Finding a relation between $P$ and $P^2$

Question

I encountered a challenging problem on a mock test today:

Let $x$ be an $n \times 1$ matrix. We define $$P = -(x^Tx)^{-1}\cdot(xx^T)$$

Now there were $4$ options, something like $P^2 - P = O$ where $O$ is the null matrix and $P^2 + P = I$, I'm afraid I don't exactly remember all of them.

My approach was to take $$x = (a_{1} , a_{2}, \dots a_{n})^T$$ and then figure out P first, which came out to be:

$$|x|^2 P = - (xx^T)$$ where $xx^T$ is a big $ n \times n$ matrix which I'll admit, I couldn't figure out the latex to type, so let's call it $A$.

($|x|^2$ here means : $\Sigma_{i = 1}^{i = n}a_{i}^2$)

$$\implies |x|^2 P = -A$$ I then tried to find out $P^2$ and I got: $$|x|^{(2n - 4)} P^2 = A$$

I fear that my calculations are incorrect as I'm unable to find any obvious relation between the two here.

It would be a great help if someone could point me in the right direction! Thanks!

Answer 1

Hint $x^\top x$ is a constant. How can we use that property when expanding our expression for $P^2$ in terms of $x$?

$$P^2 = [-(x^\top x)^{-1} x x^\top]^2 = (x^\top x)^{-2} (x x^\top) (x x^\top) = (x^\top x)^{-2} x (x^\top x) x^\top = (x^\top x)^{-2} (x^\top x) x x^\top$$