Demonstrations on Projections

1. Apr 21, 2010

Felipe_

Hi!
Studying the introductory chapters of a Operator Theory book, I have found that the author seem to find a lot of demonstrations "easy" and not worthy of demonstrations. Yet, I don't have such ease as he has... For instance, one that has bugged me (on several books) is the proof that:
range(I-E)=nullspace(E)
nullspace(I-E)=range(E)
where E is a projection and I is the identity matrix.
It always follows stating that its is simple, then, to tell the set of all surjective projections of a linear transformation L.

How are they related? I can't seem to find any way to prove/answer the above and its been a couple of days now.

Can anyone shed some light in this? I'd really appreciate it!

Thanks!

Felipe

2. Apr 21, 2010

JSuarez

First, you may acquire some intuiton about it by drawing a picture of an orthogonal projection in $\mathbb{R}^3$.

Second, consider the first equality range(I - E) = ker(E). Suppose y belongs to range(I - E): then you may write it, in terms of I - E, how? What must Ey be equal to? On the other hand, if Ey = 0, what is (E - I)y?

Last edited: Apr 21, 2010
3. Apr 21, 2010

Felipe_

Well... maybe the thing is that I don't fully understand what I-E is, as a matter of fact.
One of the things I remember is that ker(E)={0}, the null vector. Or is it just for linear spaces?

4. Apr 21, 2010

JSuarez

Consider a 2D plane P in $\matbb{R}^3$ and a vector v, not in P; then, if E is the orthogonal projection on P, then (I - E)v is the orthogonal projection on normal to the plane P.

No, this is wrong. The kernel of a linear transformation $T:V\rightarrow W$ is the set:

$$ker\left(T\right)=\left\{v\in V:Tv=0\right\}$$

Where the 0 is the null vector of W.

Here's a hint:

$$y \in R \left(I - E\right) \Leftrightarrow \exists x \in V:y=\left(I - E\right)x$$

Now remember that E is a projection, so E2=E.

5. Apr 21, 2010

Felipe_

Oh yeah, I remember E2=E... used it plenty today!

So, let me try to get things straight: $$R(I - E)$$ is the collectiong of all projections normal to the plane P (like the one in the example you gave)?

6. Apr 21, 2010

JSuarez

No, R(I - E) is the range of the projection I - E (if E is a projection, I - E is also one). In euclidian spaces, you may interpret this as the subspace orthogonal to P, but the results you are trying to prove are more general than that; they are valid for linear spaces whithout a notion of orthogonality (usually given by an inner product).