Determine Matrix

Algebra Level 2

True or False?

Let A A be an n × n n\times n complex matrix such that A k = I , A^k=I, where I I is the n × n n\times n identity matrix. Then A A is determined by tr ( A ) , tr ( A 2 ) , , tr ( A k 1 ) \text{tr}(A),\text{tr}(A^2),\dots,\text{tr}(A^{k-1}) to within similarity.

True False

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2 solutions

Brian Lie
Feb 23, 2019

Since the matrix A A satisfies the equation x k 1 , x^k-1, the minimal polynomial of A A divides x k 1. x^k-1. Since x k 1 = j = 0 k 1 ( x e 2 π i j / k ) , x^k-1=\prod_{j=0}^{k-1}(x-e^{2\pi ij/k}), the roots of x k 1 x^k-1 are all distinct. Hence the roots of the minimal polynomial are also distinct. Therefore the matrix A A is diagonalizable.

It suffices to prove that the eigenvalues λ 1 , λ 2 , , λ n \lambda_1,\lambda_2,\dots,\lambda_n of A A or p A ( t ) , p_A(t), characteristic polynomial of A , A, is uniquely determined by tr ( A ) , tr ( A 2 ) , , tr ( A k 1 ) . \text{tr}(A),\text{tr}(A^2),\dots,\text{tr}(A^{k-1}). We can determine tr ( A l ) \text{tr}(A^l) for all positive integers l . l. Since A A is always similar to its Jordan canonical form , an upper triangular matrix having λ 1 , λ 2 , , λ n \lambda_1,\lambda_2,\dots,\lambda_n on the main diagonal, we get tr ( A l ) = j = 0 n λ j l . \text{tr}(A^l)=\sum _{j=0}^n\lambda_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 ) p_A(t) by Vieta's formula .


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Mark Hennings
Mar 1, 2019

Since A k = I A^k = I , the minimum polynomial of A A is a product of distinct linear factors, and hence A A is diagonalisable. If z 1 , z 2 , . . . , z k z_1,z_2,...,z_k are the k k distinct k k th roots of 1 1 then (to within similarity) A A is determined by the respective multiplicities u 1 , u 2 , . . . , u k 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 \mathrm{Tr}(a^m) \; = \; \sum_{n=1}^k u_n z_n^m \hspace{2cm} 0 \le m \le k-1 In other words we have the matrix equation ( 1 1 1 1 z 1 z 2 z 3 z k z 1 2 z 2 2 z 3 2 z k 3 z 1 k 1 z 2 k 1 z 3 k 1 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 ) ) \left( \begin{array}{ccccc} 1 & 1 & 1 & \cdots & 1 \\ z_1 & z_2 & z_3 & \cdots & z_k \\ z_1^2 & z_2^2 & z_3^2 & \cdots & z_k^3 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ z_1^{k-1} & z_2^{k-1} & z_3^{k-1} & \cdots & z_k^{k-1} \end{array}\right) \left(\begin{array}{c} u_1 \\ u_2 \\ u_3 \\ \vdots \\ u_k \end{array}\right) \; = \; \left(\begin{array}{c} \mathrm{Tr}(I) \\ \mathrm{Tr}(A) \\ \mathrm{Tr}(A^2) \\ \vdots \\ \mathrm{Tr}(a^{k-1}) \end{array} \right) It is a standard results that this matrix is nonsingular, with determinant ± 1 m < n k ( z m z n ) \pm\prod_{1 \le m < n \le k}(z_m - z_n) and so the multiplicities u 1 , u 2 , . . . , u k u_1,u_2,...,u_k are determined by the given traces.

Just a minor typo......it should be "It is a standard RESULT"......

Aaghaz Mahajan - 2 years, 3 months ago

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