Mixing
Find a matrix given its 5th power
up vote
0
down vote
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Find a matrix $B$ such that $B^{5}$ is equal to $begin{bmatrix}
20 & 12\
19 & 13\
end{bmatrix}$.
I have tried this using the $2times2$ matrix $begin{bmatrix}
a & b\
c & d\
end{bmatrix}$ and multiplying it by itself $5$ times but that has left me with polynomials of very high degrees. Any hints on how to start this?
matrices
edited yesterday
Tianlalu
2,594632
asked yesterday
Robbie Meaney
658
add a comment |
up vote
0
down vote
favorite
Find a matrix $B$ such that $B^{5}$ is equal to $begin{bmatrix}
20 & 12\
19 & 13\
end{bmatrix}$.
I have tried this using the $2times2$ matrix $begin{bmatrix}
a & b\
c & d\
end{bmatrix}$ and multiplying it by itself $5$ times but that has left me with polynomials of very high degrees. Any hints on how to start this?
matrices
edited yesterday
Tianlalu
2,594632
asked yesterday
Robbie Meaney
658
How about computing the determinant of $B$ given the determinant of $B^5$? That gives you a constraint to work with. Not sure if that really helps though. Can't help but think that this involves diagonalisation...
– Sam Streeter
yesterday
so det(B)=ad-cb is equal to $det(B^{5})$?
– Robbie Meaney
yesterday
2
No. The determinant is a homomorphism, so $det(B^5) = det(B)^5$.
– user3482749
yesterday
It gives me that $ad-cb$=2 which is a nice constraint but still doesn't help me solve I don't think.
– Robbie Meaney
yesterday
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Find a matrix $B$ such that $B^{5}$ is equal to $begin{bmatrix}
20 & 12\
19 & 13\
end{bmatrix}$.
I have tried this using the $2times2$ matrix $begin{bmatrix}
a & b\
c & d\
end{bmatrix}$ and multiplying it by itself $5$ times but that has left me with polynomials of very high degrees. Any hints on how to start this?
matrices
edited yesterday
Tianlalu
2,594632
asked yesterday
Robbie Meaney
658
Find a matrix $B$ such that $B^{5}$ is equal to $begin{bmatrix}
20 & 12\
19 & 13\
end{bmatrix}$.
I have tried this using the $2times2$ matrix $begin{bmatrix}
a & b\
c & d\
end{bmatrix}$ and multiplying it by itself $5$ times but that has left me with polynomials of very high degrees. Any hints on how to start this?
matrices
matrices
edited yesterday
Tianlalu
2,594632
asked yesterday
Robbie Meaney
658
edited yesterday
Tianlalu
2,594632
asked yesterday
Robbie Meaney
658
edited yesterday
Tianlalu
2,594632
edited yesterday
Tianlalu
2,594632
edited yesterday
Tianlalu
2,594632
2,594632
asked yesterday
Robbie Meaney
658
asked yesterday
Robbie Meaney
658
asked yesterday
Robbie Meaney
658
658
How about computing the determinant of $B$ given the determinant of $B^5$? That gives you a constraint to work with. Not sure if that really helps though. Can't help but think that this involves diagonalisation...
– Sam Streeter
yesterday
so det(B)=ad-cb is equal to $det(B^{5})$?
– Robbie Meaney
yesterday
2
No. The determinant is a homomorphism, so $det(B^5) = det(B)^5$.
– user3482749
yesterday
It gives me that $ad-cb$=2 which is a nice constraint but still doesn't help me solve I don't think.
– Robbie Meaney
yesterday
add a comment |
How about computing the determinant of $B$ given the determinant of $B^5$? That gives you a constraint to work with. Not sure if that really helps though. Can't help but think that this involves diagonalisation...
– Sam Streeter
yesterday
so det(B)=ad-cb is equal to $det(B^{5})$?
– Robbie Meaney
yesterday
2
No. The determinant is a homomorphism, so $det(B^5) = det(B)^5$.
– user3482749
yesterday
It gives me that $ad-cb$=2 which is a nice constraint but still doesn't help me solve I don't think.
– Robbie Meaney
yesterday
How about computing the determinant of $B$ given the determinant of $B^5$? That gives you a constraint to work with. Not sure if that really helps though. Can't help but think that this involves diagonalisation...
– Sam Streeter
yesterday
How about computing the determinant of $B$ given the determinant of $B^5$? That gives you a constraint to work with. Not sure if that really helps though. Can't help but think that this involves diagonalisation...
– Sam Streeter
yesterday
so det(B)=ad-cb is equal to $det(B^{5})$?
– Robbie Meaney
yesterday
so det(B)=ad-cb is equal to $det(B^{5})$?
– Robbie Meaney
yesterday
2
2
No. The determinant is a homomorphism, so $det(B^5) = det(B)^5$.
– user3482749
yesterday
No. The determinant is a homomorphism, so $det(B^5) = det(B)^5$.
– user3482749
yesterday
It gives me that $ad-cb$=2 which is a nice constraint but still doesn't help me solve I don't think.
– Robbie Meaney
yesterday
It gives me that $ad-cb$=2 which is a nice constraint but still doesn't help me solve I don't think.
– Robbie Meaney
yesterday
add a comment |
2 Answers
2
active
oldest
votes
up vote
4
down vote
The eigenvalues of your matrix are $32$ and $1$. The vector $(1,1)$ is an eigenvector with eigenvalue $32$, whereas the vector $(-12,19)$ is an eigenvector with eigenvalue $1$. So, let$$P=begin{bmatrix}1&-12\1&19end{bmatrix}.$$Then$$P^{-1}.B^5.P=begin{bmatrix}32&0\0&1end{bmatrix}.$$So, take$$B=P.begin{bmatrix}2&0\0&1end{bmatrix}.P^{-1}=begin{bmatrix}frac{50}{31}&frac{12}{31}\frac{19}{31}&frac{43}{31}end{bmatrix}.$$
answered yesterday
José Carlos Santos
139k18111203
add a comment |
up vote
3
down vote
The characteristic polynomial is
$$
(x-20)(x-13)-12times 19 = x^2 - 33x +260 - 240 + 12= x^2 -33x + 32 = (x-32)(x-1),
$$
so $B^5$ is diagonalizable. Solve $(B^5 - 32I) x = 0$ gives a solution $(1,1)^{mathrm T}$, and solve $(B^5 - I)x = 0$ gives a solution $(12, -19)^{mathrm T}$, so there exists a matrix
$$
P = begin{bmatrix}
1 & 12 \ 1 & -19
end{bmatrix}
$$
such that
$$
P^{-1}B^5 P = mathrm {diag}(32,1).
$$
So we could take $B = Pmathrm {diag} (2, 1)P^{-1}$.
answered yesterday
xbh
4,9241421
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
The eigenvalues of your matrix are $32$ and $1$. The vector $(1,1)$ is an eigenvector with eigenvalue $32$, whereas the vector $(-12,19)$ is an eigenvector with eigenvalue $1$. So, let$$P=begin{bmatrix}1&-12\1&19end{bmatrix}.$$Then$$P^{-1}.B^5.P=begin{bmatrix}32&0\0&1end{bmatrix}.$$So, take$$B=P.begin{bmatrix}2&0\0&1end{bmatrix}.P^{-1}=begin{bmatrix}frac{50}{31}&frac{12}{31}\frac{19}{31}&frac{43}{31}end{bmatrix}.$$
answered yesterday
José Carlos Santos
139k18111203
add a comment |
up vote
4
down vote
The eigenvalues of your matrix are $32$ and $1$. The vector $(1,1)$ is an eigenvector with eigenvalue $32$, whereas the vector $(-12,19)$ is an eigenvector with eigenvalue $1$. So, let$$P=begin{bmatrix}1&-12\1&19end{bmatrix}.$$Then$$P^{-1}.B^5.P=begin{bmatrix}32&0\0&1end{bmatrix}.$$So, take$$B=P.begin{bmatrix}2&0\0&1end{bmatrix}.P^{-1}=begin{bmatrix}frac{50}{31}&frac{12}{31}\frac{19}{31}&frac{43}{31}end{bmatrix}.$$
answered yesterday
José Carlos Santos
139k18111203
add a comment |
up vote
4
down vote
up vote
4
down vote
The eigenvalues of your matrix are $32$ and $1$. The vector $(1,1)$ is an eigenvector with eigenvalue $32$, whereas the vector $(-12,19)$ is an eigenvector with eigenvalue $1$. So, let$$P=begin{bmatrix}1&-12\1&19end{bmatrix}.$$Then$$P^{-1}.B^5.P=begin{bmatrix}32&0\0&1end{bmatrix}.$$So, take$$B=P.begin{bmatrix}2&0\0&1end{bmatrix}.P^{-1}=begin{bmatrix}frac{50}{31}&frac{12}{31}\frac{19}{31}&frac{43}{31}end{bmatrix}.$$
answered yesterday
José Carlos Santos
139k18111203
The eigenvalues of your matrix are $32$ and $1$. The vector $(1,1)$ is an eigenvector with eigenvalue $32$, whereas the vector $(-12,19)$ is an eigenvector with eigenvalue $1$. So, let$$P=begin{bmatrix}1&-12\1&19end{bmatrix}.$$Then$$P^{-1}.B^5.P=begin{bmatrix}32&0\0&1end{bmatrix}.$$So, take$$B=P.begin{bmatrix}2&0\0&1end{bmatrix}.P^{-1}=begin{bmatrix}frac{50}{31}&frac{12}{31}\frac{19}{31}&frac{43}{31}end{bmatrix}.$$
answered yesterday
José Carlos Santos
139k18111203
edited yesterday
answered yesterday
José Carlos Santos
139k18111203
answered yesterday
José Carlos Santos
139k18111203
answered yesterday
José Carlos Santos
139k18111203
139k18111203
add a comment |
add a comment |
up vote
3
down vote
The characteristic polynomial is
$$
(x-20)(x-13)-12times 19 = x^2 - 33x +260 - 240 + 12= x^2 -33x + 32 = (x-32)(x-1),
$$
so $B^5$ is diagonalizable. Solve $(B^5 - 32I) x = 0$ gives a solution $(1,1)^{mathrm T}$, and solve $(B^5 - I)x = 0$ gives a solution $(12, -19)^{mathrm T}$, so there exists a matrix
$$
P = begin{bmatrix}
1 & 12 \ 1 & -19
end{bmatrix}
$$
such that
$$
P^{-1}B^5 P = mathrm {diag}(32,1).
$$
So we could take $B = Pmathrm {diag} (2, 1)P^{-1}$.
answered yesterday
xbh
4,9241421
add a comment |
up vote
3
down vote
The characteristic polynomial is
$$
(x-20)(x-13)-12times 19 = x^2 - 33x +260 - 240 + 12= x^2 -33x + 32 = (x-32)(x-1),
$$
so $B^5$ is diagonalizable. Solve $(B^5 - 32I) x = 0$ gives a solution $(1,1)^{mathrm T}$, and solve $(B^5 - I)x = 0$ gives a solution $(12, -19)^{mathrm T}$, so there exists a matrix
$$
P = begin{bmatrix}
1 & 12 \ 1 & -19
end{bmatrix}
$$
such that
$$
P^{-1}B^5 P = mathrm {diag}(32,1).
$$
So we could take $B = Pmathrm {diag} (2, 1)P^{-1}$.
answered yesterday
xbh
4,9241421
add a comment |
up vote
3
down vote
up vote
3
down vote
The characteristic polynomial is
$$
(x-20)(x-13)-12times 19 = x^2 - 33x +260 - 240 + 12= x^2 -33x + 32 = (x-32)(x-1),
$$
so $B^5$ is diagonalizable. Solve $(B^5 - 32I) x = 0$ gives a solution $(1,1)^{mathrm T}$, and solve $(B^5 - I)x = 0$ gives a solution $(12, -19)^{mathrm T}$, so there exists a matrix
$$
P = begin{bmatrix}
1 & 12 \ 1 & -19
end{bmatrix}
$$
such that
$$
P^{-1}B^5 P = mathrm {diag}(32,1).
$$
So we could take $B = Pmathrm {diag} (2, 1)P^{-1}$.
answered yesterday
xbh
4,9241421
The characteristic polynomial is
$$
(x-20)(x-13)-12times 19 = x^2 - 33x +260 - 240 + 12= x^2 -33x + 32 = (x-32)(x-1),
$$
so $B^5$ is diagonalizable. Solve $(B^5 - 32I) x = 0$ gives a solution $(1,1)^{mathrm T}$, and solve $(B^5 - I)x = 0$ gives a solution $(12, -19)^{mathrm T}$, so there exists a matrix
$$
P = begin{bmatrix}
1 & 12 \ 1 & -19
end{bmatrix}
$$
such that
$$
P^{-1}B^5 P = mathrm {diag}(32,1).
$$
So we could take $B = Pmathrm {diag} (2, 1)P^{-1}$.
answered yesterday
xbh
4,9241421
answered yesterday
xbh
4,9241421
answered yesterday
xbh
4,9241421
answered yesterday
xbh
4,9241421
4,9241421
add a comment |
add a comment |
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How about computing the determinant of $B$ given the determinant of $B^5$? That gives you a constraint to work with. Not sure if that really helps though. Can't help but think that this involves diagonalisation...
– Sam Streeter
yesterday
so det(B)=ad-cb is equal to $det(B^{5})$?
– Robbie Meaney
yesterday
2
No. The determinant is a homomorphism, so $det(B^5) = det(B)^5$.
– user3482749
yesterday
It gives me that $ad-cb$=2 which is a nice constraint but still doesn't help me solve I don't think.
– Robbie Meaney
yesterday