91Ó°ÊÓ

First, we need to recall the definition of an idempotent matrix. A matrix A is idempotent if A^2 = A. We want to show that if A is idempotent, then I - A is also idempotent. In other words, we need to show that (I - A)^2 = I - A. Let's compute (I - A)^2: \[(I - A)^2 = (I - A)(I - A) = I^2 - AI - AI + A^2\] Since A is idempotent, we know that A^2 = A. So we can substitute A for A^2: \[= I - 2AI + A\] Now, we need to show that this expression is equal to I - A. Since A is idempotent, AI = A. Using this property, we can write: \[= I - 2A + A = I - A\] Therefore, if A is idempotent, I - A is also idempotent.

To show that I + A is nonsingular, we need to find its inverse and show that it is equal to I - ½A. To find the inverse of the matrix I + A, we will multiply both sides of the equation (I + A)X = I by the inverse matrix and see if we can find the matrix X. Let's assume that the matrix X exists and is given by (I - \frac{1}{2}A). Then we need to show that (I + A)(I - \frac{1}{2}A) = I. Let's compute the product (I + A)(I - ½A): \[(I + A)(I - \frac{1}{2} A) = I^2 - \frac{1}{2} AI + AI - \frac{1}{2} A^2\] Since A is idempotent, we know that A^2 = A. So we can substitute A for A^2: \[= I - \frac{1}{2} A + A - \frac{1}{2} A\] Now we can simplify this expression: \[= I + \frac{1}{2} A\] So, the inverse of I + A is indeed I - ½A, and therefore I + A is nonsingular. In conclusion, if A is an idempotent matrix, then I - A is idempotent, I + A is nonsingular, and the inverse of I + A is given by (I - ½A).