Let $A$ be a real $m \times n$ matrix with $m \geq n$ and $\mbox{rank}(A) = n$.
In the least-squares theory, the pseudo-inverse of $A$ is defined as $$ A^\dagger = (A^T A)^{-1} A^T $$
It is easy to verify that $$ A^\dagger A = I $$
Thus, if we define $P$ as $$ P = A A^\dagger, $$ then $P$ is a real $m \times m$ matrix and $$ P^2 = P $$
Thus, $P$ is a projection matrix.
(If $\mathbf{x} = A c$ for any vector $c$, then $P \mathbf{x} = \mathbf{x}$.)
Also, if we define $M$ as $$ M = I - P, $$ then it is easy to check that $M^2 = M$. Hence, $M$ is also a projection matrix.
(For any vector $\mathbf{x} = A \mathbf{c}$, $M \mathbf{x} = (I - P) \mathbf{x} = 0$.)
Let $(\lambda, \mathbf{x})$ be any eigenpair of $P$. Then $$ P \mathbf{x} = \lambda \mathbf{x} $$
Multiplying by $P$, we get $$ P^2 \mathbf{x} = \lambda P \mathbf{x} = \lambda \cdot \lambda \mathbf{x} $$
Since $P^2 = P$, we get $$ P \mathbf{x} = \lambda \mathbf{x} = \lambda^2 \mathbf{x} $$
Since $\mathbf{x} \neq 0$, we must have $$ \lambda = \lambda^2 \ \ \mbox{or} \ \ \lambda (\lambda - 1) = 0 $$
Thus, the eigenvalues of the projection matrix $P$ are $0$ and $1$.
Similarly, we can show that the eigenvalues of the projection matrix $M = I - P$ are also $0$ and $1$.
In a text-book titled "Advanced Econometrics" by Takeshi Amemiya (1985, Harvard University Press), the author mentions the result in the Appendix of the book which states:
The eigenvalues of the projection matrix $P$ consist of $n$ ones and $m - n$ zeroes.
The eigenvalues of the projection matrix $M = I - P$ consist of $m - n$ ones and $n$ zeros.
I like to know how to prove the results 1) and 2) on the multiplicity of the eigenvalues of $P$ and $M$.
I tried to work out a proof and I outline my results for (1). [Proving (2) is similar!]
We note that $$ P = A A^\dagger $$ where $A^\dagger$ is the pseudo-inverse of $A$.
Hence, $$ \mbox{rank}(P) = \mbox{rank}(A) = n $$ ($A$ is an overdetermined $m \times n$ matrix with $m \geq n$, $\mbox{rank}(A) = n$.)
We know that $$ \mbox{rank}(P) + \mbox{nullity}(P) = m $$ which gives $$ \mbox{nullity}(P) = m - n $$
Since the null-space of $P$ coincides with the eigenspace of the eigenvalue $0$ of $P$, we conclude that the eigenvalues of $P$ has $(m - n)$ zeros.
Since $P$ has only 0 and 1 as the eigenvalues, it will follow that the eigenvalues of $P$ has $n$ ones.
In a similar way, it can be shown that the eigenvalues of $M = I - P$ has $n$ zeros and $(m - n)$ ones.
I tried this calculation as a draft for the "proof". Your comments are welcome.