I have discovered an equality that seems to be true when tested computationally, but seems false analytically.
Can anyone present a proof or counterexample?
Let $V$ be a $n$ by $(n-1)$ matrix having real entries, and let $V^{T}V$ be invertible. I claim that:
$||VV{^T} - V(V^{T}V)^{-1}V^T|| = ||I - V^TV||.$
Hint: As user1551's comment suggets, SVD is a nice approach here. Because $V^TV$ is invertible, $V$ has linearly independent columns. So, $V$ must be $m \times n$ with $m \geq n$. We note that if $V = U\Sigma W^T$ is a singular value decomposition, we have $$ I - V^TV = I - W\Sigma^T\Sigma W^T = W[I - \Sigma^T \Sigma]W^T $$ and $$ VV^T - V(V^TV)^{-1}V^T= U\Sigma \Sigma^T U^T - U\Sigma W^T (W\Sigma^T\Sigma W^T)^{-1} W\Sigma^T U^T \\ =U\Sigma \Sigma^T U^T - U\Sigma W^TW (\Sigma^T\Sigma)^{-1}W^T W\Sigma^T U^T\\ \\ =U\Sigma \Sigma^T U^T - U\Sigma(\Sigma^T\Sigma)^{-1} \Sigma^T U^T\\ = U[\Sigma \Sigma^T - \Sigma(\Sigma^T\Sigma)^{-1}\Sigma^T]U^T. $$ That is, it suffices to show that $$ \|I - \Sigma^T\Sigma\| = \|\Sigma \Sigma^T - \Sigma(\Sigma^T\Sigma)^{-1}\Sigma^T\|, $$ which can be done by direct computation since $\Sigma$ is a diagonal $m \times n$ matrix.