Failing at projection question

[reha]

let pf(x)= sum( from i=1 to k) <x, ui>ui, show pf is a projection.

Ive tried to show this fact myself but i failed. Please some one help me out. thanks

Note ui = u1...un an orthogonal basis of V where V is a vector space.

 

[Pere Callahan]

Welcome to PF, reha.

The usual procedure in showing that a given function is a projection, is to verify that is satisfies the definition of a projection. (By the way, that's true for many other things as well.)

So, what's the definition of a projection in a vector space?

 

[Architeuthis Dux]

It is understandable that you are having difficulty showing that pf(x) is a projection. However, it is important to note that pf(x) is a projection if and only if it satisfies the following properties:

1. pf(x) is idempotent, meaning that pf(pf(x)) = pf(x)
2. pf(x) is self-adjoint, meaning that <pf(x), y> = <x, pf(y)> for all x, y in V
3. pf(x) is a linear transformation

To show that pf(x) is a projection, we can use the fact that the basis {u1, u2, ..., uk} is orthogonal. This means that <ui, uj> = 0 for all i ≠ j. Using this fact, we can show that pf(x) satisfies all three properties listed above:

1. Since the basis is orthogonal, we have that pf(pf(x)) = sum(from i=1 to k) <pf(x), ui>ui = sum(from i=1 to k) <x, ui><ui, ui> = sum(from i=1 to k) <x, ui>ui = pf(x). Therefore, pf(x) is idempotent.

2. Similarly, we have that <pf(x), y> = sum(from i=1 to k) <x, ui><ui, y> = sum(from i=1 to k) <x, ui><y, ui> = <x, sum(from i=1 to k) <y, ui>ui> = <x, pf(y)>. Therefore, pf(x) is self-adjoint.

3. Finally, since pf(x) is a linear combination of the basis vectors ui, it is a linear transformation.

Therefore, we can conclude that pf(x) is indeed a projection. I hope this explanation helps you understand and show this fact yourself. If you still have any doubts or questions, please feel free to ask for further clarification. Keep up the good work!