If I have a real 3x3, symmetric and positive-definite matrix $D$ (the hessian of a convex function) and $J=\left[\array{0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 }\right]$. What properties (if any) will $JD$ have?
Properties of the matrix $JD$
0
$\begingroup$
linear-algebra
-
0thanks, you are right. corrected. – 2012-07-18
1 Answers
3
You can compute $JD$ explicitely. If $D = \begin{bmatrix}a & b & c \\ b & d & e \\ c & e & f\end{bmatrix}$, then $JD = \begin{bmatrix}-b & -d & -e \\ a & b & c \\ 0 & 0 & 0\end{bmatrix}$. Properties which come to mind are:
The trace is trivial
The determinant of the left upper $2\times 2$ block coincides with the determinant of the left upper $2\times 2$ block of $D$ and is therefore positive.
The matrix has rank $2$.
If those are helpful is up to your application.
Edit: Generally I don't expect that you have any control over $e$ and $c$ in the matrix $JD$. You have some control over them depending on $f$ (since $D$ is positive definite), but since you loose $f$ when you pass to $JD$, you also loose that control.
-
0Thanks. I was hoping to use the positive-definitiveness of $D$ to get more out. Unfortunately, I just realized I can't for example say whether $JD$ is positive-semidefinite because I don't know anything about $b$. The other two principal minors are non-negative but I would still need to know the sign of $b$. – 2012-07-18