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Let $A$, be a real $4 \times 4$ matrix such that $-1,1,2,-2$ are its eigenvalues. If $B=A^4-5A^2+5I$, then which of the following are true?

  1. $\det(A+B)=0$
  2. $\det (B)=1$
  3. $\operatorname{trace}(A-B)=0 $
  4. $\operatorname{trace}(A+B)=4$

Using Cayley-Hamilton I get $B=I$, and I know that $\operatorname{trace}(A+B)=\operatorname{trace}(A)+\operatorname{trace}(B)$. From these facts we can obtain easily about 2,3,4 but I am confused in 1. How can I verify (1)? Thanks for your help.

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    The matrix $A$ is diagonalizable, since it has $4$ pairwise distinct eigenvalues. The four conditions you consider are invariant under similarity. So work with the diagonalized form of $A$.2013-01-27

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