# Orthogonal projectors

1. Oct 1, 2009

### math8

Let P and Q be two m x m orthogonal projectors.
We show a) ||P-Q||_2 <or eq. 1
b)||P-Q||_2 < 1 implies the ranges of P and Q have equal dimensions.

I think I must use the properties of orthogonal projectors. I guess Range(P) Inters Null(P) = {0} and Range(Q) Inters Null(Q) = {0}.
Also ||P-Q||_2 = Max {|(P-Q)x|: ||x||=1|}. But I am not sure how to proceed from here.

I am seeing somewhere this lemma: ||P-Q||_2 <1 iff Range(P) Inters Null(Q) = {0} and Range(Q) Inters Null(P) = {0}.

Is it true? If yes why?

2. Oct 1, 2009

### aPhilosopher

I don't understand question (a).

I also don't know the word 'inters' in this context. Do you mean 'contains'? If by Null(Q), you mean the set of vectors that it maps to zero, then it is most certainly not trivial for any non-trivial projection operator. In fact, one has $$V = Null(Q) \oplus Range(Q)$$.

For example, in R3 for the standard basis, one has the projection operator

$$P = \begin{bmatrix}1 & 0 & 0 \\0 & 1 & 0\\0 & 0 & 0 \end{bmatrix}$$

Null(P) is generated by (0, 0, 1)T and Range(P) is generated by (1, 0, 0)T and (0, 1, 0)T.

Last edited: Oct 1, 2009
3. Oct 2, 2009

### math8

Inters means Intersection here.

4. Oct 2, 2009

### aPhilosopher

That's funny. Inters means "to bury" in common usage. That's why I thought it meant contains.

5. Oct 2, 2009

### aPhilosopher

I figured out (b)

$$v \in Ker(Q)\cap Im(P), |v| = 1 \ \Rightarrow \ (P - Q)v = v$$

One can move from there to

$$||P - Q||_{2} < 1 \ \Rightarrow Ker(Q)\cap Im(P) = \left\{0\right\}$$

Some standard projector properties and a little dimension counting close the deal.

I still don't understand question (a) though....

6. Oct 2, 2009

### jbunniii

(a) Prove that $$||P - Q||_2 \leq 1$$

Not sure how to prove it.

7. Oct 2, 2009

### aPhilosopher

Heh, I'm new to ascii math. Should have gotten that....

8. Oct 3, 2009

### aPhilosopher

To handle the first one, factorize V as

$$V = \left[ Im(P) \cap Im(Q) \right] \oplus \left[ Im(P) \cap Ker(Q) \right] \oplus \left[ Ker(P) \cap Im(Q) \right] \oplus \left[ Ker(P) \cap Ker(Q) \right]$$

So we can express a vector v = v1 + v2 + v3 +v4. We also have vivj = 0 if i isn't j by picking a basis.

From here we compare ||(P - Q)v|| to ||v|| to get the more general conclusion that ||(P - Q)v|| <or eq ||v||

I don't seem to be able to find a simpler way

Last edited: Oct 4, 2009