If $A^+$ is the Moore-Penrose pseudo inverse of $A$, that is, the matrix such that $$\tag{1} a)\;AA^+A=A, \quad
b)\;A^+AA^+=A^+, \quad
c)\;(AA^+)^*=AA^+, \quad
d)\;(A^+A)^*=A^+A,$$ then
- (i) $AA^+$ is the orthogonal projector (OP) onto the range of $A$,
- (ii) $A^+A$ is OP onto the range of $A^*$.
Note that if $Q$ is OP onto a subspace $S$, that is, $Q=Q^*$ (here, $^*$ denotes conjugate transpose), $Q^2=Q$, and $R(Q)=S$ (note that these three conditions imply that $Q$ is unique), then $I-Q$ is OP onto $S^\perp$. So, if (i) is true, then $I-AA^+$ is OP onto $R(A)^\perp=N(A^*)$ and, if (ii) is true, then $I-A^+A$ is OP onto $R(A^*)^\perp=N(A)$.
For the proof of (i) and (ii), we use only (1) and the fact that if $X=YZ$, then $R(X)\subseteq R(Y)$. We need only to show that the matrices $AA^+$ and $A^+A$ are Hermitian, idempotent, and their ranges are equal to the subspaces on which they are supposed to project.
Both $AA^+$ and $A^+A$ are obviously Hermitian; see (1c) and (1d). In addition, (1a) and/or (1b) imply that they are idempotent. It remains to show that $R(AA^+)=R(A)$ and $R(A^+A)=R(A^*)$. Clearly, $R(AA^+)\subseteq R(A)$; $R(A)\subseteq R(AA^+)$ follows from (1a). From (1d), we have $A^+A=A^*(A^+)^*$, so $R(A^+A)\subseteq R(A^*)$. From (1a) and (1d), $A^*=A^+AA^*$, so $R(A^*)\subseteq R(A^+A)$.
Summarizing, the following are OPs:
- $AA^+$ on $R(A)$,
- $I-AA^+$ on $N(A^*)$,
- $A^+A$ on $R(A^*)$,
- $I-A^+A$ on $N(A)$.
Note that this is true no matter what is the rank of $A$. While for some $A$, $A^*A$ or $AA^*$ (or both) might not be invertible depending on the rank of $A$, any $A$ has always a unique Moore-Penrose pseudoinverse and a set of unique OPs given above associated with the four subspaces of $A$.
We assume $n \geq 2$.
$A:=\frac{dd^T}{d^Td}$ is a rank-one matrix. Explanation: every $c \in d^{\perp}$ is such that $\frac{dd^T}{d^Td}c=\frac{dd^Tc}{d^Td}=\frac{d0}{d^Td}=0$. Thus $c$ is an eigenvector associated to eigenvalue $0$. In other terms, subspace $d^{\perp}$ (which is $n-1$ dimensional) is included in the kernel of $A$; it is immediate that it is in fact the kernel of $A$. From there, using the rank-nullity theorem, we can conclude to the "rank 1" property of $A$.
Thus 0 is an eigenvalue of $A$ with multiplicity $n-1$. Due to relationship $Ad=d$, the other eigenvalue is 1.
Thus $A$ is only a symmetrical semi-definite positive matrix.
Best Answer
Yes, with $W:=\operatorname{Col}(Q)$ we have for every $v$, $$ Q(Qv-v)=Q^2v-Qv=0,$$ hence $Qv-v\in\ker Q\subseteq W^\perp$, of course $Qv\in W$, and for $v\in W$ we have $Qv=v$.