Proving Properties of Linear Transformations in Jordan Canonical Form

[hitmeoff]

Homework Statement

Let T: V $$\rightarrow$$ W be a linear trans. Prove:
a) N(T) = N(-T)
b) N(T^k) = N((-T)^k)
c) If V = W and $$\lambda$$ is an eigenvalue of T, then for any positive integer k:
N((T - $$\lambda$$I_v)^k) = N(($$\lambda$$I_v - T)^k)

Homework Equations

The Attempt at a Solution

Im not sure how to start on a. I know if I can get started on that one, I can handle the rest. I like this:

-T(v) = -0
-T(v) = 0
-T(v) = T(v)

therefore N(T) = N(-T) ?

Im not sure if that's even remotely on the right track, but this question is in the first Jordan Conanical form section, and I am not sure how that concept can be applie to this very first question. I can see it needing to be applied in the other two, but not this one.

 

[Mark44]

Homework Statement

Let T: V $$\rightarrow$$ W be a linear trans. Prove:
a) N(T) = N(-T)
b) N(T^k) = N((-T)^k)
c) If V = W and $$\lambda$$ is an eigenvalue of T, then for any positive integer k:
N((T - $$\lambda$$I_v)^k) = N(($$\lambda$$I_v - T)^k)

Homework Equations

The Attempt at a Solution

Im not sure how to start on a. I know if I can get started on that one, I can handle the rest. I like this:

-T(v) = -0
-T(v) = 0
-T(v) = T(v)

therefore N(T) = N(-T) ?

How does this follow from what appears before it. You're trying to prove that the nullspace of T is the same as the nullspace of -T.

Start by assuming that v is in null(T). Can you show that v is also in null(T)?

Im not sure if that's even remotely on the right track, but this question is in the first Jordan Conanical form section, and I am not sure how that concept can be applie to this very first question. I can see it needing to be applied in the other two, but not this one.

 

[hitmeoff]

OK so if v is in N(T) then T(v) = 0

Since -T(v) is equal to T(-v), we need to show that -v is also in N(T)?

I think I am thinking what Landau was thinking in this thread:
https://www.physicsforums.com/showthread.php?t=403507
"It looks correct. Basically, you're proving that for any linear map T, T and -T have the same kernel. This is true because Tv=0 iff -Tv=0"

T(v) = 0 iff -T(v) = 0, since we are only talking about v in N, then T(v) must equal 0 then -T(v) = 0 this N(T) = N(-T), no?

 

[Mark44]

OK so if v is in N(T) then T(v) = 0

Since -T(v) is equal to T(-v), we need to show that -v is also in N(T)?

Yes, you need to show that -v is also in N(T).

Do you know why -T(v) = T(-v)?

I think I am thinking what Landau was thinking in this thread:
https://www.physicsforums.com/showthread.php?t=403507
"It looks correct. Basically, you're proving that for any linear map T, T and -T have the same kernel. This is true because Tv=0 iff -Tv=0"

T(v) = 0 iff -T(v) = 0, since we are only talking about v in N, then T(v) must equal 0 then -T(v) = 0 this N(T) = N(-T), no?