Loewner order and rank 1 matrices The Loewner order is defined over the set of Hermitian matrices as $A \leq B$ if and only if $B - A$ is positive semi-definite. If $B$ is a rank 1, positive semi-definite matrix, what are the matrices $A$ such that $0 \leq A \leq B$? Are there other solutions than $A = \lambda B$ for $0 \leq \lambda \leq 1$?
 A: No, there are no other solutions to $0 \le A \le B$.
Note that, if $X \le Y$, and $P$ is an orthogonal matrix, then $P^\top XP \le P^\top YP$, since,
$$P^\top YP - P^\top X P = P^\top (Y - X) P,$$
which is positive semi-definite, since $Y - X$ is positive semi-definite.
Since $B$ is symmetric, it is orthogonally diagonalisable. Let $P$ be an orthogonal matrix such that $P^\top BP$ is diagonal. Then,
$$0 = P^\top 0P \le P^\top AP \le P^\top BP.$$
Also note that, if we can show $P^\top A P$ is a multiple of $P^\top B P$, then $A$ is the same multiple of $B$. Thus, it suffices to consider the case where $B$ is diagonal. In this case, rank $1$ implies $B$ has a single non-zero entry, and positive-semidefiniteness implies that this single non-zero entry is positive. By rearranging the columns of $P$ appropriately, we can assume that this non-zero entry is in the top-left. In other words, we are solving,
$$0 \le A \le B = \operatorname{diag}(b, 0, 0, \ldots, 0),$$
where $b > 0$.
Let us suppose $A$ and $B$ are $n \times n$. For $v \in \operatorname{span}\{e_2, \ldots, e_n\}$, where $e_1, \ldots, e_n$ are the standard basis, we have $v^\top Bv = 0$. Since $A \ge 0$, we have $v^\top Av \ge 0$. Since $B - A \ge 0$, we have
$$v^\top(B - A)v \ge 0 \implies 0 = v^\top B v \ge v^\top A v \ge 0 \implies v^\top A v = 0.$$
Positive semi-definite matrices have unique positive semi-definite square roots, so,
$$v^\top \sqrt{A} \sqrt{A} v = 0 \implies \|\sqrt{A} v\| = 0 \implies \sqrt{A} v = 0 \implies Av = \sqrt{A}0 = 0.$$
Thus, $A$ maps $\operatorname{span}\{e_2, \ldots, e_n\}$ to $0$, just like $B$. This implies that the second through to the $n$th columns of $A$ must be $0$. Since $A^\top = A$, we know that the second through to the $n$th rows must also be $0$. Thus $A$ takes the form:
$$A = \operatorname{diag}(a, 0, 0, \ldots, 0).$$
Once again, since $A$ is positive-semidefinite, $a \ge 0$. Similarly, $B - A$ is positive-semidefinite, so $b - a \ge 0$. Thus, $0 \le a \le b$, and so your conjecture is true.
A: If $B$ has rank 1 and $A$ is a positive semidefinite matrix for which $A \preceq B$, then it follows that $A = \lambda B$ for some $0 \leq \lambda \leq 1$.
If we remove the requirement that $A$ is positive semidefinite, then any of the matrices of the form $A = \lambda B - C$ for some $0 \leq \lambda \leq 1$ and some positive semidefinite matrix $C$ also satisfy $A \preceq B$ but are not necessarily of that form.
Here's a proof. Let $B$ be $n \times n$, rank-1, and positive semidefinite. There exists a vector $x$ such that $B = xx^*$ (where $x$ is a column vector and ${}^*$ denotes a conjugate-transpose). We note that for all vectors $y$ orthogonal to $x$, we have
$$
0 \leq y^*(B-A)y = |x^*y|^2 - y^*Ay = -y^*Ay.
$$
Since $A$ is positive semidefinite, it follows that $y^*Ay = 0$. Moreover, $y^*Ay = 0 \implies Ay = 0$, so we can conclude that the kernel of $A$ contains all of $x^\perp$, which is an $(n-1)$-dimensional space. So, $A$ is also has rank $1$, and its image must be equal to $(x^\perp)^\perp$, i.e. the span of $x$.
It follows that $A$ is indeed of the form $\lambda xx^*$ for some $\lambda \in \Bbb R$. Because $A$ is positive semidefinite and $B-A$ is positive definite, we can conclude that $0 \leq \lambda \leq 1$.
