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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 $A+B$?

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Please state all conditions in the title, lest it causes confusion, thanks. P.S. I did not downvote, and knew not why it is voted, but I might surmise... – awllower May 11 '12 at 15:41
Dear sir there is limited word alloted for title. Thanks for pointing my mistakes. I will take care of that in near future. Being a new member i am not much aware of rules and regulations. But soon i wll get awre of all the rules. – srijan May 11 '12 at 16:16
up vote 6 down vote accepted

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).

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Sincerely thanks to you for your help and wonderfull way of tackling this problem. – srijan May 11 '12 at 16:54

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?

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Why my question is being down voted? I beg pardon if i did something wrong to this community. – srijan May 11 '12 at 13:22
Thanks for being lenient. – srijan May 11 '12 at 13:25
I didn't get you sir. – srijan May 11 '12 at 13:28
OK, let's start at the beginning. What does it mean to say that 2 is an eigenvalue of $A$? – Gerry Myerson May 11 '12 at 13:36
That mean there will exist at least one non zero vector, call it $v$, with corresponding eigenvalue $\lambda$ such that $Av = \lambda v$ – srijan May 11 '12 at 13:42

Forgive me if I mistake what Gerry is trying to say, but I believe it is this:

  1. 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 $A^n(v) = \lambda^nv$.

  2. From the associated eigenvalues you get for $A+B$, you should be able to explicitly state the characteristic polynomial for $A+B$.

  3. 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.

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Maybe the first point in your answer should make clear what the eigenvalues of A+B look like? And that directly solves the problem, right? – awllower May 11 '12 at 15:39

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 $A+B$.

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