1. ## General Question about Image

I'm not quite sure if I'll butcher the wording of this question, but here goes...

How would I prove that given a linear map T:V-->V, such that the composition T^2 = I, and also given P and Q s.t. P = (I+T)/2 and Q = (I-T)/2, that Ker P = Im Q, and Im P = Ker Q?

I think I'm just a little fuzzy about the whole concept of how to prove that something is in the image of P or Q. I understand Ker P is when (I+T)/2 = 0, and when I solve for T and plugin to Q, I get Q = I. Does this mean that it is in the image of Q? Or am I missing something here? .

Also, as an aside, what would P and Q be called in this question? Not a "composition," right?

2. Originally Posted by limddavid
I'm not quite sure if I'll butcher the wording of this question, but here goes...

How would I prove that given a linear map T:V-->V, such that the composition T^2 = I, and also given P and Q s.t. P = (I+T)/2 and Q = (I-T)/2, that Ker P = Im Q, and Im P = Ker Q?

I think I'm just a little fuzzy about the whole concept of how to prove that something is in the image of P or Q. I understand Ker P is when (I+T)/2 = 0, and when I solve for T and plugin to Q, I get Q = I. Does this mean that it is in the image of Q? Or am I missing something here? .

Also, as an aside, what would P and Q be called in this question? Not a "composition," right?

First notice that for any $\displaystyle v \in V$

$\displaystyle T^2(v)=T(T(v))=I(v)=v$ and

$\displaystyle P(v)=\frac{1}{2}I(v)+\frac{1}{2}T(v)$

$\displaystyle Q(v)=\frac{1}{2}I(v)-\frac{1}{2}T(v)$

$\displaystyle PQ(v)=P(\frac{1}{2}I(v)-\frac{1}{2}T(v))=\frac{1}{2}P(I(v))-\frac{1}{2}P(T(v))=$

$\displaystyle \frac{1}{4}I(I(v))-\frac{1}{4}T(I(v))+\frac{1}{4}I(T(v))-\frac{1}{4}T(T(v))$

$\displaystyle \frac{1}{4}v-\frac{1}{4}T(v)+\frac{1}{4}T(v)-\frac{1}{4}T^2(v))$

$\displaystyle \frac{1}{4}v-\frac{1}{4}T(v)+\frac{1}{4}T(v)-\frac{1}{4}v=0$

A similar computation shows

$\displaystyle QP(v)=0$

To show two sets are equal $\displaystyle A=B$

Show that

$\displaystyle A \subset B$ and $\displaystyle B \subset A$

let $\displaystyle v \in$Ker$\displaystyle (P)$ then $\displaystyle P(v)=0$ and

$\displaystyle Q(P(v))=Q(0)=0 \in$ Im$\displaystyle (Q)$

This gives Ker$\displaystyle (P) \subset$ Im$\displaystyle (Q)$

Let $\displaystyle w \in$IM$\displaystyle (Q)$ then $\displaystyle w=Q(v)$ for some $\displaystyle v \in V$

$\displaystyle PQ(v)=0=P(w) \implies w \in$Ker$\displaystyle (P)$

So the two sets are equal.

This should get you started.