Rank of product of two orthogonal matrices For $U,V\in\mathbb{R}^{n\times r}$, where each column $u_i$ satisfies that $u_i^Tu_j=0,j\ne i, u_i^Tu_i=1,\forall i=1,...,r$. So is $v_i$. Suppose we have $$\Vert UU^T-VV^T \Vert<1$$
My question is how to prove the following:
$$U^TV\text{ is full rank.}$$
Thanks!
 A: We can show the following: 

$U^TV$ is rank-deficient $\implies$ $\rho(UU^T-VV^T)\geq 1$,

where $\rho$ is the spectral radius of a matrix.
Since $UU^T-VV^T$ is symmetric, the eigenvalues of this matrix are real and for 
any symmetric matrix $X$ we have $\rho^2(X)=\rho(X^2)$.
We can expand
$$\tag{1}
(UU^T-VV^T)^2=UU^T+VV^T-U(U^TV)V^T-V(V^TU)U^T
$$
and see the $U^TV$ terms to pop out.
Since $U^TV$ is rank-deficient, there is a nonzero vector $x$ such that $U^TVx=0$. We can express this vector as a combination of rows of $V$, that is, $x=V^Ty$ for some (nonzero) $y$. We can choose $x$ and $y$ such that $y^Ty=1$.
We take this $y$ and compute the associated Rayleigh quotient of the squared matrix in (1) and see that
$$
\begin{split}
\rho^2(UU^T-VV^T)
&\geq
y^T(UU^T-VV^T)^2y
\\&=
y^TUU^Ty+x^Tx-y^TU(U^TVx)-(U^TVx)^TU^Ty
\\&=
y^TUU^Ty+x^Tx\geq x^Tx.
\end{split}
$$
So if $x^Tx\geq 1$, we are done. Indeed, we have
$$
1=y^Ty=y^T(I-VV^T)y+y^TVV^Ty\leq y^TVV^Ty=x^Tx.
$$
Hence $\rho(UU^T-VV^T)\geq 1$.
We showed the following equivalent statement:

$\rho(UU^T-VV^T)<1$ $\implies$ $U^TV$ is not rank-deficient.

Any matrix norm $\|\cdot\|$ consistent with some vector norm is an upper bound on $\rho$ so we also have

$\|UU^T-VV^T\|<1$ $\implies$ $\rho(UU^T-VV^T)<1$ $\implies$ $U^TV$ is not rank-deficient.

A: Let $A=U^TV$. As $U$ has orthonormal columns, $UU^T$ is an orthogonal projection. Therefore
\begin{aligned}
\|UU^T-UAA^TU^T\|_2
&=\|UU^T(UU^T-VV^T)UU^T\|_2\\
&\le\|UU^T-VV^T\|_2\|UU^T\|_2^2\\
&=\|UU^T-VV^T\|_2\\
&=\rho(UU^T-VV^T)\\
&\le\|UU^T-VV^T\|<1.
\end{aligned}
(Note that we have used both the induced $2$-norm $\|\cdot\|_2$ and the given norm $\|\cdot\|$ above.)
Now suppose the contrary that $A$ is singular. Then $A^T$ is singular and $A^Tx=0$ for some nonzero vector $x$. Since $U$ has full column rank, $Ux\ne0$. Thus $\|(UU^T-UAA^TU^T)Ux\|_2=\|Ux\|_2\ne0$, which is a contradiction to $\|UU^T-UAA^TU^T\|_2<1$. Hence $A=U^TV$ must be nonsingular.
