Understanding Projection in Finite-Dimensional Inner-Product Spaces

In summary, the proof shows that in a finite-dimensional inner-product space, the distance between a vector v and its projection onto a subspace W is always smaller than the distance between v and any other vector in W. If the distance is equal, then the projection must be equal to that vector in W. The proof uses Pythagoras' Theorem to show that the squared distances in (1) are equal to the squared distance in (2), which is a result of the orthogonality of the vectors involved.
  • #1
Butelle
12
0

Homework Statement



Hi - i have fully worked solutions in my notes, but i do not understand this step in the proof. Proposition 3.16. Suppose that W is a subspace of a nite-dimensional inner-product
space V and let v be an alement of V . Then ||v - projW(v)|| <= ||v - w|| for all w is an element of W. Moreover, if
w element of W and ||v - projW(v)|| = ||v - w|| then projW(v) = w.
Proof. If w is an element of W then
||v - projW(v)||^2 + ||v - projW(v)||^2 + || projW(v) - w||^2 (1)
= ||v - projW(v) + projW(v) - w||^2 (by Pythagoras' Theorem) (2)
= ||v - w||^2:

note - projW(v) is the projection of v onto W

I do not really understand how (1) implies (2)? Thanks for ur help!




The Attempt at a Solution

 
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  • #2
There is a typo in (1).

||v - projW(v)||^2 + ||v - projW(v)||^2 + || projW(v) - w||^2 (1)

should be

||v - projW(v)||^2 + || projW(v) - w||^2 (1).


Now Pythagoras says if vectors a and b are orthogonal, then

||a||^2 + ||b||^2 = ||a + b||^2.
 
  • #3
thanks!
 

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