Let n-order real square matrix A = a ^ 2 and E be n-order identity matrix. It is proved that R (a) + R (A-E) = n

Because a = a ^ 2, so a (A-E) = 0, so r (a) + R (A-E) ≤ n^_^

Find the document: let a be an invertible square matrix of order n, e be the identity matrix, the square of a = the absolute value of a * e, and prove that a * = a

Because AA * = | a | e, and a ^ 2 = | a | E
So AA * = AA
If a is invertible, a * = a is obtained by multiplying the inverse of a on both sides of the equation#

ABC is a square matrix of order n and 2E = B + e (E is the identity matrix). It is proved that a square = A and b square = E

I'm sorry I can't understand your topic
What is "a square = a condition b square = e"

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