Properties of $\det$ and $\operatorname{trace}$ given a $4\times 4$ real valued matrix 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?  


*

*$\det(A+B)=0$     

*$\det (B)=1$  

*$\operatorname{trace}(A-B)=0 $   

*$\operatorname{trace}(A+B)=4$  


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.

I had just been answering a question about "without loss of generality". This is a nice example. Since $A$ is diagonalisable (it's characteristic polynomial is $(X+1)(X-1)(X-2)(X+2)$ which is split without multiple roots), we may assume without loss of generality that $A$ is diagonal (do a base change to a basis of eigenvectors, which changes neither determinants nor traces). You can do that case explicitly by computation.
