Finding Eigenvalues of an Operator

**arcketer** · March 14th 2011, 01:10 PM

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.

**FernandoRevilla** · March 14th 2011, 01:28 PM

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

**arcketer** · March 14th 2011, 03:01 PM

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

**FernandoRevilla** · March 14th 2011, 11:49 PM

Originally Posted by arcketer

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

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

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

(contradiction) .

**arcketer** · March 15th 2011, 09:20 AM

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.

**FernandoRevilla** · March 15th 2011, 10:37 AM

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

**HallsofIvy** · March 15th 2011, 10:38 AM

I don't know where you got that. Fernando Revilla told you that the minimal polynomial must divide $(\lambda- 1)(\lambda- 2)(\lambda+ 1)$ but was NOT $\lambda- 1$ nor $(\lambda- 2)(\lambda+ 2)$ . 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.)

**arcketer** · March 15th 2011, 10:56 AM

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

**FernandoRevilla** · March 15th 2011, 11:16 AM

Originally Posted by HallsofIvy

Since the orginal equation has three distinct first degree factors, it is the minimal polynomial.

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

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

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

**HallsofIvy** · March 17th 2011, 04:39 AM

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

Thread: Finding Eigenvalues of an Operator

LinkBack

Thread Tools

Display

Finding Eigenvalues of an Operator

Similar Math Help Forum Discussions

Eigenvalues of a linear operator

Eigenvalues of Linear operator

Operator Eigenvalues

linear operator eigenvalues

Operator has no eigenvalues

Search Tags