Wait, isn't this true?

We are told that P is idempotent. First, however, we are going to show that I – P is idempotent too. In other words, we need to show that:

Expanding the left side of the equation, we get:

**(I – P)2 = (I – P)·(I – P) = I·I – 2**·**P + P2 **(3)

However, we are told that P is idempotent and we know that I is an idempotent matrix, which means P2 = P and I2=I. Thus, equation (3) can also be written as:

**(I – P)2 = (I – P)·(I – P) = I·I – 2·P + P2 = I – 2·P + P = I – P**

**or (I – P)2 = I – P **

This shows that I – P is idempotent too. Now, in part 2, we proved that an idempotent that is invertible is equal to the identity matrix. We just showed that I – P is an idempotent matrix, which suggests that it is invertible only if I – P = I. This means that P must be equal to 0 for that to hold. Thus, since P=0 is an idempotent matrix, it can be concluded that I – P will be invertible.