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?
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.