How Do Orthogonal Projectors Influence Dimensions and Norms?

[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?

 

[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.

 

[math8]

Inters means Intersection here.

 

[aPhilosopher]

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

 

[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...

 

[jbunniii]

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

I read it as

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

Not sure how to prove it.

 

[aPhilosopher]

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

 

[aPhilosopher]

To handle the first one, factorize V as

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

So we can express a vector v = v₁ + v₂ + v₃ +v₄. We also have v_iv_j = 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