Is the Null Space the Same for (T-λI)^k and (λI-T)^k in Linear Algebra?

[tinynerdi]

Homework Statement

Let T:V->W be a linear transformation. Prove that if V=W (So that T is linear operator on V) and λ is an eigenvalue on T, then for any positive integer K
N((T-λI)^k) = N((λI-T)^k)

Homework Equations

T(-v) = -T(v)
N(T) = {v in V: T(v)=0} in V hence T(v) = 0 for all v in V.

The Attempt at a Solution

we know that (T-λI)^k(-v) = -(T-λI)^k(v) = (λI-T)^k(v). So when (T-λI)^k(v) = 0 so does (λI-T)^k(v). Hence N((T-λI)^k) = N((λI-T)^k)

 

[Landau]

Homework Equations

T(-v) = -T(v)
N(T) = {v in V: T(v)=0} in V hence T(v) = 0 for all v in V.

You mean Tv=0 for all V in N.

The Attempt at a Solution

we know that (T-λI)^k(-v) = -(T-λI)^k(v) = (λI-T)^k(v). So when (T-λI)^k(v) = 0 so does (λI-T)^k(v). Hence N((T-λI)^k) = N((λI-T)^k)

You mean (T-λI)^k(v) = 0 if and only if (λI-T)^k(v) = 0.

 

[tinynerdi]

You mean Tv=0 for all V in N.

isn't N(T) is a subspace of V since V is a vectors space.

You mean (T-λI)^k(v) = 0 if and only if (λI-T)^k(v) = 0.

Yeah, that is what I am trying to prove.

 

[Landau]

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.

 

[tinynerdi]

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.

Can we just state that because Tv=0 iff and -Tv = 0 therefore N((T-λI)^k) = N((λI-T)^k) or do we have to prove that Tv=0 iff and -Tv = 0?

 

[Landau]

or do we have to prove that Tv=0 iff and -Tv = 0?

Well, there is not much to prove. The only thing you need is that -0=0.