# Thread: Finding Eigenvalues of an Operator

1. ## Finding Eigenvalues of an Operator

For a 5-dimensional vector space, I have an operator I'm working with T satisfying
(T-1)(T^2 - 4) = 0, but T =/=1 and T^2 - 4 =/= 0.

I'm not sure how to get started on this. I can find them for a matrix, but in this case I have no matrix, only an equation in T. Any push in the right direction would be great, thank you.

2. The minimal polynomial $\displaystyle \mu (\lambda)$ divides to $\displaystyle (\lambda-1)(\lambda-2)(\lambda +2)$ and $\displaystyle \mu(\lambda)\neq \lambda -1,\;\mu(\lambda)\neq (\lambda -2)(\lambda +2)$ so, ...

3. Does this imply that the minimal polynomial is (x - 2) or (x + 2) then?

4. Originally Posted by arcketer
Does this imply that the minimal polynomial is (x - 2) or (x + 2) then?

No, if $\displaystyle \mu (x)=x-2$ i s the minimal polinomial of $\displaystyle T$ then,

$\displaystyle T-2I=0\Rightarrow T^2-4I=(T-2I)\circ (T+2I)=0$

5. Well then this must mean that the minimal polynomial has to be x + 2, as I don't see any other factors dividing the original equation, and a simple check yields

T + 2I = 0 -> (T+2I)(T+2I) = T^2 + 4T + 4I = 0 is not a contradiction, so the only eigenvalue is -2? That feels right but any input would be great.

6. The minimal polynomial $\displaystyle \mu (x)$ divdes to $\displaystyle (x-1)(x-2)(x+2)$.

7. I don't know where you got that. Fernando Revilla told you that the minimal polynomial must divide $\displaystyle (\lambda- 1)(\lambda- 2)(\lambda+ 1)$ but was NOT $\displaystyle \lambda- 1$ nor $\displaystyle (\lambda- 2)(\lambda+ 2)$. Since the orginal equation has three distinct first degree factors, it is the minimal polynomial.

(The "I don't know where you got that" was in response to arcketer's last post.)

8. Oh, I see. I think I was thrown off by the wording. So having the minimal polynomial, it is obvious the corresponding eigenvalues are 1, 2, -2 each with single multiplicity, and this allows me to make some claims about the possible Jordan Canonical forms. Thank you both very much for your help

9. Originally Posted by HallsofIvy
Since the orginal equation has three distinct first degree factors, it is the minimal polynomial.

Also $\displaystyle \mu(\lambda)=(\lambda-1)(\lambda-2)$ and $\displaystyle \mu(\lambda)=(\lambda-1)(\lambda+2)$ can be minimal polynomials of $\displaystyle T$ . For example

$\displaystyle T\equiv \textrm{diag}(1,1,1,2,2)$ and $\displaystyle T\equiv \textrm{diag}(1,1,1,-2,-2)$ .

In both cases, $\displaystyle T-I\neq 0$ and $\displaystyle T^2-4I\neq 0$ .

10. Good point. I had misinterpreted "T satisfying (T-1)(T^2 - 4) = 0, but T =/=1 and T^2 - 4 =/= 0" as meaning that $\displaystyle (T-1)^2(T^2- 4)$ was the characteristic polynomial but clearly that is not necessarily true.