- #1
AcidRainLiTE
- 89
- 2
Given a matrix A (for simplicity assume 2x2) we can use the Cayley-Hamilton theorem to write:
A^k = a_0 + a_1 A
for k>= 2.
So suppose we have a given k and want to find the coefficients a_0,a_1. We can use the fact that the same equation is satisfied by the eigenvalues. That is, for any eigenvalue \lambda we have
\text{(1) } \quad \lambda^k=a_0+a_1\lambda.
If A has 2 distinct eigenvalues, we can plug them into this equation and get 2 equations which allow us to solve for a_0, a_1. And we're done.
If on, the other hand, A has a repeated eigenvalue \lambda_1, then we only get 1 equation from the above procedure. To get another equation we can differentiate (1) to get
\text{(2) } \quad k \lambda^{k-1} = a_1.
We then plug \lambda_1 into (2) to get our second equation. Now we can solve for a_0, a_1.
My question is how we know we can differentiate (1) and still get a valid equation. In order to differentiate, (1) must hold for all \lambda (or at least over some interval). But we only know that it holds for particular discrete values of \lambda (i.e. for the eigenvalues of A).
A^k = a_0 + a_1 A
for k>= 2.
So suppose we have a given k and want to find the coefficients a_0,a_1. We can use the fact that the same equation is satisfied by the eigenvalues. That is, for any eigenvalue \lambda we have
\text{(1) } \quad \lambda^k=a_0+a_1\lambda.
If A has 2 distinct eigenvalues, we can plug them into this equation and get 2 equations which allow us to solve for a_0, a_1. And we're done.
If on, the other hand, A has a repeated eigenvalue \lambda_1, then we only get 1 equation from the above procedure. To get another equation we can differentiate (1) to get
\text{(2) } \quad k \lambda^{k-1} = a_1.
We then plug \lambda_1 into (2) to get our second equation. Now we can solve for a_0, a_1.
My question is how we know we can differentiate (1) and still get a valid equation. In order to differentiate, (1) must hold for all \lambda (or at least over some interval). But we only know that it holds for particular discrete values of \lambda (i.e. for the eigenvalues of A).