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If we know that $B \geq 0 $ (positive semidefinite) and that $I-B \geq 0$, is it necessarily true that $I-B^2 \geq 0$?

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    $\begingroup$ This looks like homework. But since $B$ has a positive square root, we can pre- and post-multiply by that square root, and obtain $B^2 \leq B$, hence $B^2 \leq I$. $\endgroup$
    – David Handelman
    Commented Apr 10, 2017 at 13:22
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    $\begingroup$ Or simly note that all eigenvalues of $B$ are between 0 and 1, thus the same for $B^2$ $\endgroup$
    – Fedor Petrov
    Commented Apr 10, 2017 at 13:41

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Thanks for the answers; did not know that I could pre- and post multiply both sides of a matrix "inequality". I think of $C<D$ as just a symbol for $C-D$ being psd. The eigenvalue argument is great too. Not homework. My inverse-problems / geophysics text says that the difference between two variance matrices is $(I-B^2)S$ (for some $S$) and that since $I-B$ is psd so is $I-B^2$. I couldn't figure it out.

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    $\begingroup$ I think you can convert this (non)answer into a comment, or you can edit it into your question, provided you use the same account that you used when you asked the question. $\endgroup$
    – Andreas Blass
    Commented Apr 10, 2017 at 15:46

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