I Orthogonal projections

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Hi PF!

I've been reading and it appears that the orthogonal projection of a vector ##v## to the subspace spanned by ##e_1,...,e_n## is given by $$\sum_j\langle e_j,v \rangle e_j$$ (##e_j## are unit vectors, so ignore the usual inner product denominator for simplicity) but there is never a proof in the texts. It's always given by definition or I see "trivial" next to the proof. Surely this is something we prove, right?
 

StoneTemplePython

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it really depends on what you're trying to do here, as there are lots of flavors of this. Suppose we have vectors in ##\mathbb C^m## with ##m \geq n## and use the standard inner product.

you have some vector ##\mathbf u## and you project it to a subspace with these n mutually orthornormal vectors. You can do this explicitly with ##\mathbf v = P\mathbf u## where ##P## is a projector -- i.e. ##P^2 = P## and ##P = P^*##.

so we have a generating set for this subspace and use it to write ##\mathbf v## as a linear combination of these. (Note: ignoring the zero vector: if you have an inner product, orthogonality implies linear independence, why?)

so
##\mathbf v = \sum_{j=1}^n \alpha_j \mathbf e_j##

multiply on the left by ##\mathbf e_k^*## (conjugate transpose) to see

##\mathbf e_k^*\mathbf v = \mathbf e_k^*\sum_{j=1}^n \alpha_j \mathbf e_j = \sum_{j=1}^n \alpha_j \mathbf e_k^*\mathbf e_j = \alpha_k \mathbf e_k^*\mathbf e_k = \alpha_k##

now re-write this with general inner product notation -- same result.
 
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(Note: ignoring the zero vector: if you have an inner product, orthogonality implies linear independence, why?)
This is simple to show: take ##c_1v_1+...+c_nv_n = 0## and dot this into some ##v_j##. Then orthogonality implies for non-zero ##v##, ##c_j=0##, and thus linear independence.

And thanks! Good to see it.
 
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