# Direct sum of kernel and image of two idempotent maps

STATEMENT: Consider two linear maps $q_1,q_2:V \rightarrow V$ such that $q_1\circ q_1=q_1$ and $q_2\circ q_2=q_2$. Assume that $q_1\circ q_2=q_2\circ q_1$. Show that $q_2($ker $q_1)\subseteq$ ker $q_1$ and that $q_2(q_1(V))\subseteq q_1(V)$. Next, show that

$$V = (\text{ker } q_1 ∩ \text{ker } q_2) ⊕ (\text{ker } q_1 ∩ q_2(V )) ⊕ (q_1(V ) ∩ \text{ker }q_2) ⊕ (q_1(V ) ∩ q_2(V )).$$

QUESTION: So we have already proved another theorem that is supposed to be used to prove this one.

Theorem:Let $U, W$ be two linear subspaces of $V$. $V = U ⊕W$ if and only if there are linear maps $p_1, p_2 : V → V$such that $U = p_1(V )$ and $W = p_2(V )$ and such that $p_1 ◦ p_2 = p_2 ◦ p_1 = 0$ and $p_1 + p_2 = id_V .$

I am unsure of how to proceed given the above information to show that $$V = (\text{ker } q_1 ∩ \text{ker } q_2) ⊕ (\text{ker } q_1 ∩ q_2(V )) ⊕ (q_1(V ) ∩ \text{ker }q_2) ⊕ (q_1(V ) ∩ q_2(V )).$$

I would like some suggestions or hints to get me started in the correct direction. I will post a sketch of my ideas so far, and see if they can be put into context with the information given so far:

Proof Sketch: We note that $V=$ker $(q_1)\oplus q_1(V)$. It is therefore sufficient to show that ker$(q_1)$=$($ker $q_1 \cap$ ker $q_2)\oplus($ker $q_1 \cap q_2(V )$ and $q_1(V)=(q_1(V)\cap$ ker $q_2)\oplus (q_1(V)\cap q_2(V))$. To show the former we note that ker$(q_1\mid_{ker(q_1)})=$ker $q_1\cap$ ker $q_2$. Also $q_2\mid_{ker(q_1)}($ker$(q_1))=$ker $q_1\cap q_2(V)$. So therefore, we have show that ker$(q_1)$=$($ker $q_1 \cap$ ker $q_2)\oplus($ker $q_1 \cap q_2(V )$. Using a similar procedure we show that $q_1(V)=(q_1(V)\cap$ ker $q_2)\oplus (q_1(V)\cap q_2(V))$. We can therefore conclude that

$$V = (\text{ker } q_1 ∩ \text{ker } q_2) ⊕ (\text{ker } q_1 ∩ q_2(V )) ⊕ (q_1(V ) ∩ \text{ker }q_2) ⊕ (q_1(V ) ∩ q_2(V )).$$

Lemma 1:If $f:V\rightarrow V$ is a projective map (i.e.$f\circ f=f$), then $V=\text{ker }(f)\oplus f(V)$.
Lemma 1 proof: Since $f$ is idempotent, we have $f(v)=f(f(v))$, where $v\in V$. Thus, $f(v-f(v))=0$. This implies that $v-f(v)=w\in\text{ker}(f)$. Thus $v=w+f(v)\in \text{ker}(f)+f(V)$. Now let $m\in \text{ker}(f)\cap f(V)$. The $\exists v\in V$ such that $m=f(v)$. Applying $f$ to both sides we see that $0=f(m)=f(f(v))=f(v)=m$. Thus, we see the intersection of $\text{ker}(f)\cap f(V)$ is trivial. Therefore, $V=\text{ker }(f)\oplus f(V)$.
Proof of original problem: We note that since $p_1\circ p_1=p_1$ then $V=$ker $(q_1)\oplus q_1(V)$ by lemma 1. It is therefore sufficient to show that ker$(q_1)$=$($ker $q_1 \cap$ ker $q_2)\oplus($ker $q_1 \cap q_2(V )$ and $q_1(V)=(q_1(V)\cap$ ker $q_2)\oplus (q_1(V)\cap q_2(V))$. To show the former we note that ker$(q_1\mid_{ker(q_1)})=$ker $q_1\cap$ ker $q_2$. Also $q_2\mid_{ker(q_1)}($ker$(q_1))=$ker $q_1\cap q_2(V)$. Because the restriction of an idempotent map is idempotent, we have shown that ker$(q_1)$=$($ker $q_1 \cap$ ker $q_2)\oplus($ker $q_1 \cap q_2(V )$. Using a similar procedure we show that $q_1(V)=(q_1(V)\cap$ ker $q_2)\oplus (q_1(V)\cap q_2(V))$ by restricting domain and codomain of $q_2$ to $q_1(V)$. We can therefore conclude that
$$V = (\text{ker } q_1 ∩ \text{ker } q_2) ⊕ (\text{ker } q_1 ∩ q_2(V )) ⊕ (q_1(V ) ∩ \text{ker }q_2) ⊕ (q_1(V ) ∩ q_2(V )).$$