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.

The characteristic equation of $A$ is given by $(t-1)(t+1)(t+2)(t-2)=0 $ which implies $t^{4}-5t^{2}+4=0$. Now $A$ must satisfy its characteristic equation which gives that $A^{4}-5A^{2}+4I=0$ and so we see that $B=A^{4}-5A^{2}+4I+I=0+I=I$. Hence, the eigenvalues of $(A+B)$ is given by  $(-1+1),(1+1),(2+1),(-2+1)$  that is $0,2,3,-1.$[Without the loss of generality, one can take $A$  to be diagonal matrix which would not change trace or determinant of the matrix. ]So we see that $det(A+B)$ is the product of its eigenvalues which is $0$. . Also we see that trace of $(A+B)$ is the sum of its eigenvalues which is $(0+2+3-1)=4.$ Also, B being the identity matrix, $det(B)=1.$ So the options $(1),(2) and (4)$ are true .
