1
$\begingroup$

How do I solve for $X$ in the matrix equation (with $^T$ representing the $transpose$ operation):

$X^T~ J ~X=\Phi$,

where $X$ is a $2 \times 5$ matrix, "metric" $J_{2\times 2}=\begin{bmatrix} 0 & 1 \\ -1 & 0 \\ \end{bmatrix}$, and $\Phi_{5\times 5}$ being a known anti-symmetric/skew-symmetric square matrix with all real entries?

$\endgroup$

1 Answer 1

1
$\begingroup$

Since $J$ has rank 2, the equation in question is solvable only if the rank of $\Phi$ is either 0 or 2.

Suppose $\Phi$ is a rank-2 skew-symmetric matrix first. Then there exists a real orthogonal matrix $Q$ such that $Q^T\Phi Q=aJ\oplus 0_{3\times3}$ for some $a\ne0$. Let $Y=XQ=(\mathbf y_1,\mathbf y_2,\ldots,\mathbf y_5)$. The equation in question is then equivalent to $Y^TJY=aJ\oplus 0_{3\times3}$, which amounts to the system of equations \begin{cases} \det(\mathbf y_1,\mathbf y_2)=a,\\ \det(\mathbf y_i,\mathbf y_j)=0 \text{ when } i\notin\{1,2\} \text{ or } j\notin\{1,2\}. \end{cases} The first equation implies that $\mathbf y_1$ and $\mathbf y_2$ are linearly independent. In turn, the second one implies that $\mathbf y_3=\mathbf y_4=\mathbf y_5=0$. Therefore the general solution is given by $Y=(M,0_{2\times 3})$ or $X=(M,0_{2\times 3})\,Q^T$, where $M$ is any $2\times2$ matrix of determinant $a$.

Next we consider the case $\Phi=0$. By an analogous argument to the above, we see that the general solution is given by any matrix $X$ of rank at most 1.

$\endgroup$
6
  • $\begingroup$ Can you be more explicit on how to find those $Q$ and $M$ matrices? I have got to deal generally only with rank $2$ $\Phi$'s. $\endgroup$ Aug 25, 2016 at 8:36
  • $\begingroup$ @VineethBhaskara Do you know how to diagonalise a matrix? If $\Phi$ has rank 2, its square would be $-a^2I_2\oplus 0_{3\times3}$. The columns of $Q$ are just a set of orthonormal eigenvectors of $\Phi^2$. $\endgroup$
    – user1551
    Aug 25, 2016 at 10:24
  • $\begingroup$ Can you please explain how the fact that the metric-like $J$ being of rank $2$ restricts the rank of $\Phi$ to be at most $2$ for the solution to exist? What if you had the identity 2x2 matrix $I$ sitting there instead of $J$? $\endgroup$ Aug 25, 2016 at 22:20
  • $\begingroup$ A couple of clarifications please: 1) does $(M,0_{2 \times 3})$ mean concatenating the matrix $M$ with the 2x3 all-zero-matrix side-by-side to result in a $2\times 5$ matrix? $\endgroup$ Aug 25, 2016 at 23:03
  • $\begingroup$ 2) "$M$ can be any $2\times 2$ matrix of det = $a$'' -- means that there are an infinite solutions possible for $X$? But, we have 10 unknowns (elements of $X$) and 10 knowns (upper triangle of $\Phi$) - so having 10 unknowns with 10 equations. Why aren't we then having an unique solution for $X$ in the case where rank of $\Phi$=$2$? Thanks a bunch. $\endgroup$ Aug 25, 2016 at 23:06

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .