Cayley - Hamilton Theorem on higher powers

In summary, the conversation discusses finding the matrix C given a matrix A and using the Cayley-Hamilton theorem to solve for it. The characteristic polynomial is mentioned and it is verified that plugging in A for λ satisfies the solution. A simpler approach is suggested, using eigenvalues and polynomial functions. It is noted that for any polynomial function f(x), f(A) can be found using the matrices E_1 and E_2. The conversation ends with an explanation of how to compute f(A) for the given problem.
  • #1
KMjuniormint5
67
0

Homework Statement


Given the following matrix A = [3 -1; -1 3] Find C = (0.5*A - I)100


Homework Equations


Using the knowledge that the Cayley - Hamilton Theorem must satisfy its own characteristic polynomial.


The Attempt at a Solution



Here the characteristic polynomial is λ2 - 6*λ + 8. When we plug in A = λ we can verify that the solution A2 - 6*A + 8 = 0 which it does. If I solve by hand/brute force way, I get C = [0.5 -0.5; -0.5 0.5] regardless of what the exponential is (here the exponential is 100). As far as applying the Cayley-Hamilton theorem, I am not connecting the two. I understand how to solve for A^3 or A^4, √A, or exp(A) but not for this problem. Any help in the right direction will be greatly appreciated.
 
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  • #2
Looking at it at a different angle take w = 0.5*A - I and find the characteristic equation which is λ2 -λ = 0. We see that λ2 = λ. Plug in w for λ Now for w3 we can do w2*w but that is w*w and that is w. So for w100=w99*w= ... = w*w = w. In the end we see that w = C = [0.5 -0.5; 0.5 -0.5]. Was that the right approach?
 
  • #3
KMjuniormint5 said:

Homework Statement


Given the following matrix A = [3 -1; -1 3] Find C = (0.5*A - I)100


Homework Equations


Using the knowledge that the Cayley - Hamilton Theorem must satisfy its own characteristic polynomial.


The Attempt at a Solution



Here the characteristic polynomial is λ2 - 6*λ + 8. When we plug in A = λ we can verify that the solution A2 - 6*A + 8 = 0 which it does. If I solve by hand/brute force way, I get C = [0.5 -0.5; -0.5 0.5] regardless of what the exponential is (here the exponential is 100). As far as applying the Cayley-Hamilton theorem, I am not connecting the two. I understand how to solve for A^3 or A^4, √A, or exp(A) but not for this problem. Any help in the right direction will be greatly appreciated.

A much, much simpler way is to note that if the eigenvalues of A are distinct (which they are in this case) then for any polynomial function f(x) we have
[tex] f(A) = E_1 f(\lambda_1) + E_2 f(\lambda_2), [/tex]
with the matrices ##E_1, E_2## being the same for any function f. You can find E1 and E2 in this case by using, for example,
[tex] f(x) = x^0 = 1 \Longrightarrow f(A) = I = E_1 \lambda_1^0 + e_2 \lambda_2^0 = E_1 + E_2[/tex]
and [tex] f(x) = x \Longrightarrow f(A) = A = E_1 \lambda_1 + E_2 \lambda_2[/tex]
Now it is easy to compute f(A) for
[tex] f(x) = (0.5 x - 1)^{100}. [/tex]
 

1. What is the Cayley-Hamilton Theorem on higher powers?

The Cayley-Hamilton Theorem on higher powers is a mathematical theorem that states that every square matrix satisfies its own characteristic equation. In other words, a square matrix can be expressed as a polynomial of its own eigenvalues.

2. Why is the Cayley-Hamilton Theorem on higher powers important?

The Cayley-Hamilton Theorem on higher powers is important because it has many applications in various fields of mathematics, such as linear algebra, differential equations, and control theory. It also provides a method for computing powers of matrices without explicitly calculating them, which can be useful in certain situations.

3. Can the Cayley-Hamilton Theorem on higher powers be applied to non-square matrices?

No, the Cayley-Hamilton Theorem on higher powers only applies to square matrices. In order for a matrix to have eigenvalues and a characteristic equation, it must be a square matrix.

4. How is the Cayley-Hamilton Theorem on higher powers related to the Cayley-Hamilton Theorem?

The Cayley-Hamilton Theorem on higher powers is an extension of the original Cayley-Hamilton Theorem, which states that a square matrix satisfies its own characteristic polynomial. The higher powers theorem expands on this idea and shows that the matrix can also be expressed as a polynomial of its own eigenvalues to higher powers.

5. Are there any limitations or exceptions to the Cayley-Hamilton Theorem on higher powers?

Yes, the Cayley-Hamilton Theorem on higher powers does have some limitations and exceptions. It only applies to matrices with complex eigenvalues and does not hold for all non-square matrices. Additionally, it does not work for matrices over arbitrary fields, only for matrices over algebraically closed fields such as the complex numbers.

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