization (you have unique eigenvalues) and similar approaches are methods. See: en.wikipedia.org/wiki/Diagonalizable_matrix#Diagonalization.
– Moo
Jul 25 at 1:09
Hint: Diagonalization (you have unique eigenvalues) and similar approaches are methods. See: en.wikipedia.org/wiki/Diagonalizable_matrix#Diagonalization.
– Moo
Jul 25 at 1:09
1
1
To elaborate, supposing that $A$ can be diagonalized in the form $A=SDS^-1$ where $D$ is a diagonal matrix, one would have $A^n = (SDS^-1)^n$ which by expanding and induction you would see simplifies dramatically to be $A^n = SD^n S^-1$, remembering also that the power of a diagonal matrix is extremely easy to compute.
– JMoravitz
Jul 25 at 1:12
To elaborate, supposing that $A$ can be diagonalized in the form $A=SDS^-1$ where $D$ is a diagonal matrix, one would have $A^n = (SDS^-1)^n$ which by expanding and induction you would see simplifies dramatically to be $A^n = SD^n S^-1$, remembering also that the power of a diagonal matrix is extremely easy to compute.
– JMoravitz
Jul 25 at 1:12
2
2
No. don't use diagonalization. That is a bad advice. You waste time computing eigenvalues, eigenvectors, operations that potentially take you out of the ring of coefficients. The ideal (no pun intended) solution for this problem is the following: Compute the characteristic polynomial $p(x)$, which will have degree only $2$. By Hamilton-Cayley $p(A)=0$. Divide $x^100=p(x)q(x)+ax+b$. Then $A^100=p(A)q(A)+aA+b=aA+b$.
– user578878
Jul 25 at 1:17
No. don't use diagonalization. That is a bad advice. You waste time computing eigenvalues, eigenvectors, operations that potentially take you out of the ring of coefficients. The ideal (no pun intended) solution for this problem is the following: Compute the characteristic polynomial $p(x)$, which will have degree only $2$. By Hamilton-Cayley $p(A)=0$. Divide $x^100=p(x)q(x)+ax+b$. Then $A^100=p(A)q(A)+aA+b=aA+b$.
– user578878
Jul 25 at 1:17
Does it have to be the matrix you've written down? It isn't the best example to start with in attempting to understand this process. You may want to consider something simpler, like $$beginpmatrix1&2\2&1endpmatrix$$, which has a nice diagonal form.
– Chickenmancer
Jul 25 at 1:20
Does it have to be the matrix you've written down? It isn't the best example to start with in attempting to understand this process. You may want to consider something simpler, like $$beginpmatrix1&2\2&1endpmatrix$$, which has a nice diagonal form.
– Chickenmancer
Jul 25 at 1:20
2
2
To complete my comment above. Since $x^100=p(x)q(x)+ax+b$ and we only need $a,b$, then we can evaluate at the roots $r1,r2=frac5pmsqrt332$ of $p(x)$. We get $r_1^100=ar_1+b$ and $r_2^100=ar_2+b$. From where $a=fracr_1^100-r_2^100r_1-r_2$ and $b=fracr_1r_2^100-r_2r_1^100r_1-r_2$. Therefore, $A^100=fracr_1^100-r_2^100r_1-r_2A+fracr_1r_2^100-r_2r_1^100r_1-r_2$.
– user578878
Jul 25 at 1:34
To complete my comment above. Since $x^100=p(x)q(x)+ax+b$ and we only need $a,b$, then we can evaluate at the roots $r1,r2=frac5pmsqrt332$ of $p(x)$. We get $r_1^100=ar_1+b$ and $r_2^100=ar_2+b$. From where $a=fracr_1^100-r_2^100r_1-r_2$ and $b=fracr_1r_2^100-r_2r_1^100r_1-r_2$. Therefore, $A^100=fracr_1^100-r_2^100r_1-r_2A+fracr_1r_2^100-r_2r_1^100r_1-r_2$.
– user578878
Jul 25 at 1:34
 |Â
show 6 more comments
5 Answers
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The conventional answer is going to be to diagonalize the matrix into $A=PLambda P^-1$ and then compute $PLambda^100P^-1$, but once you have the eigenvalues $lambda_1$ and $lambda_2$ there are ways to do this without computing an eigenbasis:
- Decompose $A$ into $lambda_1P_1+lambda_2P_2$, where $P_1$ and $P_2$ are projections onto the corresponding eigenspaces with $P_1P_2=P_2P_1=0$. There’s a fairly simply formula for these projections in terms of $A$ and the two eigenvalues. If you expand $A^100$ using the binomial theorem, you’ll find that all but two terms vanish.
- Use the Cayley-Hamilton theorem to write $A^100=aI+bA$ for some undetermined coefficients $a$ and $b$. This equation is also satisfied by the eigenvalues, which gives you the system of linear equations $a+blambda_i=lambda_i^100$ for $a$ and $b$.
The above assumes that $A$ has distinct real eigenvalues. If they’re not, you will have to modify the above methods a bit.
answered Jul 25 at 1:26
amd
25.8k2943
add a comment |Â
up vote
2
down vote
You could use diagonalization.
Let $A$ be a matrix then if you diagonalize $A$ then you get $A=PBP^-1$ with $B$ as a diagonal matrix and $P^-1$ an invertible matrix. If you try for $A^2$, then it is equal to $A^2=PB^2P^-1$. Similarly for $A^n=PB^nP^-1$
$A^100=PB^100P^-1$
answered Jul 25 at 1:13
Key Flex
4,193423
add a comment |Â
up vote
2
down vote
You can first diagonalize the matrix as follows: $$A=P^-1DP,$$
where $P$ is an orthogonal matrix. The matrix $P$ is the matrix of eigenvectors $v_1,v_2$ that correspond to eigenvalues $lambda_1,lambda_2$ of $A$. Here, $$D=mathrmdiaglambda_1,lambda_2 $$
After diagonalizing, you can calculate $A^100$ as follows:
$$A^100 = (P^-1DP)^100=P^-1D^100P = P^-1mathrmdiaglambda_1^100,lambda_2^100P.
$$
edited Jul 25 at 2:09
Math Lover
12.3k21232
answered Jul 25 at 1:24
gph
216
what is value of P ?? as i can not able find the value of P
– stupid
Jul 25 at 14:01
@stupid wwwf.imperial.ac.uk/metric/metric_public/matrices/…
– gph
Jul 26 at 1:16
@stupid this is an very clear tutorial about how to compute the eigenvalues and eigenvectors of a matrix. after you have found the eigenvectors $v_1,v_2,..$,you will have to change them to orthonormal basis$u_1,u_2,...$, that is they are orthogonal to each other, and each of them have length one. This procedure can be done by "gram-schmidt orthogonalization", you can find many tutorials such as math.usm.edu/lambers/mat415/lecture3.pdf. Then $P^-1=P'=[u_1,u_2,...]$ It is solved. P is orthogonal matrix.
– gph
Jul 26 at 1:29
add a comment |Â
up vote
1
down vote
There is no need for diagonalizations.
The eigenvalues of $$A = beginbmatrix 1 &2 \ 3& 4 endbmatrix$$ are $$frac 5pmsqrt 332$$
Cayley-Hamilton Theorem indicates that $$A^100=alpha A + beta I.$$
We can find the coefficients $alpha$ and $beta $ by equations
$$ alpha lambda _1 +beta = lambda _1^100\ alpha lambda _2 +beta = lambda _2^100$$ Where $lambda _1$ and $lambda _2$ are eighenvalues of $A.$
answered Jul 25 at 2:35
Mohammad Riazi-Kermani
27.5k41852
what is correct ?? im not able to find the value of $A^100$ it is very long and difficults
– stupid
Jul 25 at 13:57
add a comment |Â
up vote
1
down vote
Hint: the characteristic function of the matrix is $$lambda^2=5lambda+2$$so according to Caylay-Hamilton theorem we have $$A^2=5A+2$$
answered Jul 25 at 11:38
Mostafa Ayaz
8,5373630
add a comment |Â
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
The conventional answer is going to be to diagonalize the matrix into $A=PLambda P^-1$ and then compute $PLambda^100P^-1$, but once you have the eigenvalues $lambda_1$ and $lambda_2$ there are ways to do this without computing an eigenbasis:
- Decompose $A$ into $lambda_1P_1+lambda_2P_2$, where $P_1$ and $P_2$ are projections onto the corresponding eigenspaces with $P_1P_2=P_2P_1=0$. There’s a fairly simply formula for these projections in terms of $A$ and the two eigenvalues. If you expand $A^100$ using the binomial theorem, you’ll find that all but two terms vanish.
- Use the Cayley-Hamilton theorem to write $A^100=aI+bA$ for some undetermined coefficients $a$ and $b$. This equation is also satisfied by the eigenvalues, which gives you the system of linear equations $a+blambda_i=lambda_i^100$ for $a$ and $b$.
The above assumes that $A$ has distinct real eigenvalues. If they’re not, you will have to modify the above methods a bit.
answered Jul 25 at 1:26
amd
25.8k2943
add a comment |Â
up vote
2
down vote
The conventional answer is going to be to diagonalize the matrix into $A=PLambda P^-1$ and then compute $PLambda^100P^-1$, but once you have the eigenvalues $lambda_1$ and $lambda_2$ there are ways to do this without computing an eigenbasis:
- Decompose $A$ into $lambda_1P_1+lambda_2P_2$, where $P_1$ and $P_2$ are projections onto the corresponding eigenspaces with $P_1P_2=P_2P_1=0$. There’s a fairly simply formula for these projections in terms of $A$ and the two eigenvalues. If you expand $A^100$ using the binomial theorem, you’ll find that all but two terms vanish.
- Use the Cayley-Hamilton theorem to write $A^100=aI+bA$ for some undetermined coefficients $a$ and $b$. This equation is also satisfied by the eigenvalues, which gives you the system of linear equations $a+blambda_i=lambda_i^100$ for $a$ and $b$.
The above assumes that $A$ has distinct real eigenvalues. If they’re not, you will have to modify the above methods a bit.
answered Jul 25 at 1:26
amd
25.8k2943
add a comment |Â
up vote
2
down vote
up vote
2
down vote
The conventional answer is going to be to diagonalize the matrix into $A=PLambda P^-1$ and then compute $PLambda^100P^-1$, but once you have the eigenvalues $lambda_1$ and $lambda_2$ there are ways to do this without computing an eigenbasis:
- Decompose $A$ into $lambda_1P_1+lambda_2P_2$, where $P_1$ and $P_2$ are projections onto the corresponding eigenspaces with $P_1P_2=P_2P_1=0$. There’s a fairly simply formula for these projections in terms of $A$ and the two eigenvalues. If you expand $A^100$ using the binomial theorem, you’ll find that all but two terms vanish.
- Use the Cayley-Hamilton theorem to write $A^100=aI+bA$ for some undetermined coefficients $a$ and $b$. This equation is also satisfied by the eigenvalues, which gives you the system of linear equations $a+blambda_i=lambda_i^100$ for $a$ and $b$.
The above assumes that $A$ has distinct real eigenvalues. If they’re not, you will have to modify the above methods a bit.
answered Jul 25 at 1:26
amd
25.8k2943
The conventional answer is going to be to diagonalize the matrix into $A=PLambda P^-1$ and then compute $PLambda^100P^-1$, but once you have the eigenvalues $lambda_1$ and $lambda_2$ there are ways to do this without computing an eigenbasis:
- Decompose $A$ into $lambda_1P_1+lambda_2P_2$, where $P_1$ and $P_2$ are projections onto the corresponding eigenspaces with $P_1P_2=P_2P_1=0$. There’s a fairly simply formula for these projections in terms of $A$ and the two eigenvalues. If you expand $A^100$ using the binomial theorem, you’ll find that all but two terms vanish.
- Use the Cayley-Hamilton theorem to write $A^100=aI+bA$ for some undetermined coefficients $a$ and $b$. This equation is also satisfied by the eigenvalues, which gives you the system of linear equations $a+blambda_i=lambda_i^100$ for $a$ and $b$.
The above assumes that $A$ has distinct real eigenvalues. If they’re not, you will have to modify the above methods a bit.
answered Jul 25 at 1:26
amd
25.8k2943
answered Jul 25 at 1:26
amd
25.8k2943
answered Jul 25 at 1:26
amd
25.8k2943
answered Jul 25 at 1:26
amd
25.8k2943
25.8k2943
add a comment |Â
add a comment |Â
up vote
2
down vote
You could use diagonalization.
Let $A$ be a matrix then if you diagonalize $A$ then you get $A=PBP^-1$ with $B$ as a diagonal matrix and $P^-1$ an invertible matrix. If you try for $A^2$, then it is equal to $A^2=PB^2P^-1$. Similarly for $A^n=PB^nP^-1$
$A^100=PB^100P^-1$
answered Jul 25 at 1:13
Key Flex
4,193423
add a comment |Â
up vote
2
down vote
You could use diagonalization.
Let $A$ be a matrix then if you diagonalize $A$ then you get $A=PBP^-1$ with $B$ as a diagonal matrix and $P^-1$ an invertible matrix. If you try for $A^2$, then it is equal to $A^2=PB^2P^-1$. Similarly for $A^n=PB^nP^-1$
$A^100=PB^100P^-1$
answered Jul 25 at 1:13
Key Flex
4,193423
add a comment |Â
up vote
2
down vote
up vote
2
down vote
You could use diagonalization.
Let $A$ be a matrix then if you diagonalize $A$ then you get $A=PBP^-1$ with $B$ as a diagonal matrix and $P^-1$ an invertible matrix. If you try for $A^2$, then it is equal to $A^2=PB^2P^-1$. Similarly for $A^n=PB^nP^-1$
$A^100=PB^100P^-1$
answered Jul 25 at 1:13
Key Flex
4,193423
You could use diagonalization.
Let $A$ be a matrix then if you diagonalize $A$ then you get $A=PBP^-1$ with $B$ as a diagonal matrix and $P^-1$ an invertible matrix. If you try for $A^2$, then it is equal to $A^2=PB^2P^-1$. Similarly for $A^n=PB^nP^-1$
$A^100=PB^100P^-1$
answered Jul 25 at 1:13
Key Flex
4,193423
edited Jul 25 at 2:00
user578878
answered Jul 25 at 1:13
Key Flex
4,193423
answered Jul 25 at 1:13
Key Flex
4,193423
answered Jul 25 at 1:13
Key Flex
4,193423
4,193423
add a comment |Â
add a comment |Â
up vote
2
down vote
You can first diagonalize the matrix as follows: $$A=P^-1DP,$$
where $P$ is an orthogonal matrix. The matrix $P$ is the matrix of eigenvectors $v_1,v_2$ that correspond to eigenvalues $lambda_1,lambda_2$ of $A$. Here, $$D=mathrmdiaglambda_1,lambda_2 $$
After diagonalizing, you can calculate $A^100$ as follows:
$$A^100 = (P^-1DP)^100=P^-1D^100P = P^-1mathrmdiaglambda_1^100,lambda_2^100P.
$$
edited Jul 25 at 2:09
Math Lover
12.3k21232
answered Jul 25 at 1:24
gph
216
what is value of P ?? as i can not able find the value of P
– stupid
Jul 25 at 14:01
@stupid wwwf.imperial.ac.uk/metric/metric_public/matrices/…
– gph
Jul 26 at 1:16
@stupid this is an very clear tutorial about how to compute the eigenvalues and eigenvectors of a matrix. after you have found the eigenvectors $v_1,v_2,..$,you will have to change them to orthonormal basis$u_1,u_2,...$, that is they are orthogonal to each other, and each of them have length one. This procedure can be done by "gram-schmidt orthogonalization", you can find many tutorials such as math.usm.edu/lambers/mat415/lecture3.pdf. Then $P^-1=P'=[u_1,u_2,...]$ It is solved. P is orthogonal matrix.
– gph
Jul 26 at 1:29
add a comment |Â
up vote
2
down vote
You can first diagonalize the matrix as follows: $$A=P^-1DP,$$
where $P$ is an orthogonal matrix. The matrix $P$ is the matrix of eigenvectors $v_1,v_2$ that correspond to eigenvalues $lambda_1,lambda_2$ of $A$. Here, $$D=mathrmdiaglambda_1,lambda_2 $$
After diagonalizing, you can calculate $A^100$ as follows:
$$A^100 = (P^-1DP)^100=P^-1D^100P = P^-1mathrmdiaglambda_1^100,lambda_2^100P.
$$
edited Jul 25 at 2:09
Math Lover
12.3k21232
answered Jul 25 at 1:24
gph
216
what is value of P ?? as i can not able find the value of P
– stupid
Jul 25 at 14:01
@stupid wwwf.imperial.ac.uk/metric/metric_public/matrices/…
– gph
Jul 26 at 1:16
@stupid this is an very clear tutorial about how to compute the eigenvalues and eigenvectors of a matrix. after you have found the eigenvectors $v_1,v_2,..$,you will have to change them to orthonormal basis$u_1,u_2,...$, that is they are orthogonal to each other, and each of them have length one. This procedure can be done by "gram-schmidt orthogonalization", you can find many tutorials such as math.usm.edu/lambers/mat415/lecture3.pdf. Then $P^-1=P'=[u_1,u_2,...]$ It is solved. P is orthogonal matrix.
– gph
Jul 26 at 1:29
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up vote
2
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up vote
2
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You can first diagonalize the matrix as follows: $$A=P^-1DP,$$
where $P$ is an orthogonal matrix. The matrix $P$ is the matrix of eigenvectors $v_1,v_2$ that correspond to eigenvalues $lambda_1,lambda_2$ of $A$. Here, $$D=mathrmdiaglambda_1,lambda_2 $$
After diagonalizing, you can calculate $A^100$ as follows:
$$A^100 = (P^-1DP)^100=P^-1D^100P = P^-1mathrmdiaglambda_1^100,lambda_2^100P.
$$
edited Jul 25 at 2:09
Math Lover
12.3k21232
answered Jul 25 at 1:24
gph
216
You can first diagonalize the matrix as follows: $$A=P^-1DP,$$
where $P$ is an orthogonal matrix. The matrix $P$ is the matrix of eigenvectors $v_1,v_2$ that correspond to eigenvalues $lambda_1,lambda_2$ of $A$. Here, $$D=mathrmdiaglambda_1,lambda_2 $$
After diagonalizing, you can calculate $A^100$ as follows:
$$A^100 = (P^-1DP)^100=P^-1D^100P = P^-1mathrmdiaglambda_1^100,lambda_2^100P.
$$
edited Jul 25 at 2:09
Math Lover
12.3k21232
answered Jul 25 at 1:24
gph
216
edited Jul 25 at 2:09
Math Lover
12.3k21232
edited Jul 25 at 2:09
Math Lover
12.3k21232
edited Jul 25 at 2:09
Math Lover
12.3k21232
12.3k21232
answered Jul 25 at 1:24
gph
216
answered Jul 25 at 1:24
gph
216
answered Jul 25 at 1:24
gph
216
216
what is value of P ?? as i can not able find the value of P
– stupid
Jul 25 at 14:01
@stupid wwwf.imperial.ac.uk/metric/metric_public/matrices/…
– gph
Jul 26 at 1:16
@stupid this is an very clear tutorial about how to compute the eigenvalues and eigenvectors of a matrix. after you have found the eigenvectors $v_1,v_2,..$,you will have to change them to orthonormal basis$u_1,u_2,...$, that is they are orthogonal to each other, and each of them have length one. This procedure can be done by "gram-schmidt orthogonalization", you can find many tutorials such as math.usm.edu/lambers/mat415/lecture3.pdf. Then $P^-1=P'=[u_1,u_2,...]$ It is solved. P is orthogonal matrix.
– gph
Jul 26 at 1:29
add a comment |Â
what is value of P ?? as i can not able find the value of P
– stupid
Jul 25 at 14:01
@stupid wwwf.imperial.ac.uk/metric/metric_public/matrices/…
– gph
Jul 26 at 1:16
@stupid this is an very clear tutorial about how to compute the eigenvalues and eigenvectors of a matrix. after you have found the eigenvectors $v_1,v_2,..$,you will have to change them to orthonormal basis$u_1,u_2,...$, that is they are orthogonal to each other, and each of them have length one. This procedure can be done by "gram-schmidt orthogonalization", you can find many tutorials such as math.usm.edu/lambers/mat415/lecture3.pdf. Then $P^-1=P'=[u_1,u_2,...]$ It is solved. P is orthogonal matrix.
– gph
Jul 26 at 1:29
what is value of P ?? as i can not able find the value of P
– stupid
Jul 25 at 14:01
what is value of P ?? as i can not able find the value of P
– stupid
Jul 25 at 14:01
@stupid wwwf.imperial.ac.uk/metric/metric_public/matrices/…
– gph
Jul 26 at 1:16
@stupid wwwf.imperial.ac.uk/metric/metric_public/matrices/…
– gph
Jul 26 at 1:16
@stupid this is an very clear tutorial about how to compute the eigenvalues and eigenvectors of a matrix. after you have found the eigenvectors $v_1,v_2,..$,you will have to change them to orthonormal basis$u_1,u_2,...$, that is they are orthogonal to each other, and each of them have length one. This procedure can be done by "gram-schmidt orthogonalization", you can find many tutorials such as math.usm.edu/lambers/mat415/lecture3.pdf. Then $P^-1=P'=[u_1,u_2,...]$ It is solved. P is orthogonal matrix.
– gph
Jul 26 at 1:29
@stupid this is an very clear tutorial about how to compute the eigenvalues and eigenvectors of a matrix. after you have found the eigenvectors $v_1,v_2,..$,you will have to change them to orthonormal basis$u_1,u_2,...$, that is they are orthogonal to each other, and each of them have length one. This procedure can be done by "gram-schmidt orthogonalization", you can find many tutorials such as math.usm.edu/lambers/mat415/lecture3.pdf. Then $P^-1=P'=[u_1,u_2,...]$ It is solved. P is orthogonal matrix.
– gph
Jul 26 at 1:29
add a comment |Â
up vote
1
down vote
There is no need for diagonalizations.
The eigenvalues of $$A = beginbmatrix 1 &2 \ 3& 4 endbmatrix$$ are $$frac 5pmsqrt 332$$
Cayley-Hamilton Theorem indicates that $$A^100=alpha A + beta I.$$
We can find the coefficients $alpha$ and $beta $ by equations
$$ alpha lambda _1 +beta = lambda _1^100\ alpha lambda _2 +beta = lambda _2^100$$ Where $lambda _1$ and $lambda _2$ are eighenvalues of $A.$
answered Jul 25 at 2:35
Mohammad Riazi-Kermani
27.5k41852
what is correct ?? im not able to find the value of $A^100$ it is very long and difficults
– stupid
Jul 25 at 13:57
add a comment |Â
up vote
1
down vote
There is no need for diagonalizations.
The eigenvalues of $$A = beginbmatrix 1 &2 \ 3& 4 endbmatrix$$ are $$frac 5pmsqrt 332$$
Cayley-Hamilton Theorem indicates that $$A^100=alpha A + beta I.$$
We can find the coefficients $alpha$ and $beta $ by equations
$$ alpha lambda _1 +beta = lambda _1^100\ alpha lambda _2 +beta = lambda _2^100$$ Where $lambda _1$ and $lambda _2$ are eighenvalues of $A.$
answered Jul 25 at 2:35
Mohammad Riazi-Kermani
27.5k41852
what is correct ?? im not able to find the value of $A^100$ it is very long and difficults
– stupid
Jul 25 at 13:57
add a comment |Â
up vote
1
down vote
up vote
1
down vote
There is no need for diagonalizations.
The eigenvalues of $$A = beginbmatrix 1 &2 \ 3& 4 endbmatrix$$ are $$frac 5pmsqrt 332$$
Cayley-Hamilton Theorem indicates that $$A^100=alpha A + beta I.$$
We can find the coefficients $alpha$ and $beta $ by equations
$$ alpha lambda _1 +beta = lambda _1^100\ alpha lambda _2 +beta = lambda _2^100$$ Where $lambda _1$ and $lambda _2$ are eighenvalues of $A.$
answered Jul 25 at 2:35
Mohammad Riazi-Kermani
27.5k41852
There is no need for diagonalizations.
The eigenvalues of $$A = beginbmatrix 1 &2 \ 3& 4 endbmatrix$$ are $$frac 5pmsqrt 332$$
Cayley-Hamilton Theorem indicates that $$A^100=alpha A + beta I.$$
We can find the coefficients $alpha$ and $beta $ by equations
$$ alpha lambda _1 +beta = lambda _1^100\ alpha lambda _2 +beta = lambda _2^100$$ Where $lambda _1$ and $lambda _2$ are eighenvalues of $A.$
answered Jul 25 at 2:35
Mohammad Riazi-Kermani
27.5k41852
edited Jul 25 at 10:28
answered Jul 25 at 2:35
Mohammad Riazi-Kermani
27.5k41852
answered Jul 25 at 2:35
Mohammad Riazi-Kermani
27.5k41852
answered Jul 25 at 2:35
Mohammad Riazi-Kermani
27.5k41852
27.5k41852
what is correct ?? im not able to find the value of $A^100$ it is very long and difficults
– stupid
Jul 25 at 13:57
add a comment |Â
what is correct ?? im not able to find the value of $A^100$ it is very long and difficults
– stupid
Jul 25 at 13:57
what is correct ?? im not able to find the value of $A^100$ it is very long and difficults
– stupid
Jul 25 at 13:57
what is correct ?? im not able to find the value of $A^100$ it is very long and difficults
– stupid
Jul 25 at 13:57
add a comment |Â
up vote
1
down vote
Hint: the characteristic function of the matrix is $$lambda^2=5lambda+2$$so according to Caylay-Hamilton theorem we have $$A^2=5A+2$$
answered Jul 25 at 11:38
Mostafa Ayaz
8,5373630
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up vote
1
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Hint: the characteristic function of the matrix is $$lambda^2=5lambda+2$$so according to Caylay-Hamilton theorem we have $$A^2=5A+2$$
answered Jul 25 at 11:38
Mostafa Ayaz
8,5373630
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Hint: the characteristic function of the matrix is $$lambda^2=5lambda+2$$so according to Caylay-Hamilton theorem we have $$A^2=5A+2$$
answered Jul 25 at 11:38
Mostafa Ayaz
8,5373630
Hint: the characteristic function of the matrix is $$lambda^2=5lambda+2$$so according to Caylay-Hamilton theorem we have $$A^2=5A+2$$
answered Jul 25 at 11:38
Mostafa Ayaz
8,5373630
answered Jul 25 at 11:38
Mostafa Ayaz
8,5373630
answered Jul 25 at 11:38
Mostafa Ayaz
8,5373630
answered Jul 25 at 11:38
Mostafa Ayaz
8,5373630
8,5373630
add a comment |Â
add a comment |Â
2
Hint: Diagonalization (you have unique eigenvalues) and similar approaches are methods. See: en.wikipedia.org/wiki/Diagonalizable_matrix#Diagonalization.
– Moo
Jul 25 at 1:09
1
To elaborate, supposing that $A$ can be diagonalized in the form $A=SDS^-1$ where $D$ is a diagonal matrix, one would have $A^n = (SDS^-1)^n$ which by expanding and induction you would see simplifies dramatically to be $A^n = SD^n S^-1$, remembering also that the power of a diagonal matrix is extremely easy to compute.
– JMoravitz
Jul 25 at 1:12
2
No. don't use diagonalization. That is a bad advice. You waste time computing eigenvalues, eigenvectors, operations that potentially take you out of the ring of coefficients. The ideal (no pun intended) solution for this problem is the following: Compute the characteristic polynomial $p(x)$, which will have degree only $2$. By Hamilton-Cayley $p(A)=0$. Divide $x^100=p(x)q(x)+ax+b$. Then $A^100=p(A)q(A)+aA+b=aA+b$.
– user578878
Jul 25 at 1:17
Does it have to be the matrix you've written down? It isn't the best example to start with in attempting to understand this process. You may want to consider something simpler, like $$beginpmatrix1&2\2&1endpmatrix$$, which has a nice diagonal form.
– Chickenmancer
Jul 25 at 1:20
2
To complete my comment above. Since $x^100=p(x)q(x)+ax+b$ and we only need $a,b$, then we can evaluate at the roots $r1,r2=frac5pmsqrt332$ of $p(x)$. We get $r_1^100=ar_1+b$ and $r_2^100=ar_2+b$. From where $a=fracr_1^100-r_2^100r_1-r_2$ and $b=fracr_1r_2^100-r_2r_1^100r_1-r_2$. Therefore, $A^100=fracr_1^100-r_2^100r_1-r_2A+fracr_1r_2^100-r_2r_1^100r_1-r_2$.
– user578878
Jul 25 at 1:34