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Mar 29 at 17:20 comment added Nathaniel Johnston @Derek - $Q_0$ being positive definite (and hence invertible) does not imply that $Q$ is invertible. For example, suppose $Q_0$ is $1 \times 1$ and equal to the scalar $1$, and $Q$ is the $2\times 2$ all-ones matrix. The inequality Q > Q0+Q0+...Q0 is not true in general.
Mar 29 at 15:35 history edited YCor CC BY-SA 4.0
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S Mar 29 at 14:03 review First questions
Mar 29 at 17:22
S Mar 29 at 14:03 history asked Benjamin Tennyson CC BY-SA 4.0