Example of a Projection Operator in $\mathbb{R^3}$

Question

I'm looking for an operator $\hat P$ in $\mathbb{R^3}$ such that $\hat P^2=\hat P$ that is also Hermitian

Answer 1

A standard example for a projection operator in $\mathbb{R}^3$ is the projection onto the subspace spanned by a single vector:

Let $u = (u_1,u_2,u_3)^T \in \mathbb{R}^3$, then $P_u x = \frac{x \cdot u}{u \cdot u} u$. Note that the linearity of the dot product makes $P_u$ linear, and $P_u u= u$.

Now $$P_u ( P_u x ) = P_u \left( \frac{x \cdot u}{ u \cdot u} u \right) = \frac{x \cdot u}{ u \cdot u} P_u u = \frac{x \cdot u}{ u \cdot u} u$$ holds for all $x$, so $P_u^2 = P_u$.

Finally, $(P_u x) \cdot y = \frac{(x \cdot u)(y \cdot u)}{u \cdot u} = x \cdot (P_u y)$, so $P_u$ is self-adjoint (i.e. Hermitian).

---

We can express $P_u$ as a matrix using the standard basis in $\mathbb{R}^3$. Let $e_1 = (1,0,0)^T$, $e_2 = (0,1,0)^T$ and $e_3=(0,0,1)^T$, and let $\beta = \{ e_1, e_2, e_3\}$ be the ordered set of basis vectors.

$$P_u e_i = \frac{u_i}{u\cdot u} u = \frac{1}{u \cdot u} (u_i u_1, u_i u_2, u_i u_3)^T.$$

Thus $$[P_u]_\beta = \frac{1}{u \cdot u} \begin{pmatrix} u_1 u_1 & u_2 u_1 & u_3 u_1\\ u_1 u_2 & u_2 u_2 & u_3 u_2\\ u_1 u_3 & u_2 u_3 & u_3 u_3\end{pmatrix}.$$

You can see directly from the matrix representation that the operator $P_u$ is self-adjoint. If you were so inclined you could also demonstrate through matrix multiplication that this matrix satisfies $[P_u]_\beta = [P_u]_\beta^2$.

Answer 2

The matrix all of whose elements are equal to $1/3$ is Hermitian and a projection, though "Hermitian" is silly in a real vector space.

Answer 3

The identity operator is an example.