# If V is a vector space why is T^2(V) = T(V) iff ker(T^2) = ker(T)?

• Eclair_de_XII
In summary, we have shown that if ##\ker(T^2)=\ker(T)##, then ##T## and ##T^2## are both one-to-one and onto. This means that ##T(V)=V## and ##T^2(V)=V##, and in turn, ##\ker(T)=\{0\}##.
Eclair_de_XII
Homework Statement
Let ##V## be a finite-dimensional vector space. Prove that for a linear operator ##T##, that ##T^2(V)=T(V)## iff ##ker(T^2)=ker(T)##.
Relevant Equations
##ker(T)##: The set of ##v\in V## such that ##T(v)=0##
##T(V)##: The image of ##v\in V## through ##T##.
--##ker(T^2)=ker(T)## if ##T(V)=T^2(V)##--

Suppose that ##T^2(V)=T(V)##. So ##T:T(V)\mapsto T^2(V)=T(V)##. Hence, ##T## is one-to-one and so ##ker(T)=\{0\}##. Suppose that ##T^2(w)=0## for some ##w\in ker(T^2)##. Then ##T^2(w)=T(T(w))=0## which implies that ##T(w)\in ker(T)## and so ##T(w)=0##, so ##w=0##. Hence, ##ker(T^2)=ker(T)=\{0\}##.

--##T^2(V)=T(V)## if ##ker(T^2)=ker(T)##--

Suppose that ##ker(T^2)=ker(T)##. Let ##v\in ker(T^2),ker(T)## so that ##T^2(v)=0##. We have that ##T^2:ker(T)\mapsto \{0\}## and that ##T:T(ker(T))=\{0\}\mapsto \{0\}##. Hence, since ##ker(T)=ker(T^2)##, we have that ##ker(T)=\{0\}##. In turn, ##T## is surjective. And so ##T(V)=V##, and furthermore, ##T^2(V)=T(V)=V##.

All in all, I'm not really too sure about the last step.

Your conclusion that ##\operatorname{ker}(T) = \{0\}## is wrong, ##T(V)## is not necessarily ##V##. All that ##T^2(V) = T(V)## tells you is that they are the same subspace of ##V##. All you can say is that ##T## is bijective from ##T(V)## to ##T(V)##.

As a counter example, consider the zero mapping ##T(x) = 0## for all ##x \in V##. This map is clearly linear and its kernel is all of ##V##. You also have ##T(T(x)) = T(0) = T(x)## for all ##x## so ##T^2 = T## and they therefore obviously have the same kernel.

Orodruin said:
All you can say is that ##T## is bijective from ##T(V)## to ##T(V)##.

Does that mean that ##\ker(T|_{T(V)})=\{0\}##? If so, then I don't know how that will relate to ##T## or its kernel on the whole domain of ##V##.

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By definition, if $v \in \ker T^2$ then $T^2(v) = 0$. Now that could be because $T(v) = 0$ so that $v \in \ker T$, but could also be because $T(v)$ is some other member of $\ker T$ so that $v \notin \ker T$.

Okay, let me think, here...

Let ##v\in \ker(T)##. Then ##T(v)=0##. Also, ##T^2(v)=T(T(v))=T(0)=0##, which means that ##v\in \ker(T^2)##. Hence ##\ker(T)\subset \ker(T^2)##.

Let ##v \in \ker(T^2)##. Then ##T^2(v)=T(T(v))=0##, which can mean either that:

(1) ##T(v)=0## so that ##v\in \ker(T)##, and ##\ker(T^2)\subset \ker(T)##, which means that we are done in this case.
(2) ##T(v)\in \ker(T)##.

In the latter case, it is implied that ##T(V)=T^2(V)\subset \ker(T)##. So ##T(T^2(u))=0## for some ##u\in V##. So my question then becomes that of figuring out if there is a ##u\in V## such that ##T^2(u)=v##.

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you should be able to streamline this by observing that T + its complement creates the identity map. This is a general point about idempotence. (equivalently the identity minus T is the complement...)

StoneTemplePython said:
T + its complement creates the identity map

Okay, let me take a quick stab at this...

Let ##T^c## be the complement of ##T## such that ##T+T^c=I_V##. Let ##v\in \ker(I_V-T^c)##. With this equality, we have that: ##T=I_V-T^c##. And so if we apply this to ##v##, ##T(v)=(I_V-T^c)(v)=0##, so ##v\in \ker(T)##. On the other hand, ##T^2(V)=T(V)##, so ##T^2=I_V-T^c##, and ##T^2(v)=(I_V-T^c)(v)=0##. Hence, ##v\in \ker(T^2)##. Equality follows, I think?

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## 1. What is a vector space?

A vector space is a mathematical structure that consists of a set of elements (vectors) and a set of operations (addition and scalar multiplication) that satisfy certain properties, such as closure, associativity, and distributivity. Examples of vector spaces include the space of real numbers, the space of polynomials, and the space of matrices.

## 2. What does T^2(V) mean?

T^2(V) refers to the composition of a linear transformation T with itself. This means applying the transformation T to the vectors in the vector space V, and then applying T again to the resulting vectors.

## 3. Why is T^2(V) = T(V) if and only if ker(T^2) = ker(T)?

This statement is known as the "fundamental theorem of linear algebra." It states that the composition of a linear transformation with itself is equal to the original transformation if and only if the kernel (null space) of the composition is equal to the kernel of the original transformation. In other words, the composition of T with itself does not change the kernel of T.

## 4. What is the significance of this theorem?

This theorem is important because it relates the properties of a linear transformation to its kernel. It allows us to simplify calculations and proofs by focusing on the kernel of a transformation rather than the entire vector space. It also helps us understand the structure of vector spaces and how they are affected by linear transformations.

## 5. Can this theorem be extended to higher powers of T?

Yes, this theorem can be extended to higher powers of T. In general, for any positive integer n, T^n(V) = T(V) if and only if ker(T^n) = ker(T). This means that if a linear transformation is composed with itself n times, the resulting transformation will be equal to the original transformation if and only if their kernels are equal.

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