1. ## Eigenvalue related question.

if T: V -> V has the property that T^2 has a non negative eigenvalue (lambda)^2, prove that at least one of (lambda) or -(lambda) is an eigenvalue for T.

The Hint that it gives is the equality:

T^2 - (lambda)^2 * I = (T + lambda * I)*(T - lambda * I)

where I is the identity matrix.

Any help?

2. Originally Posted by hashshashin715
if T: V -> V has the property that T^2 has a non negative eigenvalue (lambda)^2, prove that at least one of (lambda) or -(lambda) is an eigenvalue for T.

The Hint that it gives is the equality:

T^2 - (lambda)^2 * I = (T + lambda * I)*(T - lambda * I)

where I is the identity matrix.

Any help?
What is T?

3. T is the linear transformation. Like as in T(x) = (lambda)x. But we are given T^2(x), which I think is T(T(x)) = (lambda)^2 * x

4. Originally Posted by hashshashin715
if T: V -> V has the property that T^2 has a non negative eigenvalue (lambda)^2, prove that at least one of (lambda) or -(lambda) is an eigenvalue for T.

The Hint that it gives is the equality:

T^2 - (lambda)^2 * I = (T + lambda * I)*(T - lambda * I)
By hypothesis,

$\exists x\in V\;(x\neq 0):T^2(x)=\lambda^2 x$

equivalently

$\exists x\in V\;(x\neq 0)T^2-\lambda^2 I)(x)=0" alt="\exists x\in V\;(x\neq 0)T^2-\lambda^2 I)(x)=0" />.

Now, use:

$(T+\lambda I)[ (T-\lambda I)(x)]=0$

and analyze what would happen if

$(T-\lambda I)(x)\neq 0$ .

Fernando Revilla