Orthogonal Projections and Orthogonal Bases I have come across this idea over and over but cannot understand why it is true.
For an inner product space $A$ over the field $K$, where $B \subset A$ and a fixed $\{b_1 ... b_n\}$ is an orthogonal basis for $B$, the projection of any $a \in A$ onto $B$ is independent of the choice of $\{b_1 ... b_n\}$.
The explanations I've found use these properties of $A$ and $a$.
$A = B \oplus B^{\perp}$
$a = b + (a-proj_{B}a)$
I'm confused by how a discussion about the basis of a subspace relates to the idea of larger inner product space. Would you be able to point me in the right direction?
 A: Recall from Calculus that you defined the orthogonal projection of a point $\vec{p}$ onto a line or a plane is the unique point $\vec{q}$ on that line or plane such that $\vec{q}-\vec{p}$ is orthogonal to the line or plane. The orthogonal projection and the closest point projection are the same.
Orthorthogonal projection generalizes to any dimension of inner product space $X$. Specifically, if $M$ is a subspace of $X$, then the orthogonal projection of $x \in X$ onto $M$ is the unique $m\in M$ such that $(x-m)\perp M$. And a closest point projection $m'\in M$ of $x$ onto $M$ is the unique $m'\in M$ such that $\|x-m'\| \le \|x-m\|$ holds for all $m\in M$. It turns out that the closest point projection of $x$ onto $M$ exists iff the orthogonal projection of $x$ onto $M$ exists, and these two are the same if they exist. If $M$ is finite-dimensional, then the orthogonal projection of $x$ onto $M$ exists for all $x\in X$ because it may be constructed using an orthonormal basis $\{ e_1,e_2,\cdots,e_n \}$ of $M$; indeed, the orthogonal projection of $x$ onto $M$ is $P_Mx = \sum_{j=1}^{n}\langle x,e_j\rangle e_j$. More generally, if $M$ is a complete subspace of $X$, then $P_M$ exists.
