Proving Orthogonal Projection and Norm using Inner Products

[Zoe-b]

Homework Statement

Let U be the orthogonal complement of a subspace W of a real inner product space V.
Have already shown that T is a projection along a subspace W onto U, and that V is the direct sum of W and U.

The questions now says: show
||T(v)|| = inf (w in W) || v - w ||

Homework Equations

I have some vague notion that in R^3 say, an orthogonal projection can be used to find the shortest distance between a plane and a point. I have absolutely no idea how to prove this using inner products though.

The Attempt at a Solution

if we write T' for the projection along U onto W, then we have:

v = (T + T')(v)

T(v) = v - T'(v)

now T'(v) is in W, but I don't know how to show it is the w that minimises || v - w ||

Any suggestions for resources would also be welcome- this is not in my notes at all and google hasn't been that helpful :P

 

[susskind_leon]

There is a theorem called 'projection theorem' which gives you exactly that, but it only works in Hilbert spaces, so I'm not sure if it's general enough.

 

[micromass]

Let w_0=v-T(v), then w is supposed to be the point in W which is closest to v. Can you prove that for each w in W holds that

\|v-w_0\|\leq \|v-w\|

Hint: draw a picture and see if you get something perpendicular. Use Pythagoras theorem.

 

[Zoe-b]

Thanks for the replies- Hilbert spaces aren't on my course yet so I don't think that's the way to go. I'm trying to use the second hint and do it with a diagram but not getting that far- also this is in any real inner product space not necessarily R^n..

 

[micromass]

T(v) is orthogonal to W, right??

So v-w₀, w-w₀ and v-w forms a right triangle.

 

[Zoe-b]

Yeah done it now, thanks! I guess I did use pythagoras- Just conceptually not that happy with drawing triangles when the elements I'm using aren't necessarily vectors? Anyway, done it using just inner product notation and now happy :)