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Rank Relations


Suppose we have a matrix A \in \mathcal{M}_n(\mathbb{R}) such that A^3=A. Prove that rank(A) +rank (A- I_n)+rank (A + I_n)=2n.

As a generalization, we can state the following:
Given a matrix A with p(A)=0, where p is a real polynomial with distinct roots, we have \sum_{j=1}^k rank(A- x_j I) =(k-1)\cdot n, and x_i are the distinct roots of p.

Solution:
Since p(A)=0 and p has distinct real roots it follows that the minimal polynomial of A has distinct linear factors, implying that A is diagonalizable. Therefore, there exists a non-singular matrix P such that A=PDP^{-1}, where D is a diagonal matrix having the eigenvalues of A as entries on the diagonal.
Since p(A)=0 we can see that any root of p is an eigenvalue for A. Therefore, on the diagonal of D we have only roots of p.
Furthermore, we know that if B is a matrix and C is a nonsingular matrix, then rank(B)=rank(CB)=rank(BC), so we can say that rank(A-x_j I)=rank(PDP^{-1}-x_j I)=rank(D-x_j I)= number of diagonal entries not equal to x_j.
Adding up these relations we get the desired result.

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Categories: Algebra, Higher Algebra
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