Let $A$ be $4\times 4$ matrix with real entries such that $-1$, $1$, $2$, and $-2$ are its eigenvalues.
If $B = A^4 - 5A^2+5I$, where $I$ denotes $4\times 4$ identity matrix, then what would be determinant and trace of matrix $A+B$?
Let $A$ be $4\times 4$ matrix with real entries such that $-1$, $1$, $2$, and $-2$ are its eigenvalues.
If $B = A^4 - 5A^2+5I$, where $I$ denotes $4\times 4$ identity matrix, then what would be determinant and trace of matrix $A+B$?
Since $A$ is $4\times 4$ and its eigenvalues are $2$, $-2$, $1$, and $-1$, the minimal and characteristic polynomials of $A$ agree and are both equal to $(t-1)(t+1)(t-2)(t+2) = (t^2-1)(t^2-4) = t^4 - 5t^2 + 4.$ In particular, by the Cayley-Hamilton Theorem, $A^4 - 5A^2 + 4I = 0,$ and therefore $B+A = A^4 - 5A^2 + 5I + A = (A^4-5A^2+4I) + (A+I) = A+I.$
Now notice that $\lambda$ is an eigenvalue of $A$ if and only if $\alpha\lambda+\beta$ is an eigenvalue of $\alpha A+\beta I$, to conclude that the eigenvalues of $B+A=A+I$ are $0$, $-1$, $2$, and $3$. Therefore, the trace is $0-1+2+3 = 4$, and the determinant is $0$ (since $A+I$ is not invertible, or since the determinant is the product of the eigenvalues).
Pick one of the eigenvectors of $A$, call it $\bf v$, with corresponding eigenvalue $\lambda$. Can you work out $(A+B){\bf v}$? From that, can you work out the determinant and trace?
Forgive me if I mistake what Gerry is trying to say, but I believe it is this:
If $v$ is an eigenvector of $A$, then $v$ is an eigenvector of $A+B$ (what is the associated eigenvalue, then?). Use the formula for $B$, and note that $A^n(v) = \lambda^nv$.
From the associated eigenvalues you get for $A+B$, you should be able to explicitly state the characteristic polynomial for $A+B$.
For a 4x4 matrix, the trace is the negative of the coefficient of the cubic term in the characteristic polynomial, and the determinant is the constant term.
Start by finding the eigenvalues of $B$, using the eigenvalues of $A$. Find the eigenvalues of $A+B$ using this.
Then recall that the determinant here will be the product of the eigenvalues of $A+B$ and the trace will be the negative of the sum of the eigenvalues of $A+B$.