Prove 9 is eigenvalue of ##T^2\iff## 3 or -3 eigenvalue of ##T##.

Suppose ##9## is an eigenvalue of ##T^2##.

Then ##T^2v=9v## for certain vectors in ##V##, namely the eigenvectors of eigenvalue ##9##.

Then

##(T^2-9I)v=0##

##(T+3I)(T-3I)v=0##

There seem to be different ways to go about continuing the reasoning here.

My question will be about the associated eigenvectors.

One proof is the following

- Since ##(T^2-9I)=(T+3I)(T-3I)## is not injective then it can be shown that one of ##T+3I## or ##T-3I## must also not be injective.

- But this means that either 3 or ##-3## is an eigenvalue of ##T##.

Here is another proof

- Let ##p(x)=x+3## and ##q(x)=x-3##.

- Then ##p(T)=T+3I## and ##q(T)=T-3I##.

- Now, the product of the polynomials is ##(pq)(T)=p(T)q(T)=(T+3I)(T-3I)## and it can be shown that we can change the order of the factors. That is ##(pq)(T)=q(T)p(T)=(T-3I)(T+3I)##.

- Thus, if ##(T+3I)(T-3I)v=(T-3I)(T+3I)v=0## then let's consider all (four) possible cases.

Case 1: 3 is an eigenvalue of ##T##.

Case 1.1: -3 is an eigenvalue of ##T##.

The equation is satisfied.

Case 1.2: -3 is not an eigenvalue of ##T##.

The equation is satisfied.

Case 2: 3 is not an eigenvalue of ##T##.

Subcase 2.1: -3 is not an eigenvalue of ##T##.

The equation is not satisfied since no matter what ##(T-3I)v## is, when we apply ##T+3I## to it we will not get ##0##.

Subcase 2.2: -3 is an eigenvalue of ##T##.

The equation is satisfied because ##(T-3I)(T+3I)v=(T-3I)0=0##.

So a this point we've considered all possibilities. The conclusion is that we must either have Case 1.1, Case 1.2, or Case 2.2.

Thus, 3 is an eigenvalue of ##T## or ##-3## is an eigenvalue of ##T##.

My question is about the eigenvectors.

Notice that ##(T^2-9I)v=0## is true for eigenvectors of eigenvalue 9 of ##T^2##.

It seems that in case 1 an implicit result is that the eigenvectors of ##T^2## for eigenvalue 9 are the same eigenvectors of ##T## for eigenvalue 3.

Similarly, in case 2.2 these are also the eigenvectors for eigenvalue -3 of ##T##.

But what happens in case 1.1, when both -3 and 3 are eigenvalues?

If there exists [itex]v \neq 0[/itex] such that [tex](T^2 - 9I)v = (T - 3I)(T + 3I)v = (T + 3I)(T - 3I)v = 0[/tex] then either there exists [itex]u \neq 0[/itex] such that [itex](T - 3I)u = 0[/itex] or there exists [itex]u \neq 0[/itex] such that [itex](T + 3I)u = 0[/itex] (in both cases, [itex]u[/itex] could be [itex]v[/itex].) These cannot both be true for the same [itex]u[/itex], but that does not exclude the possibility that there exist distinct [itex]u_1 \neq 0[/itex] and [itex]u_2 \neq 0[/itex] such that [itex](T - 3I)u_1 = 0[/itex] and [itex](T + 3I)u_2 = 0[/itex].

pasmith said:

If there exists [itex]v \neq 0[/itex] such that [tex](T^2 - 9I)v = (T - 3I)(T + 3I)v = (T + 3I)(T - 3I)v = 0[/tex] then either there exists [itex]u \neq 0[itex] such that [itex](T - 3I)u = 0[/itex] or there exists [itex]u \neq 0[/itex] such that [itex](T + 3I)u = 0[/itex] (in both cases, [itex]u[/itex] could be [itex]v[/itex].) These cannot both be true for the same [itex]u[/itex], but that does not exclude the possibility that there exist distinct [itex]u_1 \neq 0[/itex] and [itex]u_2 \neq 0[/itex] such that [itex](T - 3I)u_1 = 0[/itex] and [itex](T + 3I)u_2 = 0[/itex].

We start with the equation ##(T^2-9I)v=0## which by assumption is true for the eigenspace corresponding to eigenvalue 9 of ##T^2##.

Then, from the initial equation we reach ##(T+3I)(T-3I)v=0##. This equation is implied by the initial equation and it seems to me it is true for the same eigenvectors as before.

This is precisely the thing I am confused about. We start with a specific linear map ##T^2## with a specific eigenvalue ##9## and we write an equation that is true for very specific vectors, the eigenvectors of eigenvalue ##9##.

When we write the equation differently by expanding the ##(T^2-9I)## term, the equation seems to still be true only for the eigenvectors of ##T^2## of eigenvalue 9.

Am I wrong about that?

Next, if these same eigenvectors also satisfy ##(T-3I)v=0## then they are also in the eigenspace of eigenvalue ##3## of ##T##.

But then the eigenspace of eigenvalue ##3## of ##T## is the same as the eigenspace of eigenvalue ##9## of ##T^2##.

Is this correct so far?

zenterix said:

We start with the equation ##(T^2-9I)v=0## which by assumption is true for the eigenspace corresponding to eigenvalue 9 of ##T^2##.

Then, from the initial equation we reach ##(T+3I)(T-3I)v=0##. This equation is implied by the initial equation and it seems to me it is true for the same eigenvectors as before.

This is precisely the thing I am confused about. We start with a specific linear map ##T^2## with a specific eigenvalue ##9## and we write an equation that is true for very specific vectors, the eigenvectors of eigenvalue ##9##.

When we write the equation differently by expanding the ##(T^2-9I)## term, the equation seems to still be true only for the eigenvectors of ##T^2## of eigenvalue 9.

Am I wrong about that?

Next, if these same eigenvectors also satisfy ##(T-3I)v=0## then they are also in the eigenspace of eigenvalue ##3## of ##T##.

This doesn't have to be true.

What you have is [tex]
(T + 3I)(T - 3I)v = 0.[/tex] Set [itex](T - 3I)v = u[/itex]. Then [itex](T + 3I)u = 0[/itex]. Now there are two options: If [itex]u \neq 0[/itex] then [itex]u[/itex] is an eigenvector of [itex]T[/itex] with eigenvalue -3, and [itex]v[/itex] is not a eigenvector of [itex]T[/itex] with eigenvalue 3; it's something which is mapped by [itex](T - 3I)[/itex] to the -3 eigenspace. On the other hand, if [itex]u = 0[/itex] then [itex]v[/itex] is an eigenvector with eigenvalue 3.

Suppose your ##u=0##. Then, as you said, ##v## is an eigenvector with eigenvalue 3. But then the eigenspace ##E(3, T)## is the same as the eigenspace ##E(9, T^2)##.

Now suppose ##u\neq 0## (that is, 3 is not an eigenvalue). Can we not rewrite the equation as ##(T-3I)(T+3I)v=0## in which case ##v## must be in the eigenspace ##E(-3,T)=E(9,T^2)##?

If we stick to the order ##(T+3I)(T-3I)v=0## and ##-3## is an eigenvalue but ##3## is not, then it seems that ##T-3I## maps ##v\in E(-3, T)## to ##E(-3, T)##, as follows

##(T-3I)v=Tv-3v=-3v-3v=-6v##

##(T+3I)(-6v)=-6(T+3I)v=0##

In fact, this makes intuitive sense since ##T-3I## is just maps an eigenvector of ##E(-3,T)## to a multiple of itself and then subtracts another multiple of itself. So ##T-3I## does map an eigenvector of ##E(-3,T)## to ##E(-3,T)##.

Now that I think of it, it also makes intuitive sense that ##E(-3,T)=E(9,T^2)##.

##T^2## takes a vector ##v## from ##E(-3,T)## to ##-3v## and then takes that vector to ##9v##. So every vector in ##E(-3,T)## must be in ##E(9,T^2)##.

Now, is it possible for ##T## to have an additional eigenvalue ##3##?

If it does, then by the same reasoning, every vector in ##E(3,T)## must be in ##E(9,T^2)##.

But in this case we must have two different eigenspaces ##E(3,T)## and ##E(-3,T)## that share only the zero vector, and then ##E(9,T^2)## seems to be the direct sum of those two.

Suppose ##v_1\in E(3,T)## and ##v_2\in E(-3,T)##.

Then

##(T^2-9I)(v_1+v_2)=(T-3I)(T+3I)(v_1+v_2)##
##=(T-3I)(3v_1-3v_1)##
##=(T-3I)(6v_1)##
##=0##

Thus, ##v_1+v_2\in E(9,T^2)##.

By assumption ##\det (T^2-\lambda^2E) = 0##. Write
[tex]
\det (T^2-\lambda^2E) = \det ((T-\lambda E)(T+\lambda E)) = \det (T-\lambda E)\det (T+\lambda E) = 0.
[/tex]
Conclude what is required. Conversely, if you apply a polynomial to ##T##, then its eigenvalues are converted exactly the same way.