# Prove $p^2=p$ and $qp=0$

I am not really aware what's going on in this question. I appreciate your help.

Let $U$ be a vector space over a field $F$ and $p, q: U \rightarrow U$ linear maps. Assume $p+q = \text{id}_U$ and $pq=0$. Let $K=\ker(p)$ and $L=\ker(q)$.

Prove $p^2=p$ and $qp=0$.

$$p(p+q) = p(\text{id}_U) \Rightarrow p^2+pq=p \Rightarrow p^2 =p.$$ I actually first found this by letting $(\text{id}_U) =1$, although probably a wrong to do it.

Since $p^2 =p$ and $q$ is also a linear map defined by $q: U \rightarrow U$ then $q^2=q$. Again, I'm not sure if this actually is the right way to do it.

Then we have $$q(p+q) = q(\text{id}_U) \Rightarrow q^2+qp=q \Rightarrow q^2+ qp= q^2 \Rightarrow qp=0.$$

Prove $K=\text{im}(q)$.

For this question, I honestly do not know how to tackle it from the following definitions.

$K = \ker(p)=\left\{u \in U \ |\ p(u)=0_U\right\}$ and $\text{im}(q)=\left\{q(u) \ | \ u \in U \right\}$.

What's the correct way of doing these questions? Thank you for your time.

• I can't really tell when you're multiplying functions and when you're composing them. – Git Gud May 30 '13 at 1:01
• First treating $id_U=1$ is correct. Because when you multiply a matrix by identity matrix, you got the matrix. So you can just ignore identity matrix during matrix multiplication, as you ignore 1 during multiplication. – 1LiterTears May 30 '13 at 1:03
• @MathSnail That would depended on what the OP means by $1$. – Git Gud May 30 '13 at 1:04
• @GitGud Hi, I simply answered the earlier question "I actually first found this by letting (idU)=1, although probably a wrong to do it." – 1LiterTears May 30 '13 at 1:05
• To answer the following (sub) question, this statement is correct: "Since $p^2 =p$ and $q$ is also a linear map defined by $q: U \rightarrow U$ then $q^2=q$. But I believe the reasoning is the following computation similar to the previous one. The claim you made "because $q$ is also a linear map" justified the commutativity of addition for linear maps, a.k.a $p+q = q+p$: $$q(q+p) = q(p+q) = q(\text{id}_U) \Rightarrow qp + q^2=q \Rightarrow q^2 =q.$$ – 1LiterTears May 30 '13 at 1:16

## 3 Answers

Since $\mbox{id}_U$ is the unit of the ring $L(U)$, it is standard to denote it by $1$.

So you know that $p$ is idempotent ($p^2=p$). It follows automatically that $q=1-p$ is idempotent as $q^2=(1-p)^2=1^2-p-p+p^2=1-2p+p=1-p=q$. And then $qp=(1-p)p=p-p^2=0$ as well.

Such a pair $(p,q)$ is the $n=2$ case of what is called an orthogonal decomposition of the unit. When such a thing happens, we have the Peirce decomposition which is very handy for many purposes. In operator theory/algebras in particular.

For any given idempotent $p$, the range and the nullspace yield a direct sum decomposition of the vector space $V=\mbox{im} p\oplus \ker p$. Conversely, a direct sum decomposition of $V=I\oplus K$ defines a unique idempotent such that $\mbox{im} p=I$ and $\ker p=K$. Cf. remark below. Now let us answer your question.

The relation between the idempotents $p$ and $q=1-p$ is $$\mbox{im }(1-p)=\ker p\qquad \ker (1-p)=\mbox{im} \,p.$$

Proof: by symmetry, it suffices to prove one of these two properties. Let us do the lhs. Since $0=p-p^2=p(1-p)$, we get $p(1-p)x=0$ for every $x$ whence $\mbox{im}(1-p)\subseteq \ker p$. Now take $x\in \ker p$. Then $x=x-px=(1-p)x$ belongs to $\mbox{im} (1-p)$. QED.

Remark: it is also useful to remember that the idempotents are exactly the diagonalizable operators with spectrum in $\{0,1\}$. If $p$ is nontrivial ($p\ne 0,1$), we have $\ker p$ the eigenspace of $0$, and $\mbox{im} p=\ker(1-p)$ the eigenspace of $1$. That is $\mbox{im} p=\{x\in V\,;\,px=x\}$, as opposed to $\ker p=\{x\in V\,;\,px=0$.

• Thank you for your answer. Would it be a better way to answer the first question if I write something as follows? Let $u$ be some vector where $u \in U$. Then start the proof by writing $(p(p+q))(u)$ or $p(u)(p(u)+q(u))$? Is that necessary, and which of them is the correct way to write it? – user4167 May 30 '13 at 1:35
• @user4167 If you don't need to take vectors and can do it algebraically with the operators, that's better in general. So $p=p1=p(p+q)=p^2+pq=p^2$ is the best possible argument here. Nicely done. – Julien May 30 '13 at 1:37

First treating id$_U$=1 is correct. Because when you multiply a matrix by identity matrix, you got the matrix. So you can just ignore identity matrix during matrix multiplication, as you ignore 1 during multiplication.

Since $p^2 =p$ and $q$ is also a linear map defined by $q: U \rightarrow U$ then $q^2=q$.

But I believe the reasoning is the following computation similar to the previous one. The claim you made "because $q$ is also a linear map" justified the commutativity of addition for linear maps, a.k.a $p+q = q+p$:

$$q(q+p) = q(p+q) = q(\text{id}_U) \Rightarrow qp + q^2=q \Rightarrow q^2 =q.$$

Then following your own computation we're done.

You did the questions correctly (see below). It might be slightly more common to capitalize $P$ and $Q$, and it would be useful to see these as projections onto orthogonal subspaces to ge geometric intution. For instance, you could take $P : R^2 \rightarrow R^2$ to be $P(x,y) = (x,0)$ and $Q(x,y) = (0,y)$. In this case, the vector space is finite dimensional and the operators are defined on a basis. I find the best way to convince myself of the kind of operations you're carrying out is to say, "if $P$ and $Q$ both do the same thing to any $u \in U$, then $P = Q$."

To finish your question, we show $im(Q) = K = ker(P)$.

i) $im(Q) \subset ker(P)$. Let $v \in im(Q)$. Then $v = Qw$ for some $w \in U$. Then $Pv = PQw = 0w = 0.$

ii) $ker(P) \subset im(Q)$. Let $v \in ker(P)$. Then $v = (P+Q)v = Pv + Qv = 0 + Qv$, so in fact $v = Qv$.

To address something mentioned in the comments: you might be treating the linear maps as matrices and implicitly using the fact that matrix multiplication corresponds to operator composition. In other words, I can define an operator $P$ on $\mathbb{R}^2$ that, say, rotates clockwise by $\pi/4.$ This is a linear map on $\mathbb{R}^2$, so if I pick a basis $\mathcal{B}$, I can represent $P$ by its standard matrix $[P]_\mathcal{B}$. Then if I have any other operator $Q$, $(P \circ Q)(v) = [P]_\mathcal{B}[B]_\mathcal{B}v_\mathcal{B}$.