Square root of a projection matrix

Question

For a Projection matrix $P$ we know that it equals to its square, i.e. $P^2=P$.

Can we say similar things hold for its square root? i.e. does $P^{\frac{1}{2}}=P$ hold?

What seems to me is that there exists at least one square root of $P$ that equals to itself, but I don't know if a projection matrix can have multiple square roots or not.

Answer 1

Consider $$Q=\begin{pmatrix}0&-1\\-1&0\end{pmatrix}$$

Let $P=Q^2$. Since $P=I$, $P$ is a projection, but $Q$ is not.

Answer 2

The most trivial example would be the identity matrix - it is a projection on the whole space. The matrices with $\pm 1$ on the main diagonal and $0$ outside this diagonal are square roots of the identity. In other words, you have at least $2^n$ different square roots.

Answer 3

If $P^2=P$ then $P\in\mathbb{R}^{n\times n}$ is root of equation $$ P^2-P=0. $$ But we have a bit more. If $P^2=P$ then $$ \cdots=P^{2\cdot n}=\cdots=P^{4}=P^{2}=P^{} $$ Then for any fixed $n$ and $n$ real numbers $a_1,a_2,\ldots, a_n$ such that $ a_1+a_2+\ldots+a_n=1$ we have that $P$ is root of equation $$ a_n\cdot P^{2\cdot n}+a_{n-1}\cdot P^{2\cdot (n-1)}+\ldots+a_1\cdot P=P $$