Proving Equivalence of Statements Involving Endomorphisms - Can We Take $v=x$?

[mathmari]

Hey!

Let $V$ be a finite dimensional vector space and $\Phi \in \text{End}(V)$.
I want to show that the following statements are equivalent:

1. $V=\ker (\Phi)\oplus \text{im} (\Phi )$
2. $\ker (\Phi )=\ker (\Phi^2)$
3. $\text{im} (\Phi )=\text{im} (\Phi^2)$

I have done the following:

We suppose that the statement 1. holds.
Let $v\in V$ then we have that $v=k+\Phi (x)$, where $k\in \ker (\Phi )$, so $\Phi (k)=0$, and $\Phi (x)\in \text{im} (\Phi )$.
We have that $y\in \ker (\Phi)\cap \text{im} (\Phi )\Rightarrow y=0$.

Can we take $v=x$ ? (Wondering)

If yes, then we would have the following:

$$x=k+\Phi (x) \Rightarrow \Phi (x)=\Phi (k+\Phi (x)) \Rightarrow \Phi (x)=\Phi (k)+\Phi^2(x) \Rightarrow \Phi (x)=\Phi^2(x)$$

From this, we get the statements 2. and 3., or not? (Wondering)

 

[I like Serena]

Hey mathmari! ;)

I don't think we can assume that v=x.
Instead we need to prove that $\Phi(v)=0$ implies that $\Phi^2(v)=0$ and the converse (for which we can use that the intersection of kernel and image is empty)...

 

[mathmari]

Instead we need to prove that $\Phi(v)=0$ implies that $\Phi^2(v)=0$ and the converse (for which we can use that the intersection of kernel and image is empty)...

Suppose that the statement 1. holds. So, $v\in V$, $v=k+\Phi (x)$, where $k\in \ker \Phi $ and $\Phi (x)\in \text{im}\Phi$.

Let $y\in \ker \Phi \cap \text{im}\Phi $ then $\Phi (y)=0$ and $\exists w: y=\Phi (w)$. Since the intersection is trivial , we have that this holds only for $y=0$. Suppose that $v\in \ker \Phi $, so $\Phi (v)=0$. Therefore, $\Phi (k+\Phi (x))=0 \Rightarrow \Phi (k)+\Phi ^2(x)=0\Rightarrow \Phi^2(x)=0 \Rightarrow \Phi (\Phi (x))=0 \Rightarrow \Phi (x)\in \ker\Phi$.

So, we have that $\Phi (x)\in \text{im}\Phi$ and $\Phi (x)\in \ker\Phi$, that implies that $\Phi(x)=0$.

Replacing this at the initial equation we get that $v=k$. So, we have that $\Phi (v)=0 \Rightarrow \Phi^2(v)=\Phi (0)=0$.

This means that $v\in \ker \Phi^2$.

Therefore, $\ker \Phi\subseteq \ker \Phi^2$.

Is this direction correct? (Wondering) For the other direction, we assume that $v\in \ker \Phi^2 $, so $\Phi^2 (v)=0$.

Therefore, we have that $\Phi^2 (k+\Phi (x))=0 \Rightarrow \Phi (\Phi (k)+\Phi^2(x))=0 \Rightarrow \Phi(\Phi^2(x))=0$.

From that we get either that $\Phi^2(x)\in \ker \Phi$ or that $\Phi \in \ker \Phi^2$, right? (Wondering)

How could we continue? (Wondering)

 

[I like Serena]

I think we can simplify the first one a bit. (Thinking)

For any $v\in V$ we have that if $\Phi(v)=0$ it follows that $\Phi(\Phi(v))=\Phi(0)=0$.
So if $v\in \operatorname{Ker}\Phi$ we also have that $v\in \operatorname{Ker}\Phi^2$.
We don't even need statement (1) for that.

As for the converse, if $\Phi^2(v)=\Phi(\Phi(v))=0$ it follows that $\Phi(v) \in \operatorname{Ker}\Phi$.
But by definition we also have that $\Phi(v) \in \operatorname{Im}\Phi$.
So $\Phi(v) \in \operatorname{Ker}\Phi \cap \operatorname{Im}\Phi = \{0\}$.
And thus $v \in \operatorname{Ker}\Phi$. (Cool)

 

[mathmari]

I see! (Happy) For $(2)\Rightarrow (3)$ do we do the following:

We have that $\dim (V)=\dim \ker \Phi +\dim \text{im}\Phi$.

We consider the mapping $\Phi^2$.

Then we have that $\dim (V)=\dim \ker \Phi^2 +\dim \text{im}\Phi^2$.

Since $\ker \Phi=\ker \Phi^2$ we have that

$\dim (V)=\dim \ker \Phi +\dim \text{im}\Phi^2 \\ \Rightarrow \dim \ker \Phi +\dim \text{im}\Phi=\dim \ker \Phi +\dim \text{im}\Phi^2 \\ \Rightarrow \dim \text{im}\Phi=\dim \text{im}\Phi^2$

Is it correct so far? (Wondering)

When the dimensions are equal, does it follow that $\text{im}\Phi=\text{im}\Phi^2$ ? (Wondering)

 

[I like Serena]

Is it correct so far? (Wondering)

When the dimensions are equal, does it follow that $\text{im}\Phi=\text{im}\Phi^2$ ? (Wondering)

It's correct, but no, that does not follow.

Consider for instance the sets $A=\{1,2,3\}$ and $B=\{4,5,6\}$.
We can see that $|A|=|B|$, but we do not have that $A=B$.To prove statement 3, we need to prove that:
$$\forall v \in V: v \in\operatorname{im} \Phi^2 \overset{?}\to v \in \operatorname{im} \Phi$$
It means that $\exists x: v=\Phi^2(x) = \Phi(\Phi(x))$.
Can we conclude that $v \in \operatorname{im} \Phi$? (Wondering)

 

[mathmari]

To prove statement 3, we need to prove that:
$$\forall v \in V: \operatorname{im} \Phi^2 \overset{?}\to v \in \operatorname{im} \Phi$$
It means that $\exists x: v=\Phi^2(x) = \Phi(\Phi(x))$.
Can we conclude that $v \in \operatorname{im} \Phi$? (Wondering)

Ah yes! (Nerd)

For the other direction:
Let $v\in \operatorname{im} \Phi$, then there exists a $x\in V$ such that $v=\Phi (x)$.
Do we conclude from here that $v\in \operatorname{im} \Phi^2$.
Do we apply $\Phi$ at the equation $v=\Phi (x)$ ? (Wondering)

 

[I like Serena]

Ah yes! (Nerd)

For the other direction:
Let $v\in \operatorname{im} \Phi$, then there exists a $x\in V$ such that $v=\Phi (x)$.
Do we conclude from here that $v\in \operatorname{im} \Phi^2$.
Do we apply $\Phi$ at the equation $v=\Phi (x)$ ? (Wondering)

That's what we want to conclude but we can't just yet.
We need to prove that $v = \Phi(x) \to \exists y \in V: v=\Phi(\Phi(y))$. (Thinking)

 

[mathmari]

We need to prove that $v = \Phi(x) \to \exists y \in V: v=\Phi(\Phi(y))$. (Thinking)

How could we do that? Could you give me a hint? (Wondering)

 

[mathmari]

Having that

$\dim \text{im}\Phi=\dim \text{im}\Phi^2$

and

$$\forall v \in V: \operatorname{im} \Phi^2 \to v \in \operatorname{im} \Phi$$

so $\operatorname{im} \Phi^2 \subseteq \in \operatorname{im} \Phi$

we get that $ \operatorname{im}\Phi= \operatorname{im}\Phi^2$, or not? (Wondering)

 

[I like Serena]

How could we do that? Could you give me a hint? (Wondering)

Well, if we use statement (1) and write $\Phi(x) = k + \Phi(y)$ with $k\in\ker\Phi$, it follows that:
$$v=\Phi(x)=\Phi(k + \Phi(y)) = \Phi(k) + \Phi^2(y) = \Phi^2(y)$$
That would complete the proof for (1) -> (3).
So if we prove (2)->(1) as well, we have proven (2)->(3).

For $(2)\Rightarrow (3)$ do we do the following:

We have that $\dim (V)=\dim \ker \Phi +\dim \text{im}\Phi$.

We consider the mapping $\Phi^2$.

Then we have that $\dim (V)=\dim \ker \Phi^2 +\dim \text{im}\Phi^2$.

Since $\ker \Phi=\ker \Phi^2$ we have that

$\dim (V)=\dim \ker \Phi +\dim \text{im}\Phi^2 \\ \Rightarrow \dim \ker \Phi +\dim \text{im}\Phi=\dim \ker \Phi +\dim \text{im}\Phi^2 \\ \Rightarrow \dim \text{im}\Phi=\dim \text{im}\Phi^2$

I think the first statement is not generally true based on only statement (2).

Consider for instance $\Phi=\begin{pmatrix}1&1\\0&0\end{pmatrix}$ with $V=\mathbb R^2$.
I has kernel $\{0\}$ and image $\left\{\lambda \begin{pmatrix}1\\0\end{pmatrix}\right\}$.
So $2=\dim V\ne\dim\ker\Phi + \dim\operatorname{im}\Phi$. (Worried)