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So I have this matrix I am working with that looks like $$S=\begin{pmatrix}0 & -a & b\\ a & 0 & -c\\ -b & c& 0 \end{pmatrix} $$ and upon asking mathematica for the eigenvalues I get Eigenvlaues[S] $=0, \sqrt{-a^2+-b^2+-c^2},-\sqrt{-a^2+-b^2+-c^2} $. Yes the eigenvalues are imaginary, but why doesn't it simplify it to be imaginary, and how can I force it to replace those minus sines and slap an $i$ in there instead? When I go to find the Eigenvectors does it recognize the terms I have as imaginary eigenvalues and simplify as such, or do I need to force it somehow.

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    $\begingroup$ People here generally like users to post code as Mathematica code instead of just images or TeX, so they can copy-paste it. It makes it convenient for them and more likely you will get someone to help you. You may find the meta Q&A, How to copy code from Mathematica so it looks good on this site, helpful. -- Also, keep in mind a, b, and c are treated as complex numbers. -- Finally, something like this might work: PowerExpand[Sqrt[-a^2]] $\endgroup$
    – Michael E2
    Commented May 30, 2022 at 22:59
  • $\begingroup$ Oh wow I didn't know that! Okay thank you for the heads up. Could I just make the assumption in the code that a,b,c are all real? $\endgroup$
    – Joey
    Commented May 30, 2022 at 23:02
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    $\begingroup$ Try ComplexExpand. $\endgroup$ Commented May 30, 2022 at 23:04
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    $\begingroup$ Compare: Simplify[Sqrt[-a^2 - b^2], a > 0 && b > 0], Simplify[Sqrt[-a^2 - b^2], a >= 0 && b >= 0], Simplify[Sqrt[-a^2 - b^2], {a, b} \[Element] Reals], Simplify[Sqrt[-a^2 - b^2], a != 0 && b != 0 && {a, b} \[Element] Reals] -- That the case a == b == 0 seems important here looks like a weakness of Mathematica. It probably does not check whether the expression under the radical can change sign or not, probably for efficiency's sake. $\endgroup$
    – Michael E2
    Commented May 30, 2022 at 23:14
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    $\begingroup$ Mma seems to want to treat 0 as a non-imaginary number and keep it separate from the (nonzero) imaginary numbers: Assuming[a^2 + b^2 >= 0, FullSimplify@PiecewiseExpand@ComplexExpand[Sqrt[-a^2 - b^2]] ] $\endgroup$
    – Michael E2
    Commented May 30, 2022 at 23:28

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