Find a matrix B such that $B^5 = A$ [duplicate]
$begingroup$
This question already has an answer here:
Find a matrix $B$ such that $B^3 = A$
4 answers
I am being asked to find a matrix $B$ where $B^5 = A$
$$A = begin{bmatrix} 1 & 3 \ 3 & 1 end{bmatrix}$$
In the first part of the question I was asked to find the eigenvalues & eigenvectors for the matrix which I found successfully.
If someone could help me finish this then that would be great.
matrices eigenvalues-eigenvectors matrix-calculus
edited Dec 5 '18 at 22:18
Robert Lewis
45.6k23065
asked Dec 5 '18 at 21:32
Ray FitzgeraldRay Fitzgerald
1112
$endgroup$
marked as duplicate by Math Lover, amd, Jean-Claude Arbaut, Chinnapparaj R, KReiser Dec 6 '18 at 3:11
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.
add a comment |
$begingroup$
This question already has an answer here:
Find a matrix $B$ such that $B^3 = A$
4 answers
I am being asked to find a matrix $B$ where $B^5 = A$
$$A = begin{bmatrix} 1 & 3 \ 3 & 1 end{bmatrix}$$
In the first part of the question I was asked to find the eigenvalues & eigenvectors for the matrix which I found successfully.
If someone could help me finish this then that would be great.
matrices eigenvalues-eigenvectors matrix-calculus
edited Dec 5 '18 at 22:18
Robert Lewis
45.6k23065
asked Dec 5 '18 at 21:32
Ray FitzgeraldRay Fitzgerald
1112
$endgroup$
marked as duplicate by Math Lover, amd, Jean-Claude Arbaut, Chinnapparaj R, KReiser Dec 6 '18 at 3:11
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.
$begingroup$
So what did you find? For the eigenvalues +vectors?
$endgroup$
– Mason
Dec 5 '18 at 21:47
1
$begingroup$
I edited your post to more properly $LaTeX$ify it. Cheers!
$endgroup$
– Robert Lewis
Dec 5 '18 at 22:18
$begingroup$
@Mason i got λ=4,λ=-2 for eigenvals. Corresponding to $$A = begin{bmatrix} 1 \ 1 end{bmatrix}$$ and $$A = begin{bmatrix} -1 \ 1 end{bmatrix}$$
$endgroup$
– Ray Fitzgerald
Dec 13 '18 at 18:52
add a comment |
$begingroup$
This question already has an answer here:
Find a matrix $B$ such that $B^3 = A$
4 answers
I am being asked to find a matrix $B$ where $B^5 = A$
$$A = begin{bmatrix} 1 & 3 \ 3 & 1 end{bmatrix}$$
In the first part of the question I was asked to find the eigenvalues & eigenvectors for the matrix which I found successfully.
If someone could help me finish this then that would be great.
matrices eigenvalues-eigenvectors matrix-calculus
edited Dec 5 '18 at 22:18
Robert Lewis
45.6k23065
asked Dec 5 '18 at 21:32
Ray FitzgeraldRay Fitzgerald
1112
$endgroup$
This question already has an answer here:
Find a matrix $B$ such that $B^3 = A$
4 answers
I am being asked to find a matrix $B$ where $B^5 = A$
$$A = begin{bmatrix} 1 & 3 \ 3 & 1 end{bmatrix}$$
In the first part of the question I was asked to find the eigenvalues & eigenvectors for the matrix which I found successfully.
If someone could help me finish this then that would be great.
This question already has an answer here:
Find a matrix $B$ such that $B^3 = A$
4 answers
matrices eigenvalues-eigenvectors matrix-calculus
matrices eigenvalues-eigenvectors matrix-calculus
edited Dec 5 '18 at 22:18
Robert Lewis
45.6k23065
asked Dec 5 '18 at 21:32
Ray FitzgeraldRay Fitzgerald
1112
edited Dec 5 '18 at 22:18
Robert Lewis
45.6k23065
asked Dec 5 '18 at 21:32
Ray FitzgeraldRay Fitzgerald
1112
edited Dec 5 '18 at 22:18
Robert Lewis
45.6k23065
edited Dec 5 '18 at 22:18
Robert Lewis
45.6k23065
edited Dec 5 '18 at 22:18
Robert Lewis
45.6k23065
45.6k23065
asked Dec 5 '18 at 21:32
Ray FitzgeraldRay Fitzgerald
1112
asked Dec 5 '18 at 21:32
Ray FitzgeraldRay Fitzgerald
1112
asked Dec 5 '18 at 21:32
Ray FitzgeraldRay Fitzgerald
1112
1112
marked as duplicate by Math Lover, amd, Jean-Claude Arbaut, Chinnapparaj R, KReiser Dec 6 '18 at 3:11
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 Math Lover, amd, Jean-Claude Arbaut, Chinnapparaj R, KReiser Dec 6 '18 at 3:11
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.
$begingroup$
So what did you find? For the eigenvalues +vectors?
$endgroup$
– Mason
Dec 5 '18 at 21:47
1
$begingroup$
I edited your post to more properly $LaTeX$ify it. Cheers!
$endgroup$
– Robert Lewis
Dec 5 '18 at 22:18
$begingroup$
@Mason i got λ=4,λ=-2 for eigenvals. Corresponding to $$A = begin{bmatrix} 1 \ 1 end{bmatrix}$$ and $$A = begin{bmatrix} -1 \ 1 end{bmatrix}$$
$endgroup$
– Ray Fitzgerald
Dec 13 '18 at 18:52
add a comment |
$begingroup$
So what did you find? For the eigenvalues +vectors?
$endgroup$
– Mason
Dec 5 '18 at 21:47
1
$begingroup$
I edited your post to more properly $LaTeX$ify it. Cheers!
$endgroup$
– Robert Lewis
Dec 5 '18 at 22:18
$begingroup$
@Mason i got λ=4,λ=-2 for eigenvals. Corresponding to $$A = begin{bmatrix} 1 \ 1 end{bmatrix}$$ and $$A = begin{bmatrix} -1 \ 1 end{bmatrix}$$
$endgroup$
– Ray Fitzgerald
Dec 13 '18 at 18:52
$begingroup$
So what did you find? For the eigenvalues +vectors?
$endgroup$
– Mason
Dec 5 '18 at 21:47
$begingroup$
So what did you find? For the eigenvalues +vectors?
$endgroup$
– Mason
Dec 5 '18 at 21:47
1
1
$begingroup$
I edited your post to more properly $LaTeX$ify it. Cheers!
$endgroup$
– Robert Lewis
Dec 5 '18 at 22:18
$begingroup$
I edited your post to more properly $LaTeX$ify it. Cheers!
$endgroup$
– Robert Lewis
Dec 5 '18 at 22:18
$begingroup$
@Mason i got λ=4,λ=-2 for eigenvals. Corresponding to $$A = begin{bmatrix} 1 \ 1 end{bmatrix}$$ and $$A = begin{bmatrix} -1 \ 1 end{bmatrix}$$
$endgroup$
– Ray Fitzgerald
Dec 13 '18 at 18:52
$begingroup$
@Mason i got λ=4,λ=-2 for eigenvals. Corresponding to $$A = begin{bmatrix} 1 \ 1 end{bmatrix}$$ and $$A = begin{bmatrix} -1 \ 1 end{bmatrix}$$
$endgroup$
– Ray Fitzgerald
Dec 13 '18 at 18:52
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
HINT: If you have found the eigenvalues and eigenvectors, then you should easily be able to diagonalize this matrix as $A=PDP^{-1}$, where $D$ is diagonal. Then use the fact that $A^n=PD^n P^{-1}$, and the fact that the powers of a diagonal matrix are the matrices consisting of the powers of its entries.
answered Dec 5 '18 at 21:36
FrpzzdFrpzzd
22.9k841109
$endgroup$
3
$begingroup$
Maybe, use $D$ instead of $B$ to avoid the confusion as the OP is asking to find $B$ such that $B^5 = A$. Also, this problem could be considered as a duplicate, I think.
$endgroup$
– Math Lover
Dec 5 '18 at 21:39
$begingroup$
@MathLover Oh, sorry. I will change it to avoid confusion.
$endgroup$
– Frpzzd
Dec 5 '18 at 21:41
$begingroup$
@Frpzzd any chance you could expand on the latter part of that suggestion. I know how to do matrix diagonalisation but what do you mean by "Then use the fact that An=PDnP−1, and the fact that the powers of a diagonal matrix are the matrices consisting of the powers of its entries." Thanks
$endgroup$
– Ray Fitzgerald
Dec 13 '18 at 18:56
add a comment |
$begingroup$
We can also use projector decomposition. The left and right eigenvectors are equal, because of the symmetry. So the projector decomposition is
$$A=lambda_1 frac{v_1 circ v_1}{v_1 cdot v_1}+lambda_2 frac{v_2 circ v_2}{v_2 cdot v_2}$$
And
$$f(A)=f(lambda_1) frac{v_1 circ v_1}{v_1 cdot v_1}+f(lambda_2) frac{v_2 circ v_2}{v_2 cdot v_2}$$
Where $(a circ b)_{ij}=a_i b_j$.
answered Dec 5 '18 at 21:53
BotondBotond
5,6832732
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
HINT: If you have found the eigenvalues and eigenvectors, then you should easily be able to diagonalize this matrix as $A=PDP^{-1}$, where $D$ is diagonal. Then use the fact that $A^n=PD^n P^{-1}$, and the fact that the powers of a diagonal matrix are the matrices consisting of the powers of its entries.
answered Dec 5 '18 at 21:36
FrpzzdFrpzzd
22.9k841109
$endgroup$
3
$begingroup$
Maybe, use $D$ instead of $B$ to avoid the confusion as the OP is asking to find $B$ such that $B^5 = A$. Also, this problem could be considered as a duplicate, I think.
$endgroup$
– Math Lover
Dec 5 '18 at 21:39
$begingroup$
@MathLover Oh, sorry. I will change it to avoid confusion.
$endgroup$
– Frpzzd
Dec 5 '18 at 21:41
$begingroup$
@Frpzzd any chance you could expand on the latter part of that suggestion. I know how to do matrix diagonalisation but what do you mean by "Then use the fact that An=PDnP−1, and the fact that the powers of a diagonal matrix are the matrices consisting of the powers of its entries." Thanks
$endgroup$
– Ray Fitzgerald
Dec 13 '18 at 18:56
add a comment |
$begingroup$
HINT: If you have found the eigenvalues and eigenvectors, then you should easily be able to diagonalize this matrix as $A=PDP^{-1}$, where $D$ is diagonal. Then use the fact that $A^n=PD^n P^{-1}$, and the fact that the powers of a diagonal matrix are the matrices consisting of the powers of its entries.
answered Dec 5 '18 at 21:36
FrpzzdFrpzzd
22.9k841109
$endgroup$
3
$begingroup$
Maybe, use $D$ instead of $B$ to avoid the confusion as the OP is asking to find $B$ such that $B^5 = A$. Also, this problem could be considered as a duplicate, I think.
$endgroup$
– Math Lover
Dec 5 '18 at 21:39
$begingroup$
@MathLover Oh, sorry. I will change it to avoid confusion.
$endgroup$
– Frpzzd
Dec 5 '18 at 21:41
$begingroup$
@Frpzzd any chance you could expand on the latter part of that suggestion. I know how to do matrix diagonalisation but what do you mean by "Then use the fact that An=PDnP−1, and the fact that the powers of a diagonal matrix are the matrices consisting of the powers of its entries." Thanks
$endgroup$
– Ray Fitzgerald
Dec 13 '18 at 18:56
add a comment |
$begingroup$
HINT: If you have found the eigenvalues and eigenvectors, then you should easily be able to diagonalize this matrix as $A=PDP^{-1}$, where $D$ is diagonal. Then use the fact that $A^n=PD^n P^{-1}$, and the fact that the powers of a diagonal matrix are the matrices consisting of the powers of its entries.
answered Dec 5 '18 at 21:36
FrpzzdFrpzzd
22.9k841109
$endgroup$
HINT: If you have found the eigenvalues and eigenvectors, then you should easily be able to diagonalize this matrix as $A=PDP^{-1}$, where $D$ is diagonal. Then use the fact that $A^n=PD^n P^{-1}$, and the fact that the powers of a diagonal matrix are the matrices consisting of the powers of its entries.
answered Dec 5 '18 at 21:36
FrpzzdFrpzzd
22.9k841109
edited Dec 5 '18 at 21:41
answered Dec 5 '18 at 21:36
FrpzzdFrpzzd
22.9k841109
answered Dec 5 '18 at 21:36
FrpzzdFrpzzd
22.9k841109
answered Dec 5 '18 at 21:36
FrpzzdFrpzzd
22.9k841109
22.9k841109
3
$begingroup$
Maybe, use $D$ instead of $B$ to avoid the confusion as the OP is asking to find $B$ such that $B^5 = A$. Also, this problem could be considered as a duplicate, I think.
$endgroup$
– Math Lover
Dec 5 '18 at 21:39
$begingroup$
@MathLover Oh, sorry. I will change it to avoid confusion.
$endgroup$
– Frpzzd
Dec 5 '18 at 21:41
$begingroup$
@Frpzzd any chance you could expand on the latter part of that suggestion. I know how to do matrix diagonalisation but what do you mean by "Then use the fact that An=PDnP−1, and the fact that the powers of a diagonal matrix are the matrices consisting of the powers of its entries." Thanks
$endgroup$
– Ray Fitzgerald
Dec 13 '18 at 18:56
add a comment |
3
$begingroup$
Maybe, use $D$ instead of $B$ to avoid the confusion as the OP is asking to find $B$ such that $B^5 = A$. Also, this problem could be considered as a duplicate, I think.
$endgroup$
– Math Lover
Dec 5 '18 at 21:39
$begingroup$
@MathLover Oh, sorry. I will change it to avoid confusion.
$endgroup$
– Frpzzd
Dec 5 '18 at 21:41
$begingroup$
@Frpzzd any chance you could expand on the latter part of that suggestion. I know how to do matrix diagonalisation but what do you mean by "Then use the fact that An=PDnP−1, and the fact that the powers of a diagonal matrix are the matrices consisting of the powers of its entries." Thanks
$endgroup$
– Ray Fitzgerald
Dec 13 '18 at 18:56
3
3
$begingroup$
Maybe, use $D$ instead of $B$ to avoid the confusion as the OP is asking to find $B$ such that $B^5 = A$. Also, this problem could be considered as a duplicate, I think.
$endgroup$
– Math Lover
Dec 5 '18 at 21:39
$begingroup$
Maybe, use $D$ instead of $B$ to avoid the confusion as the OP is asking to find $B$ such that $B^5 = A$. Also, this problem could be considered as a duplicate, I think.
$endgroup$
– Math Lover
Dec 5 '18 at 21:39
$begingroup$
@MathLover Oh, sorry. I will change it to avoid confusion.
$endgroup$
– Frpzzd
Dec 5 '18 at 21:41
$begingroup$
@MathLover Oh, sorry. I will change it to avoid confusion.
$endgroup$
– Frpzzd
Dec 5 '18 at 21:41
$begingroup$
@Frpzzd any chance you could expand on the latter part of that suggestion. I know how to do matrix diagonalisation but what do you mean by "Then use the fact that An=PDnP−1, and the fact that the powers of a diagonal matrix are the matrices consisting of the powers of its entries." Thanks
$endgroup$
– Ray Fitzgerald
Dec 13 '18 at 18:56
$begingroup$
@Frpzzd any chance you could expand on the latter part of that suggestion. I know how to do matrix diagonalisation but what do you mean by "Then use the fact that An=PDnP−1, and the fact that the powers of a diagonal matrix are the matrices consisting of the powers of its entries." Thanks
$endgroup$
– Ray Fitzgerald
Dec 13 '18 at 18:56
add a comment |
$begingroup$
We can also use projector decomposition. The left and right eigenvectors are equal, because of the symmetry. So the projector decomposition is
$$A=lambda_1 frac{v_1 circ v_1}{v_1 cdot v_1}+lambda_2 frac{v_2 circ v_2}{v_2 cdot v_2}$$
And
$$f(A)=f(lambda_1) frac{v_1 circ v_1}{v_1 cdot v_1}+f(lambda_2) frac{v_2 circ v_2}{v_2 cdot v_2}$$
Where $(a circ b)_{ij}=a_i b_j$.
answered Dec 5 '18 at 21:53
BotondBotond
5,6832732
$endgroup$
add a comment |
$begingroup$
We can also use projector decomposition. The left and right eigenvectors are equal, because of the symmetry. So the projector decomposition is
$$A=lambda_1 frac{v_1 circ v_1}{v_1 cdot v_1}+lambda_2 frac{v_2 circ v_2}{v_2 cdot v_2}$$
And
$$f(A)=f(lambda_1) frac{v_1 circ v_1}{v_1 cdot v_1}+f(lambda_2) frac{v_2 circ v_2}{v_2 cdot v_2}$$
Where $(a circ b)_{ij}=a_i b_j$.
answered Dec 5 '18 at 21:53
BotondBotond
5,6832732
$endgroup$
add a comment |
$begingroup$
We can also use projector decomposition. The left and right eigenvectors are equal, because of the symmetry. So the projector decomposition is
$$A=lambda_1 frac{v_1 circ v_1}{v_1 cdot v_1}+lambda_2 frac{v_2 circ v_2}{v_2 cdot v_2}$$
And
$$f(A)=f(lambda_1) frac{v_1 circ v_1}{v_1 cdot v_1}+f(lambda_2) frac{v_2 circ v_2}{v_2 cdot v_2}$$
Where $(a circ b)_{ij}=a_i b_j$.
answered Dec 5 '18 at 21:53
BotondBotond
5,6832732
$endgroup$
We can also use projector decomposition. The left and right eigenvectors are equal, because of the symmetry. So the projector decomposition is
$$A=lambda_1 frac{v_1 circ v_1}{v_1 cdot v_1}+lambda_2 frac{v_2 circ v_2}{v_2 cdot v_2}$$
And
$$f(A)=f(lambda_1) frac{v_1 circ v_1}{v_1 cdot v_1}+f(lambda_2) frac{v_2 circ v_2}{v_2 cdot v_2}$$
Where $(a circ b)_{ij}=a_i b_j$.
answered Dec 5 '18 at 21:53
BotondBotond
5,6832732
edited Dec 5 '18 at 22:26
answered Dec 5 '18 at 21:53
BotondBotond
5,6832732
answered Dec 5 '18 at 21:53
BotondBotond
5,6832732
answered Dec 5 '18 at 21:53
BotondBotond
5,6832732
5,6832732
add a comment |
add a comment |
$begingroup$
So what did you find? For the eigenvalues +vectors?
$endgroup$
– Mason
Dec 5 '18 at 21:47
1
$begingroup$
I edited your post to more properly $LaTeX$ify it. Cheers!
$endgroup$
– Robert Lewis
Dec 5 '18 at 22:18
$begingroup$
@Mason i got λ=4,λ=-2 for eigenvals. Corresponding to $$A = begin{bmatrix} 1 \ 1 end{bmatrix}$$ and $$A = begin{bmatrix} -1 \ 1 end{bmatrix}$$
$endgroup$
– Ray Fitzgerald
Dec 13 '18 at 18:52