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Apr 24 at 7:08 vote accept respectableuser1
Apr 24 at 7:03 comment added respectableuser1 Yes I see it. Thank you very much for your time.
Apr 24 at 1:24 comment added Iosif Pinelis @respectableuser1 : It is shown here that $Q(B_*)\lesssimeq Q(B)$ for all $B$, where $\lesssimeq$ is the Loewner order. That $\lesssimeq$ is a partial order does not diminish this fact. Somewhat similarly, $0\lesssimeq B^T B$ for all $B$, even though $\lesssimeq$ is only a partial order. Do you see it now?
Apr 23 at 7:13 comment added respectableuser1 Thank you. Since the Loewner order is only a partial ordering, can we actually make the conclusion that $B_{*}$ is a minimizer (wrt. the ordering) over all $k \times p$ matrices $B$?
Apr 22 at 16:01 history edited Iosif Pinelis CC BY-SA 4.0
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Apr 22 at 15:48 history answered Iosif Pinelis CC BY-SA 4.0