# To find trace and determinant of matrix [duplicate]

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?

1. $\det (A+B)=0$
2. $\det B=1$
3. $\text{trace}(A-B)=0$.
4. $\text{trace}(A+B)=4$.

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?

-
Have you seen that the eigenvalues of $p(A)$ are $p(\text{the eigenvalues of }A)$ when $p$ is a polynomial? If not, you could prove this, and use it. Note that $B$, $A+B$, and $A-B$ are polynomials in $A$. –  Jonas Meyer May 22 '12 at 5:45
This is essentially this question! –  Arturo Magidin May 22 '12 at 5:45

## marked as duplicate by Arturo Magidin, copper.hat, Chris Eagle, Alex Becker♦, t.b.May 22 '12 at 14:16

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 $\mathrm{trace}(A)+\mathrm{trace}(B) = 0 + \mathrm{trace}(I_4) = 4$.

-
Thanks! Sorry i don't know this question is already asked. And thanks once again that now i understand this question properly. –  Kns May 22 '12 at 5:51