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.

Well, perhaps we can pass on to the Jordan Canonical form of $\,A\,$:
$$J_A=\begin{pmatrix}1&0&0&0\\0&\!\!\!-1&0&0\\0&0&2&0\\0&0&0&\!\!\!-2\end{pmatrix}\Longrightarrow B=J_A^4-5J_A^2+5I=\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}$$
We can do the above since determinant and trace are invariant under similarity.
