Let $m \times n$ full-column matrix be $A$. Define an orthogonal projection onto the column space of $A$ as $P(A)$, which is $P(A)=A(A'A)^{-1}A'$. Denote an annihilator matrix (or residual maker) as $M(A)$, where $M(A)=I_m - p(A) = I_m - A(A'A)^{-1}A'$. $I_m$ denote $m\times m$ identity matrix.
I'm looking for what condition of $A$ guarantees $M(A)=D$, for a GIVEN diagonal matrix $D$.
What I've tried so far is like this: Use QR decomposition of $A$. $A=QR$, where $Q$ is $m\times m$ unitary matrix, $R$ is $m\times n$ triangular matrix. Then, $A=QR=[Q_1 , Q_2]R$, where $Q_1$ is the first $n$ column of $Q$, $Q_2$ is the last $(m-n)$ column of $Q$.
Then the projection matrix $P(A)=Q_1 Q_1'$. Using the definition of annihilator matrix, $M(A)=I_m -P(A) = I_m - Q_1 Q_1'=Q_2 Q_2'$. This is because of the fact $Q_1 Q_1' + Q_2 Q_2'=I_m$. Hence I'm looking for the CONDITION OF A that guarantees $M(A)=Q_2 Q_2' = D$, for a given diagonal matrix $D$.
I guess $A$ is somehow related to $D$, but doesn't know how it is related, where $A$ is $m\times n$, $D$ is $m \times m$. In a nutshell, I want to show that first, given invertible $D$, such a matrix $A$ exists. Also I need to suggest an example of such a matrix for a given $D$.
First of all, $D$ has to have only $1$'s and zeros. I also assume that is given that $A$ has full column-rank (perhaps this is what you mean by a "full-column" matrix).
With that being the case, let $D$ be given, and let $i_1,\dots,i_n$ be the diagonal entries such that $D(i_n,i_n) = 1$. Then we will have $M(A) = D$ if and only if the rows $i_1,\dots,i_n$ from $A$ contain only zeros.
So, for instance: with $$ D = \pmatrix{0\\&1\\&&0} $$ we could have $$ A = \pmatrix{1&2\\0&0\\3&4} $$