Rank nullity inequality fom T^2

[rsa58]

Homework Statement

Let T be a linear operator on a vector space V (finite dimensional). show that
nullity(T^2)<= 2nullity(T)

Homework Equations

the rank nullity theorem.

The Attempt at a Solution

i take T to be a lin. transf. from ker(T^2) to V, that is i am restricting the domain of T.

well, i show kerT is a subset of ker(T^2) then i write
nullity(T^2)= nullity(T(Ker(T^2)) + dimT(ker(T^2))= dimkerT +dimT(ker(T^2))

and this last equality i am not sure of. then i show T(kerT^2) is a subset of kerT and the theorem follows from the equality which i am not sure of. the proff. said to use the fact that kerT is a subset of ker(T^2). but i don't see how this helps. i started by examining T(kerT^2) and said that oh... okay i get it now. since kerT is a subset of kerT^2 there could be some vectors ouside of the kerT in this latter set. but that still makes the kernel of T(kerT^2) equal to the kerT since it is the kernel of {0, T(v1), ... T(vn)}. is this correct, please respond i have wasted so much time on this damn thing.

 

[mighty2000]

Thank you for your question. I understand your confusion about the equality in your solution, but I believe you are on the right track.

To start, let's recall the rank-nullity theorem, which states that for a linear operator T on a finite-dimensional vector space V, we have:

dim(V) = rank(T) + nullity(T)

Now, let's consider the linear operator T^2. By the definition of a linear operator, we have:

T^2(v) = T(T(v)) for all v in V

Now, let's consider the nullity of T^2. This is the dimension of the kernel of T^2, denoted as nullity(T^2). By definition, the kernel of T^2 is the set of all vectors v in V such that T^2(v) = 0. But since T^2(v) = T(T(v)), this means that the kernel of T^2 is the set of all vectors v in V such that T(T(v)) = 0. In other words, the kernel of T^2 is a subset of the kernel of T.

Now, let's consider the nullity of T. This is the dimension of the kernel of T, denoted as nullity(T). By definition, the kernel of T is the set of all vectors v in V such that T(v) = 0. But since the kernel of T^2 is a subset of the kernel of T, this means that the kernel of T contains all the vectors in the kernel of T^2, plus possibly some additional vectors. In other words, the dimension of the kernel of T is greater than or equal to the dimension of the kernel of T^2.

Using the rank-nullity theorem, we can write:

dim(V) = rank(T) + nullity(T)
dim(V) = rank(T^2) + nullity(T^2)

Since the dimension of the kernel of T is greater than or equal to the dimension of the kernel of T^2, we can write:

dim(V) = rank(T) + nullity(T)
dim(V) = rank(T^2) + nullity(T^2)
dim(V) >= rank(T^2) + nullity(T)

This shows that the rank of T^2 is less than or equal to the rank of T, and the nullity of T^2 is less than or equal