How to find the matrix represented by the polynomials $A^{12}-5A^{11}+…+3I$?
I need to find the characteristic equation of the matrix $ A = begin{bmatrix}
2&1&1\
0&1&0\
1&1&2\
end{bmatrix} $ and find the matrix represented by $ A^{12}-5A^{11}+7A^{10}-3A^{9}-A^8+5A^7-7A^6+3A^5+A^4-5A^3+8A^2-4A+3I $
My attempt:
I started by finding the character equation which was $ lambda^3-5lambda^2+7lambda-3=0 $ where $lambda$ is the eigenvalue, and found $lambda = 5,frac{9+sqrt{69}}{2},frac{9-sqrt{69}}{2} $
$lambda = 5$ satisfies the equation and I would take $lambda = 5$ let $lambda = A$ and $ A^3-5 A^2+7A -3=0 $.
I don't know how to proceed further. All help would be appreciated.
Edit: What I can do, is multiply the matrix $A$ by itself and consequently the resultant matrix would be $A^2$ and so on. However, I'm not sure if this will give an accurate answer.
linear-algebra matrices
asked Dec 9 '18 at 21:30
tNotr
92
add a comment |
I need to find the characteristic equation of the matrix $ A = begin{bmatrix}
2&1&1\
0&1&0\
1&1&2\
end{bmatrix} $ and find the matrix represented by $ A^{12}-5A^{11}+7A^{10}-3A^{9}-A^8+5A^7-7A^6+3A^5+A^4-5A^3+8A^2-4A+3I $
My attempt:
I started by finding the character equation which was $ lambda^3-5lambda^2+7lambda-3=0 $ where $lambda$ is the eigenvalue, and found $lambda = 5,frac{9+sqrt{69}}{2},frac{9-sqrt{69}}{2} $
$lambda = 5$ satisfies the equation and I would take $lambda = 5$ let $lambda = A$ and $ A^3-5 A^2+7A -3=0 $.
I don't know how to proceed further. All help would be appreciated.
Edit: What I can do, is multiply the matrix $A$ by itself and consequently the resultant matrix would be $A^2$ and so on. However, I'm not sure if this will give an accurate answer.
linear-algebra matrices
asked Dec 9 '18 at 21:30
tNotr
92
1
I suspect the characteristic equation is $lambda^3 - 5lambda^2 + 7lambda - 3 = 0$. You might want to double check your work on that part.
– JimmyK4542
Dec 9 '18 at 21:43
Yes, you're right about the characteristic equation. I took a different problem's matrix for this one. My bad.
– tNotr
Dec 9 '18 at 21:48
add a comment |
I need to find the characteristic equation of the matrix $ A = begin{bmatrix}
2&1&1\
0&1&0\
1&1&2\
end{bmatrix} $ and find the matrix represented by $ A^{12}-5A^{11}+7A^{10}-3A^{9}-A^8+5A^7-7A^6+3A^5+A^4-5A^3+8A^2-4A+3I $
My attempt:
I started by finding the character equation which was $ lambda^3-5lambda^2+7lambda-3=0 $ where $lambda$ is the eigenvalue, and found $lambda = 5,frac{9+sqrt{69}}{2},frac{9-sqrt{69}}{2} $
$lambda = 5$ satisfies the equation and I would take $lambda = 5$ let $lambda = A$ and $ A^3-5 A^2+7A -3=0 $.
I don't know how to proceed further. All help would be appreciated.
Edit: What I can do, is multiply the matrix $A$ by itself and consequently the resultant matrix would be $A^2$ and so on. However, I'm not sure if this will give an accurate answer.
linear-algebra matrices
asked Dec 9 '18 at 21:30
tNotr
92
I need to find the characteristic equation of the matrix $ A = begin{bmatrix}
2&1&1\
0&1&0\
1&1&2\
end{bmatrix} $ and find the matrix represented by $ A^{12}-5A^{11}+7A^{10}-3A^{9}-A^8+5A^7-7A^6+3A^5+A^4-5A^3+8A^2-4A+3I $
My attempt:
I started by finding the character equation which was $ lambda^3-5lambda^2+7lambda-3=0 $ where $lambda$ is the eigenvalue, and found $lambda = 5,frac{9+sqrt{69}}{2},frac{9-sqrt{69}}{2} $
$lambda = 5$ satisfies the equation and I would take $lambda = 5$ let $lambda = A$ and $ A^3-5 A^2+7A -3=0 $.
I don't know how to proceed further. All help would be appreciated.
Edit: What I can do, is multiply the matrix $A$ by itself and consequently the resultant matrix would be $A^2$ and so on. However, I'm not sure if this will give an accurate answer.
linear-algebra matrices
linear-algebra matrices
asked Dec 9 '18 at 21:30
tNotr
92
asked Dec 9 '18 at 21:30
tNotr
92
edited Dec 9 '18 at 21:59
asked Dec 9 '18 at 21:30
tNotr
92
asked Dec 9 '18 at 21:30
tNotr
92
asked Dec 9 '18 at 21:30
tNotr
92
92
1
I suspect the characteristic equation is $lambda^3 - 5lambda^2 + 7lambda - 3 = 0$. You might want to double check your work on that part.
– JimmyK4542
Dec 9 '18 at 21:43
Yes, you're right about the characteristic equation. I took a different problem's matrix for this one. My bad.
– tNotr
Dec 9 '18 at 21:48
add a comment |
1
I suspect the characteristic equation is $lambda^3 - 5lambda^2 + 7lambda - 3 = 0$. You might want to double check your work on that part.
– JimmyK4542
Dec 9 '18 at 21:43
Yes, you're right about the characteristic equation. I took a different problem's matrix for this one. My bad.
– tNotr
Dec 9 '18 at 21:48
1
1
I suspect the characteristic equation is $lambda^3 - 5lambda^2 + 7lambda - 3 = 0$. You might want to double check your work on that part.
– JimmyK4542
Dec 9 '18 at 21:43
I suspect the characteristic equation is $lambda^3 - 5lambda^2 + 7lambda - 3 = 0$. You might want to double check your work on that part.
– JimmyK4542
Dec 9 '18 at 21:43
Yes, you're right about the characteristic equation. I took a different problem's matrix for this one. My bad.
– tNotr
Dec 9 '18 at 21:48
Yes, you're right about the characteristic equation. I took a different problem's matrix for this one. My bad.
– tNotr
Dec 9 '18 at 21:48
add a comment |
2 Answers
2
active
oldest
votes
The characteristic polynomial of $A$ is $p(lambda) = det(lambda I - A) = lambda^3-5lambda^2+7lambda-3$.
So by the Cayley-Hamilton theorem $p(A) = A^3-5A^2+7A-3I = 0$ (*).
Multiplying (*) by $A^9$ yields $A^{12}-5A^{11}+7A^{10}-3A^9 = 0$ (1).
Multiplying (*) by $-A^5$ yields $-A^8+5A^7-7A^6+3A^5 = 0$ (2).
Multiplying (*) by $A$ yields $A^4-5A^3+7A^2-3A = 0$ (3).
If we add (1), (2), and (3), we get $$A^{12}-5A^{11}+7A^{10}-3A^9-A^8+5A^7-7A^6+3A^5+A^4-5A^3+7A^2-3A = 0$$
Can you take it from here?
answered Dec 9 '18 at 21:39
JimmyK4542
40.5k245105
I find ${}+8lambda$ for the characteristic polynomial.
– Bernard
Dec 9 '18 at 21:53
yes absolutely! However, I am not allowed to multiply the matrix A by any matrix greater than $A^3$ in the exam for some reason. Is this an attempt to baffle the student into making a mistake? I will never know but nonetheless I will try your solution though regardless of the risks.
– tNotr
Dec 9 '18 at 21:55
You don't have to actually compute $A^3$ or any other higher power of $A$. Just add $A^2-A+3I$ to both sides of the last equation I wrote. Then, the left side is the expression you are looking for, and the right side is something that should be easy enough to compute.
– JimmyK4542
Dec 9 '18 at 22:25
add a comment |
Check you computation for the characteristic polynomial $chi_A$.
Hint:
Divide the polynomial $p(x)=x^{12}-5x^{11}+dots+3$ by the characteristic polynomial:
$$p(x)=q(x)chi_A(x)+r(x)qquad (deg r le 2),$$ to get
$$p(A)=q(A)chi_A(A)+r(A)=r(A).$$:
answered Dec 9 '18 at 22:00
Bernard
118k639112
add a comment |
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2 Answers
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2 Answers
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The characteristic polynomial of $A$ is $p(lambda) = det(lambda I - A) = lambda^3-5lambda^2+7lambda-3$.
So by the Cayley-Hamilton theorem $p(A) = A^3-5A^2+7A-3I = 0$ (*).
Multiplying (*) by $A^9$ yields $A^{12}-5A^{11}+7A^{10}-3A^9 = 0$ (1).
Multiplying (*) by $-A^5$ yields $-A^8+5A^7-7A^6+3A^5 = 0$ (2).
Multiplying (*) by $A$ yields $A^4-5A^3+7A^2-3A = 0$ (3).
If we add (1), (2), and (3), we get $$A^{12}-5A^{11}+7A^{10}-3A^9-A^8+5A^7-7A^6+3A^5+A^4-5A^3+7A^2-3A = 0$$
Can you take it from here?
answered Dec 9 '18 at 21:39
JimmyK4542
40.5k245105
I find ${}+8lambda$ for the characteristic polynomial.
– Bernard
Dec 9 '18 at 21:53
yes absolutely! However, I am not allowed to multiply the matrix A by any matrix greater than $A^3$ in the exam for some reason. Is this an attempt to baffle the student into making a mistake? I will never know but nonetheless I will try your solution though regardless of the risks.
– tNotr
Dec 9 '18 at 21:55
You don't have to actually compute $A^3$ or any other higher power of $A$. Just add $A^2-A+3I$ to both sides of the last equation I wrote. Then, the left side is the expression you are looking for, and the right side is something that should be easy enough to compute.
– JimmyK4542
Dec 9 '18 at 22:25
add a comment |
The characteristic polynomial of $A$ is $p(lambda) = det(lambda I - A) = lambda^3-5lambda^2+7lambda-3$.
So by the Cayley-Hamilton theorem $p(A) = A^3-5A^2+7A-3I = 0$ (*).
Multiplying (*) by $A^9$ yields $A^{12}-5A^{11}+7A^{10}-3A^9 = 0$ (1).
Multiplying (*) by $-A^5$ yields $-A^8+5A^7-7A^6+3A^5 = 0$ (2).
Multiplying (*) by $A$ yields $A^4-5A^3+7A^2-3A = 0$ (3).
If we add (1), (2), and (3), we get $$A^{12}-5A^{11}+7A^{10}-3A^9-A^8+5A^7-7A^6+3A^5+A^4-5A^3+7A^2-3A = 0$$
Can you take it from here?
answered Dec 9 '18 at 21:39
JimmyK4542
40.5k245105
I find ${}+8lambda$ for the characteristic polynomial.
– Bernard
Dec 9 '18 at 21:53
yes absolutely! However, I am not allowed to multiply the matrix A by any matrix greater than $A^3$ in the exam for some reason. Is this an attempt to baffle the student into making a mistake? I will never know but nonetheless I will try your solution though regardless of the risks.
– tNotr
Dec 9 '18 at 21:55
You don't have to actually compute $A^3$ or any other higher power of $A$. Just add $A^2-A+3I$ to both sides of the last equation I wrote. Then, the left side is the expression you are looking for, and the right side is something that should be easy enough to compute.
– JimmyK4542
Dec 9 '18 at 22:25
add a comment |
The characteristic polynomial of $A$ is $p(lambda) = det(lambda I - A) = lambda^3-5lambda^2+7lambda-3$.
So by the Cayley-Hamilton theorem $p(A) = A^3-5A^2+7A-3I = 0$ (*).
Multiplying (*) by $A^9$ yields $A^{12}-5A^{11}+7A^{10}-3A^9 = 0$ (1).
Multiplying (*) by $-A^5$ yields $-A^8+5A^7-7A^6+3A^5 = 0$ (2).
Multiplying (*) by $A$ yields $A^4-5A^3+7A^2-3A = 0$ (3).
If we add (1), (2), and (3), we get $$A^{12}-5A^{11}+7A^{10}-3A^9-A^8+5A^7-7A^6+3A^5+A^4-5A^3+7A^2-3A = 0$$
Can you take it from here?
answered Dec 9 '18 at 21:39
JimmyK4542
40.5k245105
The characteristic polynomial of $A$ is $p(lambda) = det(lambda I - A) = lambda^3-5lambda^2+7lambda-3$.
So by the Cayley-Hamilton theorem $p(A) = A^3-5A^2+7A-3I = 0$ (*).
Multiplying (*) by $A^9$ yields $A^{12}-5A^{11}+7A^{10}-3A^9 = 0$ (1).
Multiplying (*) by $-A^5$ yields $-A^8+5A^7-7A^6+3A^5 = 0$ (2).
Multiplying (*) by $A$ yields $A^4-5A^3+7A^2-3A = 0$ (3).
If we add (1), (2), and (3), we get $$A^{12}-5A^{11}+7A^{10}-3A^9-A^8+5A^7-7A^6+3A^5+A^4-5A^3+7A^2-3A = 0$$
Can you take it from here?
answered Dec 9 '18 at 21:39
JimmyK4542
40.5k245105
answered Dec 9 '18 at 21:39
JimmyK4542
40.5k245105
answered Dec 9 '18 at 21:39
JimmyK4542
40.5k245105
answered Dec 9 '18 at 21:39
JimmyK4542
40.5k245105
40.5k245105
I find ${}+8lambda$ for the characteristic polynomial.
– Bernard
Dec 9 '18 at 21:53
yes absolutely! However, I am not allowed to multiply the matrix A by any matrix greater than $A^3$ in the exam for some reason. Is this an attempt to baffle the student into making a mistake? I will never know but nonetheless I will try your solution though regardless of the risks.
– tNotr
Dec 9 '18 at 21:55
You don't have to actually compute $A^3$ or any other higher power of $A$. Just add $A^2-A+3I$ to both sides of the last equation I wrote. Then, the left side is the expression you are looking for, and the right side is something that should be easy enough to compute.
– JimmyK4542
Dec 9 '18 at 22:25
add a comment |
I find ${}+8lambda$ for the characteristic polynomial.
– Bernard
Dec 9 '18 at 21:53
yes absolutely! However, I am not allowed to multiply the matrix A by any matrix greater than $A^3$ in the exam for some reason. Is this an attempt to baffle the student into making a mistake? I will never know but nonetheless I will try your solution though regardless of the risks.
– tNotr
Dec 9 '18 at 21:55
You don't have to actually compute $A^3$ or any other higher power of $A$. Just add $A^2-A+3I$ to both sides of the last equation I wrote. Then, the left side is the expression you are looking for, and the right side is something that should be easy enough to compute.
– JimmyK4542
Dec 9 '18 at 22:25
I find ${}+8lambda$ for the characteristic polynomial.
– Bernard
Dec 9 '18 at 21:53
I find ${}+8lambda$ for the characteristic polynomial.
– Bernard
Dec 9 '18 at 21:53
yes absolutely! However, I am not allowed to multiply the matrix A by any matrix greater than $A^3$ in the exam for some reason. Is this an attempt to baffle the student into making a mistake? I will never know but nonetheless I will try your solution though regardless of the risks.
– tNotr
Dec 9 '18 at 21:55
yes absolutely! However, I am not allowed to multiply the matrix A by any matrix greater than $A^3$ in the exam for some reason. Is this an attempt to baffle the student into making a mistake? I will never know but nonetheless I will try your solution though regardless of the risks.
– tNotr
Dec 9 '18 at 21:55
You don't have to actually compute $A^3$ or any other higher power of $A$. Just add $A^2-A+3I$ to both sides of the last equation I wrote. Then, the left side is the expression you are looking for, and the right side is something that should be easy enough to compute.
– JimmyK4542
Dec 9 '18 at 22:25
You don't have to actually compute $A^3$ or any other higher power of $A$. Just add $A^2-A+3I$ to both sides of the last equation I wrote. Then, the left side is the expression you are looking for, and the right side is something that should be easy enough to compute.
– JimmyK4542
Dec 9 '18 at 22:25
add a comment |
Check you computation for the characteristic polynomial $chi_A$.
Hint:
Divide the polynomial $p(x)=x^{12}-5x^{11}+dots+3$ by the characteristic polynomial:
$$p(x)=q(x)chi_A(x)+r(x)qquad (deg r le 2),$$ to get
$$p(A)=q(A)chi_A(A)+r(A)=r(A).$$:
answered Dec 9 '18 at 22:00
Bernard
118k639112
add a comment |
Check you computation for the characteristic polynomial $chi_A$.
Hint:
Divide the polynomial $p(x)=x^{12}-5x^{11}+dots+3$ by the characteristic polynomial:
$$p(x)=q(x)chi_A(x)+r(x)qquad (deg r le 2),$$ to get
$$p(A)=q(A)chi_A(A)+r(A)=r(A).$$:
answered Dec 9 '18 at 22:00
Bernard
118k639112
add a comment |
Check you computation for the characteristic polynomial $chi_A$.
Hint:
Divide the polynomial $p(x)=x^{12}-5x^{11}+dots+3$ by the characteristic polynomial:
$$p(x)=q(x)chi_A(x)+r(x)qquad (deg r le 2),$$ to get
$$p(A)=q(A)chi_A(A)+r(A)=r(A).$$:
answered Dec 9 '18 at 22:00
Bernard
118k639112
Check you computation for the characteristic polynomial $chi_A$.
Hint:
Divide the polynomial $p(x)=x^{12}-5x^{11}+dots+3$ by the characteristic polynomial:
$$p(x)=q(x)chi_A(x)+r(x)qquad (deg r le 2),$$ to get
$$p(A)=q(A)chi_A(A)+r(A)=r(A).$$:
answered Dec 9 '18 at 22:00
Bernard
118k639112
answered Dec 9 '18 at 22:00
Bernard
118k639112
answered Dec 9 '18 at 22:00
Bernard
118k639112
answered Dec 9 '18 at 22:00
Bernard
118k639112
118k639112
add a comment |
add a comment |
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1
I suspect the characteristic equation is $lambda^3 - 5lambda^2 + 7lambda - 3 = 0$. You might want to double check your work on that part.
– JimmyK4542
Dec 9 '18 at 21:43
Yes, you're right about the characteristic equation. I took a different problem's matrix for this one. My bad.
– tNotr
Dec 9 '18 at 21:48