$\text{Ker P $ ,$ Im P}$ and Direct Sum

Question

• Let $V$ be a vector space. Let $P:V\rightarrow V$ be a linear map such that $P^2=P$. Show that

$$V=\text{Ker P + Im P}.$$ and $$\text{Ker P}\cap\text{Im P}=\left\{0\right\}.$$

HINT. an element $v\in V$ is $v=v-P(v)+P(v)$.

Proof. Since $P(v-P(v))=P(v)-P^2(v)=0$, we see that $V=\text{Ker P + Im P}.$ How did we see this, I couldn't understand As for intersection, note that if $w$ lies in the set $\text{Ker P}\cap\text{Im P}$, then there exist a vector $v$ such that $P(v)=w$ I couldn't understand this $P(v)=w$ **, so that $P(v)=P(w)$. But since $P(w)=0$, we conclude that $w=P(v)=0$ **Why?. Hence

$$V=\text{Ker P + Im P}.$$ and $$\text{Ker P}\cap\text{Im P}=\left\{0\right\}.$$

that is it is direct sum.

I have a question that I wrote phrases in bold

Answer 1

Since $P(v−P(v))=P(v)−P^2(v)=0$, we see that $V=\operatorname{Ker} P + \operatorname{Im} P$. How did we see this, I couldn't understand

What they're showing here is that $v - P(v) \in \operatorname{Ker} P$. Just consider the definition of $\operatorname{Ker} P$; it's the set of vectors that $P$ sends to $0$. Since they're applying $P$ to $v - P(v)$ and getting $0$, this means $v - P(v) \in \operatorname{Ker} P$.

Couple this with the simple observation that $P(v) \in \operatorname{Im} P$ (it is, after all, the result of applying $P$ to an element of $V$), and this implies that $$v = \underbrace{P(v)}_{\in \; \operatorname{Im} P} + \underbrace{v - P(v)}_{\in \; \operatorname{Ker} P} \in \operatorname{Im} P + \operatorname{Ker} P.$$

As for intersection, note that if $w$ lies in the set $\operatorname{Ker} P \cap \operatorname{Im} P$, then there exist a vector $v$ such that $P(v)=w$ I couldn't understand this

If $w \in \operatorname{Ker} P \cap \operatorname{Im} P$, then $w \in \operatorname{Im} P$ (it's also true that $w \in \operatorname{Ker} P$, but we don't need this fact just yet). The image of $P$ is the set of all vectors of the form $P(v)$, where $v \in V$. This is literally the definition. To say $w \in \operatorname{Im} P$ is exactly the same thing as saying there is some $v \in V$ such that $w = P(v)$.

so that $P(v)=P(w)$. But since $P(w)=0$, we conclude that $w=P(v)=0$ Why?

We know that $w = P(v)$, by construction of $v$. Applying $P$ to both sides, using the fact that $P^2 = P$, we get $$P(w) = P^2(v) = P(v).$$ We know that $P(w) = 0$ because $w \in \operatorname{Ker} P$ (remember, $w$ is in both $\operatorname{Ker} P$ and $\operatorname{Im} P$ by assumption!) and the definition of the kernel is the set of vectors that $P$ sends to $0$. So, $P$ sends $w$ to $0$. By the above equality, this means $P(v) = 0$. But, our construction of $v$ was a vector in $V$ such that $w = P(v)$, so joining these equalities up, $$w = P(v) = 0.$$ That is, the only vector in $V$ that lies both in $\operatorname{Ker} P$ and $\operatorname{Im} P$ is the zero vector!