Let n square matrix a satisfy a ^ 2 = A and E be the identity matrix of order n, and prove that R (a) + R (A-E) = n

Let n square matrix a satisfy a ^ 2 = A and E be the identity matrix of order n, and prove that R (a) + R (A-E) = n

Because a * a = a, so a (A-E) = 0; so every column vector of A-E is the solution of the equation AX = 0, because the column vector in A-E may not form the base of the solution space, so r (a) + R (A-E) is less than or equal to n; then R (a) + R (b) > = R (a + b); immediately R (a) + R (A-E) = R (a) + R (e-A) > = R (a + e-A) = R (E) = n; so r (a) +

Let a be a square matrix of order n and E be an identity matrix of order n. It is proved that R (a + e) + R (A-E) "n,

It is proved that: let a and B be square matrices of the same order, A1, A2... Ar be maximal linearly independent vector group of a, then: R (a) = R, similarly, let B1, B2,... BS be maximal linearly independent vector group of B, then: R (b) = s, and a + B and a and B are square matrices of the same order, its maximal linearly independent group cannot be greater than R + s, that is: R (a) + R (b) ≥ R (a + b)

Given n-order square matrix A, satisfy a ^ 3 + A ^ 2-2a = 0, I is n-order unit matrix, prove that matrix A + I must be invertible

A^3+A^2-2A=0
A^2(A+I)-2A-2I=-2I
(A^2-2I)(A+I)=-2I
-1/2(A^2-2I)(A+I)=I
So a + I is reversible
The inverse matrix is
-1/2(A^2-2I)