$A$ is a non singular $n \times n$ matrix with all eigenvalues real. The trace $T(A^2)= T(A^3)= T(A^4)$. Find $T(A)$.

I have tried using the fact that trace is a linear functional and tried to explore its kernel, but have made no progress. Can I get some help?

My knowledge of linear algebra is limited to the first $5$ chapters of Hoffman and Kunze, basic properties of eigenvalues, annihilating polynomials and invariant subspaces. Is it sufficient to do this problem?

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  • that the trace is the sum of the eigenvalues and that the eigenvalues of $A^n$ are the eigenvalues of $A$ raised to the nth power may be useful – operatorerror Apr 04 '17 at 04:06
  • isnt the trace sum of eigenvalues only when A is diagonalizable? – tony Apr 04 '17 at 04:18
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    @tony No, it always holds. Every matrix can be made upper triangular, and the diagonal entries of an upper triangular matrix are the eigenvalues (with their algebraic multiplicities). – Aaron Apr 04 '17 at 04:21
  • ok, i will check it out. – tony Apr 04 '17 at 04:23
  • @tony see here: http://math.stackexchange.com/questions/546155/proof-that-the-trace-of-a-matrix-is-the-sum-of-its-eigenvalues – operatorerror Apr 04 '17 at 04:23
  • consider the matrix over the field of reals with characteristic polynomial (x^2)+1. It has no eigenvalue, but trace is zero. How is trace equal to sum of eigenvalues? – tony Apr 04 '17 at 04:46
  • @tony the assumption (which should have been stated above) is that the field you are working over is algebraically closed (that's when you get every matrix is similar to an upper triangular matrix). In your problem, you assume the eigenvalues exist in the field you are working over, the reals, so you are ok. – operatorerror Apr 04 '17 at 04:55

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