- #1

- 90

- 2

[tex]A^k = a_0 + a_1 A[/tex]

for k>= 2.

So suppose we have a given k and want to find the coefficients [itex]a_0,a_1[/itex]. We can use the fact that the same equation is satisfied by the eigenvalues. That is, for any eigenvalue [itex]\lambda[/itex] we have

[tex]\text{(1) } \quad \lambda^k=a_0+a_1\lambda.[/tex]

If A has 2 distinct eigenvalues, we can plug them into this equation and get 2 equations which allow us to solve for [itex] a_0, a_1[/itex]. And we're done.

If on, the other hand, A has a repeated eigenvalue [itex]\lambda_1[/itex], then we only get 1 equation from the above procedure. To get another equation we can differentiate (1) to get

[tex]\text{(2) } \quad k \lambda^{k-1} = a_1.[/tex]

We then plug [itex]\lambda_1[/itex] into (2) to get our second equation. Now we can solve for [itex]a_0, a_1[/itex].

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 [itex]\lambda[/itex] (or at least over some interval). But we only know that it holds for particular discrete values of [itex] \lambda[/itex] (i.e. for the eigenvalues of A).