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Given that: $detA > 0$ and $detB > 0$, is it the case that $det(A+B) \ge 0$?

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    $\begingroup$ why do you think that is a possibility?? the most basic example is... $A=I$ $B=-I$.. $\endgroup$
    – user87543
    Nov 23, 2013 at 5:52
  • $\begingroup$ Good point, but would det(A+B) >= 0? (Would it be positive?) $\endgroup$ Nov 23, 2013 at 5:57
  • $\begingroup$ i am not sure who upvoted my comment but, I did not mention the size of the matrix $I$ which is necessary for $\det B >0$... :D but i hope the user got the idea $\endgroup$
    – user87543
    Nov 23, 2013 at 6:00
  • $\begingroup$ that is up to you to check for positiveness.... $\endgroup$
    – user87543
    Nov 23, 2013 at 6:02
  • $\begingroup$ Use a basic two by two example (i.e. the identity and the negative identity as suggested.) Then add them. See what happens. $\endgroup$
    – Vladhagen
    Nov 23, 2013 at 6:11

1 Answer 1

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$\det\left(\begin{bmatrix}2&1\\ 1&2\end{bmatrix}+\begin{bmatrix}-1&1\\ 0&-1\end{bmatrix}\right)=\det\begin{bmatrix}1&2\\ 1&1\end{bmatrix}=-1$.

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