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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?

  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?

Please help me out and explain the method.


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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ 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 $A$. $\endgroup$ – Jonas Meyer May 22 '12 at 5:45
  • $\begingroup$ This is essentially this question! $\endgroup$ – Arturo Magidin May 22 '12 at 5:45

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

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

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