Subspaces of $\mathbb{R}^n$, related to Grassmannian

Question

This question arose when I was trying to understand a proof related to the Grassmannian of $\mathbb{R}^n$.

Let $P=\langle e_1,\ldots,e_k\rangle$ and $Q=\langle e_{k+1},\ldots,e_n\rangle$, where $\{e_1,\ldots,e_n\}$ is the standard basis of $\mathbb{R}^n$. Then $P\oplus Q=\mathbb{R}^n$.

Define projections $\pi_P:\mathbb{R}^n\to P$, $\pi_Q:\mathbb{R}^n\to Q$.

Suppose $S\subset\mathbb{R}^n$ is a $k$-dimensional subspace which intersects $Q$ trivially.

I know that $\pi_P|_S$ is an isomorphism.

Why is it true that there are $a_1,\ldots,a_k\in Q$ such that $e_i+a_i\in S$ for all $i\in\{1,\ldots,k\}$ ?

Answer 1

Hint The key observation is that $S$ is a complement of $Q$ in $\Bbb R^n$, so that any vector in $\Bbb R^n$ (in particular, the basis vectors $e_1, \ldots e_k$) can be written as a sum of a vector in $Q$ and vector in $S$.

Answer 2

To help develop intuition, take a small example: the $x$-axis, $y$-axis for $P,Q$, and $S=\{t(1,1):t\in\Bbb R\}$. Then $e_1+e_2=(1,1)$.

Further, let $S_\theta=\{t(\cos\theta, \sin\theta):t\in \Bbb R$. Then any $\theta \ne\pi/2$ works. Then $e_1+\tan\theta\cdot e_2\in S_\theta$.

So this appears true.

Since $P\oplus Q=\Bbb R^n$, there have to be such $a_i$. If $S=P$, they can all be zero.

I guess I have done the base case for a proof by induction.

Answer 3

Since $f = \pi_P\mid_S$ is isomrophism let $a_i = f^{-1}\left(e_i\right) - e_i$. So $\pi_P\left(a_i\right)=0$ this proves that $a_i \in Q$. Your claim follows immediately.