Projections of standard basis

Question

Suppose that we are given the standard basis $e_1,\ldots, e_n$ is $\mathbb{R}^n$, where $n$ is odd. Is there an orthogonal projection $P$ to two dimensional subspace and a positive integer $k$ such that $Pe_i=u$ for $1\leq i\leq k$ and $Pe_j=v$ for $k+1\leq j\leq n$ where $u,v$ are orthogonal and have the same length?

Answer 1

When $k\in\{1,\dots,n-1\}$, the image of the proposed linear map would be $\operatorname{span}(u,v)$. Let's find the kernel. For $x=\lambda_1 e_1+\dots+\lambda_n e_n$ we have $$ Px = (\lambda_1+\dots+\lambda_k) u + (\lambda_{k+1}+\dots+\lambda_n)v, $$ so that $Px=0$ is equivalent to $\lambda_1+\dots+\lambda_k=0$ and $\lambda_{k+1}+\dots+\lambda_n=0$, since $u$ and $v$ are linearly independent. We conclude that $$ \ker(P) = \operatorname{span}(e_i-e_{i+1} : i=1,\dots,k-1,k+1,\dots,n-1). $$ The orthogonal complement of $\ker(P)$ consists of all vectors of the form $$ \lambda(e_1+\dots+e_k)+\mu(e_{k+1}+\dots+e_n), $$ so in order for kernel and image to be orthgonal, $u$ and $v$ have to be of this form. Let $u$ be given by choosing $\lambda_u, \mu_u$ and $v$ by choosing $\lambda_v,\mu_v$ in the expression above. In order for $P$ to be a projection, we need to have \begin{align*} u &= Pu = k\lambda_u u + (n-k)\mu_u v, \\ v &= Pv = k\lambda_v u + (n-k)\mu_v v. \end{align*} Since $u,v$ are linearly independent, this uniquely determines $u$ and $v$ as \begin{align*} u &=\frac 1 k (e_1+\dots+e_k), \\ v &=\frac 1 {n-k} (e_{k+1}+\dots+e_k). \end{align*} Those vectors are in fact orthogonal, but their square norms are $\frac 1 k$ and $\frac 1 {n-k}$ respectively, so they can only have the same length when $n=2k$.

Hence, there is no such projection when $n$ is odd.