True or False?
Let A be an n × n complex matrix such that A k = I , where I is the n × n identity matrix. Then A is determined by tr ( A ) , tr ( A 2 ) , … , tr ( A k − 1 ) to within similarity.
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Since A k = I , the minimum polynomial of A is a product of distinct linear factors, and hence A is diagonalisable. If z 1 , z 2 , . . . , z k are the k distinct k th roots of 1 then (to within similarity) A is determined by the respective multiplicities u 1 , u 2 , . . . , u k of its eigenvalues, and we have T r ( a m ) = n = 1 ∑ k u n z n m 0 ≤ m ≤ k − 1 In other words we have the matrix equation ⎝ ⎜ ⎜ ⎜ ⎜ ⎜ ⎛ 1 z 1 z 1 2 ⋮ z 1 k − 1 1 z 2 z 2 2 ⋮ z 2 k − 1 1 z 3 z 3 2 ⋮ z 3 k − 1 ⋯ ⋯ ⋯ ⋱ ⋯ 1 z k z k 3 ⋮ z k k − 1 ⎠ ⎟ ⎟ ⎟ ⎟ ⎟ ⎞ ⎝ ⎜ ⎜ ⎜ ⎜ ⎜ ⎛ u 1 u 2 u 3 ⋮ u k ⎠ ⎟ ⎟ ⎟ ⎟ ⎟ ⎞ = ⎝ ⎜ ⎜ ⎜ ⎜ ⎜ ⎛ T r ( I ) T r ( A ) T r ( A 2 ) ⋮ T r ( a k − 1 ) ⎠ ⎟ ⎟ ⎟ ⎟ ⎟ ⎞ It is a standard results that this matrix is nonsingular, with determinant ± 1 ≤ m < n ≤ k ∏ ( z m − z n ) and so the multiplicities u 1 , u 2 , . . . , u k are determined by the given traces.
Just a minor typo......it should be "It is a standard RESULT"......
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Since the matrix A satisfies the equation x k − 1 , the minimal polynomial of A divides x k − 1 . Since x k − 1 = j = 0 ∏ k − 1 ( x − e 2 π i j / k ) , the roots of x k − 1 are all distinct. Hence the roots of the minimal polynomial are also distinct. Therefore the matrix A is diagonalizable.
It suffices to prove that the eigenvalues λ 1 , λ 2 , … , λ n of A or p A ( t ) , characteristic polynomial of A , is uniquely determined by tr ( A ) , tr ( A 2 ) , … , tr ( A k − 1 ) . We can determine tr ( A l ) for all positive integers l . Since A is always similar to its Jordan canonical form , an upper triangular matrix having λ 1 , λ 2 , … , λ n on the main diagonal, we get tr ( A l ) = j = 0 ∑ n λ j l . Using Newton's identities to express the elementary symmetric polynomials in terms of the power sums, we can find the coefficients of p A ( t ) by Vieta's formula .
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