<s>[INST] To find trace and determinant of matrix 
Possible Duplicate:
Computing the trace and determinant of $A+B$, given eigenvalues of $A$ and an expression for $B$ 

Let $A$ be a $4\times 4$ matrix with real entries such that $-1,1,2,-2$ are its eigenvalues. If $B=A^{4}-5A^{2}+5I$, where $I$ denotes the $4\times 4$ identity matrix, then which of the following statements are correct?


*

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

*$\det B=1$ 

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

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


NOTE: There may one or more options correct.
I know that trace of matrix means sum of eigenvalues of matrix and determinant means product of eigenvalues. But i dont know how to apply these things in this question?
Please help me out and explain the method. 
 [/INST] The characteristic polynomial of $A$ is
$$(t+1)(t-1)(t+2)(t-2) = (t^2-1)(t^2-4) = t^4 - 5t^2 + 4.$$
Therefore, by the Cayley-Hamilton Theorem,
$$A^4 - 5A^2 + 4I = 0.$$
In particular, $B= A^4 - 5A^2 + 5I = (A^4-5A^2+4I)+I = I$.
So $B=I$, $A+B=A+I$, and $A-B=A-I$.
The eigenvalues of $A+\mu I$ are of the form $\lambda+\mu$, where $\lambda$ is an eigenvalue of $A$.
So: Since $-1$ is an eigenvalue of $A$, then $0=-1+1$ is an eigenvalue of $A+I=A+B$, so $\det(A+B)=0$.
Since $B=I$, $\det(B)=1$.
The trace of $A-B$ is the sum of the eigenvalues of $A-B$; the sum of the eigenvalues of $A-B$ is $(-1-1) + (1-1) + (2-1) + (-2-1) = -2+0+1-3 = -4$. Alternatively, it is the sum of the trace of $A$ (which is $0$, since its eigenvalues add up to $0$) and the trace of $-B$, which is $-I$, hence the trace is $-4$. 
And the trace of $A+B$ is the sum of the eigenvalues of $A+B$, which is $(-1+1) + (1+1) + (2+1) + (-2+1) = 4$. Or it is $\mathrm{trace}(A)+\mathrm{trace}(B) = 0 + \mathrm{trace}(I_4) = 4$. 
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