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For example, consider the specific question: Given $a_{11},a_{12},a_{21},a_{22}$ does that uniquely determine

$A=\begin{bmatrix} a_{11}&a_{12}&a_{13} \\ a_{21}&a_{22}&a_{23} \\ a_{31}&a_{32}&a_{33} \end{bmatrix}$

where $A\in SO(3)$.

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    @MartinSleziak, unless the $(n-1) \times (n-1)$ block is already orthogonal, we can negate both the final row and the final column to get a second element of $SO_n.$ In this process, $a_{nn}$ remains the same.2012-06-09

2 Answers 2

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Hint 1: The sum of the squares of each column is $1$.

Hint 2: The third column is $\pm$ the cross product of the first two.

Hint 1 is immediately applicable to $n\times n$ matrices.

There is an $n$-dimensional analog of the cross product that extends Hint 2.

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    Ok I see that if $a_{11}^2+a_{12}^2 \neq 1$ then there are two choices for $a_{13}$. Each such value uniquely determines $a_{23}$ according to column 1 and column 2 being orthogonal. Assuming this value of $a_{23}$ is consistent with $a_{12}^2+a_{22}^2+a_{32}= 1$, then there is a 3rd column that ensures $\det A = 1$ and its negative which gives $\det A = -1$2012-06-09
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Note that if $A \in SO_n(\mathbb R)$ with

$ A \; = \; \left( \begin{array}{rr} E & F \\ G & H \end{array} \right) , $
and submatrices $F,G$ rectangular and $E,H$ both perfectly square, then $ \det E = \det H. $ I'm just saying. In your case, if you have the upper left square, it turns out that $ a_{33} = \; a_{11} a_{22} \; - \; a_{12} a_{21}. $ Put another way, if you are given the upper left block and complete it, another choice for $A \in SO_n(\mathbb R)$ is to negate column $n$ and row $n,$ which means that $a_{nn}$ is negated twice and actually stays put.

Proof: $ \left( \begin{array}{cc} E & F \\ 0 & I \end{array} \right) \left( \begin{array}{cc} E^t & G^t \\ F^t & H^t \end{array} \right) = \left( \begin{array}{cc} I & 0 \\ F^t & H^t \end{array} \right). $ Note that this is more general as both $E,H$ are allowed to have size larger than one, if $n > 3$ anyway.