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Let W be a finite dimensional subspace of V and let {u_1,...,u_k}be an orthonormal basis of Q. The orthogonal projection of v onto the subspace of W is $$proj_w(v)=<v,u_1>u_1+...<v,u_k>u_k$$ Show that proj_w(v) is independent of the choice of orthonormal basis.

I started by showing that if {a_1,...,a_k} and {b_1,...,b_k} are two orthonormal basis then $$proj_w(v)=<v,a_1>a_1+...<v,a_k>a_k =proj_w(v)=<v,b_1>b_1+...<v,b_k>b_k$$and I wanted to use things like $$<u_i,u_j>=0 <u_i,u_i>=1$$ but I can't see how to use them to prove the above equation.

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  • $\begingroup$ $\text{proj}_W(v)$ is the nearest point in $W$ to $v$. $\endgroup$ – Lord Shark the Unknown Sep 17 '17 at 12:28
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Without any reference to bases, $\operatorname{proj}_W(v)$ is uniquely characterized by the property that $\operatorname{proj}_W(v)\in W$ and $\langle \operatorname{proj}_W(v)-v,w\rangle=0$ for all $w\in W$

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