TITLE: How much redundancy resides in an $n \times n$ orthogonal matrix?
QUESTION [19 upvotes]: Suppose one has an $n \times n$ orthogonal matrix $M$:
$$
\left(
\begin{array}{ccc}
 0.239326 & 0.846726 &
   0.475161 \\
 0.768893 & 0.13356 &
   -0.625272 \\
 0.592897 & -0.514992 &
   0.619077 \\
\end{array}
\right)
$$
Because it is orthogonal, $M^T M = I$.
Suppose one entry of $M$ is erased, say $M(2,2)$:
$$
\left(
\begin{array}{ccc}
 0.239326 & 0.846726 &
   0.475161 \\
 0.768893 & \color{red}{x} &
   -0.625272 \\
 0.592897 & -0.514992 &
   0.619077 \\
\end{array}
\right)
$$
It can be recovered from $M^TM=I$. For example, we must have the $(2,2)$ entry 
of $M^TM$ equal to $1$:
\begin{eqnarray}
0.982162 + x^2 &=& 1 \\
x &=& \pm 0.13356
\end{eqnarray}
and then, e.g., entry $(2,1)$ of $M^TM$ disambiguates (or determines on its own):
\begin{eqnarray}
-0.102694 + 0.768893 x &=& 0 \\
x &=& 0.13356
\end{eqnarray}
My question is:

Q. What is the maximum number $k$ of entries of an 
  $n \times n$ orthogonal matrix $M$
  that can be erased and then uniquely recovered, 
  knowing only that $M$ is orthogonal?

It could well be that $k$ depends on which entries are erased, which itself
could be interesting. But I am at the moment seeking the maximum of $k$ over
all possible entries that permits exact recovery.

REPLY [2 votes]: The fact that $\frac{n(n-1)}2$ well chosen coefficients determine the other ones is supported by the Schur parametrization : let $U$ be unitary (it works as well for real orthogonal matrices) and Hessenberg (that is $i\ge j+2$ implies $u_{ij}=0$). Up to multiplication by a diagonal unitary matrix, we may assume that the diagonal of $U$ is real non-negative. Then there are unique pairs $(a_k,b_k)$ with $b_k$ real and $|a_k|^2+b_k^2=1$, such that $U=G_1\cdots G_{n-1}$ with
$$G_k={\rm diag}\left(I_{k-1},\begin{pmatrix} -a_k & b_k \\ b_k & \overline{a_k} \end{pmatrix},I_{n-k-1}\right).$$
It is a case where one gives the sub-diagonal part of $U$ (with many zeroes).The pairs are determined by the entries $u_{i,i-1}$.
Notice that in the unitary case, the dimension of the tangent space $SK_n$ over $\mathbb R$ is $n^2$, and we are given $n^2$ real data, namely  $\frac{n(n-1)}2$ complex numbers (sub-diagonal entries) and $n$ real numbers (the imaginary parts of the diagonal entries).