I would appreciate assistance on the following question posed in an exam paper.
Let $\textsf{V}$ be a finite-dimensional vector space over a field $F$. Let $\alpha\in\operatorname{End}(\textsf{V})$. Then prove that the following conditions are equivalent :
$$\begin{align}
(1) &\quad \textsf{V} = \ker(\alpha)\oplus\operatorname{im}(\alpha) \\
(2) &\quad \operatorname{im}(\alpha) = \operatorname{im}(\alpha^2) \\
(3) &\quad \ker(\alpha) = \ker(\alpha^2)
\end{align}$$
The solution is given as:
(1)$\Leftrightarrow(2):\;\; V\;=\;$ker($\alpha$)$\; \oplus\; $im($\alpha$)$\;\Leftrightarrow\;$ ker($\alpha$)$\;\cap\;$im($\alpha$) = {0}
$\Leftrightarrow\; \alpha|im(\alpha):im(\alpha)\rightarrow im$($\alpha^{2})\;\subseteq \;im(\alpha$) is one to one………………(A)
$\Leftrightarrow \;$ im($\alpha)\;=\;\;$im($\alpha^{2}$).
I do not understand line (A), specifically the first part of line (A) and why is it a one to one transformation?
Likewise, the second part of the proof states that:
(2)$\Leftrightarrow(3):\;\;$ since $dim(V)\;=\;dim(im(\alpha))+dim(ker(\alpha))$
and$dim(V)\;=\;dim(im(\alpha^{2}))+dim(ker(\alpha^{2}))$,
then
$im(\alpha)=im(\alpha^{2})\;
\Leftrightarrow\; dim(im(\alpha))\;=\;dim(im(\alpha^{2})),\;$ since $\;im(\alpha^{2}) \subseteq im(\alpha)$…………(B)
$\Leftrightarrow\;\; dim(ker(\alpha))\;=\;dim(ker(\alpha^{2}))$
$\Leftrightarrow\;\; ker(\alpha)\;=\;ker(\alpha^{2})$, since $ker(\alpha)\;\subseteq\;ker(\alpha^{2})$…………(C)
I also do not understand the second part of equations (B) and (C), i.e. why is $\;im(\alpha^{2}) \subseteq im(\alpha)$ and $\;ker(\alpha)\;\subseteq\;ker(\alpha^{2})$
Thank you for your patience with this old lady.
Best Answer
Recall that $\operatorname{im}(\alpha) = \{\alpha(v) : v \in \mathsf V\}$, and $\alpha|_{\operatorname{im}(\alpha)}$ denotes the restriction of $\alpha$ to the image of $\alpha$. Now, note that $$ \operatorname{im}(\alpha|_{\operatorname{im}(\alpha)}) = \{\alpha(w): w \in \operatorname{im}(\alpha)\} = \{\alpha(\alpha(v)): v \in \mathsf V\} = \{\alpha^2(v): v \in \mathsf V\} = \operatorname{im}(\alpha^2). \tag{$\star$} $$ That is, $\operatorname{im}(\alpha|_{\operatorname{im}(\alpha)})$ is the same set as $\operatorname{im}(\alpha^2)$. So, we may think of $\alpha|_{\operatorname{im}(\alpha)}$ as a map (in fact an onto map) from $\operatorname{im}(\alpha)$ to $\operatorname{im}(\alpha^2)$, and we see that $\operatorname{im}(\alpha^2)$ is indeed a subspace of $\operatorname{im}(\alpha)$.
Now, $\alpha|_{\operatorname{im}(\alpha)}$ will be one to one if and only if its kernel is trivial, but what does the kernel of this map look like? Note that $$ \ker(\alpha|_{\operatorname{im}(\alpha)}) = \{w \in \operatorname{im}(\alpha) : \alpha(w) = 0\} = \{w \in \mathsf V: w \in \operatorname{im}(\alpha) \text{ and } w \in \ker(\alpha)\} = \operatorname{im}(\alpha) \cap \ker(\alpha). $$ That is, $\ker(\alpha|_{\operatorname{im}(\alpha)}) = \operatorname{im}(\alpha) \cap \ker(\alpha)$. So, $\alpha|_{\operatorname{im}(\alpha)}$ will be one to one if and only if its kernel is trivial, which is to say that $\operatorname{im}(\alpha) \cap \ker(\alpha) = \{0\}$. That explains line (A) and its equivalence to the line before.
I have stated above that $\operatorname{im}(\alpha) \subseteq \operatorname{im}(\alpha^2)$, but let's make this relationship more explicit. As I say in the equations marked $(\star)$ above, we have $$ \operatorname{im}(\alpha^2) = \{\alpha(w) : w \in \operatorname{im}(\alpha)\}. $$ Because $\alpha: \mathsf V \to \mathsf V$, we have $\operatorname{im}(\alpha) \subseteq \mathsf V$. Thus, it is clear that we have $$ \operatorname{im}(\alpha^2) = \{\alpha(w) : w \in \operatorname{im}(\alpha)\} \subseteq \{\alpha(w) : w \in \mathsf V\} = \operatorname{im}(\alpha). $$ We can argue that $\ker(\alpha) \subset \ker(\alpha^2)$ by noting that we have $v \in \ker(\alpha) \implies v \in \ker(\alpha^2)$. In particular, if $v \in \ker(\alpha)$, then $\alpha(v) = 0$. It follows that $$ \alpha^2(v) = \alpha(\alpha(v)) = \alpha(0) = 0. $$