# Special orthogonal matrix uniquely determined by $n-1 \times n-1$ entries?

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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Both $\begin{pmatrix}1&0&0\\0&1&0\\0&0&\pm1\end{pmatrix}$ are orthogonal, so the matrix is not determined uniquely. – Martin Sleziak Jun 9 '12 at 4:41
@MartinSleziak With $-1$ in the lower right corner, the determinant is $-1$ and hence we do not have an element of $SO$. – anon Jun 9 '12 at 4:44
I see. I was thinking $O(3)$ instead of $SO(3)$. The title only mentioned orthogonal matrices - I should have read the body of the question more carefully. – Martin Sleziak Jun 9 '12 at 4:46
I changed the headline accordingly. I assumed this is okay with the OP. – Hauke Strasdat Jun 9 '12 at 10:34
@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. – Will Jagy Jun 9 '12 at 16:05

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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For the last point: given $n-1$ vectors $v_1,\cdots,v_{n-1}$, we can pick a $w$ from the orthogonal complement of the generated subspace, ie $w\in\langle v_1,\cdots,v_{n-1}\rangle^\perp$, such that $\|w\|$ is equal to the magnitude of the Grammian determinant of the $v_i$'s. – anon Jun 9 '12 at 4:38
@anon: if the vectors $v_1,\dots,v_{n-1}$ are orthonormal, then yes, let $v_n$ be column $n$ of the cofactor matrix (which does not depend on column $n$). – robjohn Jun 9 '12 at 4:52
@anon: oh, duh! You were making a statement, not asking a question. :-) – robjohn Jun 9 '12 at 5:04
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 $\det A = -1$ – user782220 Jun 9 '12 at 5:35

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.

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