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

[zenterix]

Homework Statement

Suppose $T\in\mathcal{L}(V)$.
Prove that 9 is an eigenvalue of $T^2\iff$ $3$ or $-3$ is an eigenvalue of $T$.

Relevant Equations

Let's consider just one direction of the biconditional. Let's suppose 9 is an eigenvalue of $T^2$ and prove that 3 or $-3$ is an 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?

 

[pasmith]

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

 

[zenterix]

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

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?

 

[pasmith]

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

 

[zenterix]

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

 

[nuuskur]

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