Suppose $L_1 \succeq L_2$, where $L_1,L_2$ are positive semidefinite matrices (actually combinatorial Laplacians). Is the following inequality true, and if no, under which conditions?
$$L_1^2 \succeq L_2^2$$
Asked
Active
Viewed 162 times
1
-
The answer for general PSD matrices is "no". See If $A^2\succ B^2$, then necessarily $A\succ B$. Need to think about the Laplacian case, though. – user1551 Aug 22 '17 at 05:38
2 Answers
2
It's not always true. Counterexample: $$ \begin{align*} &L_1=\pmatrix{1&-1&0\\ -1&2&-1\\ 0&-1&1}, \ L_2=\pmatrix{1&-1&0\\ -1&1&0\\ 0&0&0},\\ &L_1^2-L_2^2=\pmatrix{0&-1&1\\ -1&4&-3\\ 1&-3&2}. \end{align*} $$ I'm not sure if there is any good (non-restrictive) sufficient condition for the inequality to hold.
user1551
- 139,064
0
It is also not true in general in the $2\times 2$ case: consider for instance $$ L_1=\begin{bmatrix} 3&0\\0&0\end{bmatrix},\ \ \ \ \ L_2=\begin{bmatrix}4&2\\2&4\end{bmatrix}. $$ Then $$ L_2^2-L_1^2=\begin{bmatrix} 11&16\\16&20\end{bmatrix}, $$ which is not positive semidefinite.
Martin Argerami
- 205,756