Rank nullity inequality fom T^2

• rsa58
In summary: T. In summary, by using the fact that the kernel of T^2 is a subset of the kernel of T, we can show that nullity(T^2) <= 2nullity(T). I hope this helps clarify your understanding.
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.

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

1. What is the Rank-Nullity Inequality formula for T^2?

The Rank-Nullity Inequality formula for T^2 states that for a linear transformation T: V → W, the rank of T plus the nullity of T must be less than or equal to the dimension of the domain V.

2. How is the Rank-Nullity Inequality formula useful in linear algebra?

The Rank-Nullity Inequality formula is useful in linear algebra because it helps determine the dimensions of the range and null space of a linear transformation. It also provides valuable information about the injectivity and surjectivity of a linear transformation.

3. Can you provide an example of how to use the Rank-Nullity Inequality formula for T^2?

Sure! Let's say we have a linear transformation T: R^3 → R^2. The rank-nullity inequality formula tells us that the rank of T plus the nullity of T must be less than or equal to 3 (since the domain is R^3). So, if we find that the rank of T is 2, then we know that the nullity of T must be 1, since 2+1=3.

4. How does the Rank-Nullity Inequality formula relate to the dimensions of vector spaces?

The Rank-Nullity Inequality formula relates to the dimensions of vector spaces by providing a limit on the possible dimensions of the range and null space of a linear transformation. It also helps determine the number of independent vectors that span the range and null space.

5. Is the Rank-Nullity Inequality formula applicable to all linear transformations?

Yes, the Rank-Nullity Inequality formula is applicable to all linear transformations. This is because it is based on the fundamental properties of linear transformations and vector spaces, such as the preservation of linear combinations and the dimensionality of vector spaces.

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