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Given that

  1. $A$ is a $n \times n$ random orthogonal matrix, such that $A^TA=I$, $A^T1_n=1_n$, where $1_n$ is the vector of ones of size $n$.

  2. $X = (X_1;X_2)$ is an $n \times p$ matrix with full column rank.

  3. $X_1$ and $X_2$ are $\frac{n}{2} \times p$ matrices.

  4. $A$, $X_2$ are unknown, while $X_1$ and $B$ are known.

Can we solve $X_2$ given $A(X_1;X_2) = B$?

Rachel Woods
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  • What have you tried? What do you mean by $A^T1_n=1_n$? Is $1_n=I$? – Dietrich Burde Feb 25 '20 at 16:10
  • Welcome to Math.SE! What are your thoughts on this? You'll find that questions here receive a lot more positive attention and attract more helpful answers if make an effort to show some understanding of the problem and/or an attempt to solve it. – Brian61354270 Feb 25 '20 at 16:11
  • @Fakemistake, sorry for the confusion, I was thinking of matrix with full column rank. – Rachel Woods Feb 25 '20 at 16:16
  • @DietrichBurde, $1_n$ is the vector of ones. – Rachel Woods Feb 25 '20 at 16:17
  • Since $X$ has full column rank, $p \le n$. But how is $p$ related to $n/2$? Certainly if $p=1$ and $n$ is large you shouldn't expect to determine $X_2$. – Robert Israel Feb 25 '20 at 16:37
  • @RobertIsrael, thanks for your thoughts. What if $p=\frac{n}{2}$? – Rachel Woods Feb 25 '20 at 16:48
  • An orthogonal matrix preserves the Euclidean norm of vectors, so a necessary condition for a solution to exist is that the norm of each column of $X_1$ is at most the norm of the corresponding column of $B$. – Robert Israel Feb 25 '20 at 20:31

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Here is an example in the case $n=4$, $p=2$.

$$ X_1 = \pmatrix{1 & 0\cr 0 & 1\cr}, B = \pmatrix{1 & 0 \cr 0 & 1\cr 1 & 2\cr 3 & 4\cr} $$ Among the possibilities are

$$ A = \pmatrix{1 & 0 & 0 & 0\cr 0 & 1 & 0 & 0\cr 0 & 0 & 1 & 0\cr 0 & 0 & 0 & 1\cr},\ X_2 = \pmatrix{1 & 2\cr 3 & 4\cr} $$ and $$A = \pmatrix{1 & 0 & 0 & 0\cr 0 & 1 & 0 & 0\cr 0 & 0 & 0 & 1\cr 0 & 0 & 1 & 0\cr},\ X_2 = \pmatrix{3 & 4\cr 1 & 2\cr} $$

So no, you can't solve for $X_2$.

Robert Israel
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