Projecting points onto a plane

Question

Let $P$ be a plane (not necessarily a hyperplane) in $\mathbb{R}^n$ passing through the origin. Let $\{e_i\}_{i=1}^n$ be the standard coordinate basis for $\mathbb{R}^n$ and $\{e_i^*\}_{i=1}^n$ be their orthogonal projections onto $P$. For any $p\in P$ do we always have the expansion $$p=\sum_{i=1}^n\langle p,e_i^*\rangle e_i^*$$ where $\langle\cdot,\cdot\rangle$ is the Euclidean inner product?

As an example the line $f(x)=-x$ in $\mathbb{R}^2$ yields the orthogonal projections $e_1^*=(1/2,-1/2)$ and $e_2^*=(-1/2,1/2)$ from $e_1=(1,0)$ and $e_2=(0,1)$ respectively. In this example it is easy to check that the above formula holds for any $p$ on that line.

Answer 1

For each $i$ write $e_i=e_i^*+h_i$ where $h_i \perp P$. Since $e_i$ is an orthogonal basis, then you can write $$p=\sum_{i=1}^n\langle p,e_i\rangle e_i$$ If you assume that $p\in P$ then $\langle p,e_i\rangle=\langle p,e_i^*\rangle+\langle p,h_i\rangle=\langle p,e_i^*\rangle$. Projecting the equation above to $P$ will give you $$p=\pi(p)=\sum_{i=1}^n\langle p,e_i\rangle \pi(e_i)=\sum_{i=1}^n\langle p,e_i^*\rangle e_i^*$$