What is the relationship between range and nullspace for projections?

[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

 

[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?

 

[Felipe_]

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?

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?

 

[JSuarez]

Well... maybe the thing is that I don't fully understand what I-E is, as a matter of fact.

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.

One of the things I remember is that ker(E)={0}

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.

 

[Felipe_]

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.

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)?

 

[JSuarez]

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)?

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).