Here is a proof of the fact that any transitive matrix $M\in\mathbb{R}
^{d\times d}$ is symmetric. Together with the argument in the answer by Keith
Kearnes, this proves that any transitive matrix $M\in\mathbb{R}^{d\times d}$
is a positive-semidefinite symmetric matrix of rank $\leq1$.
We first prove a lemma: If $p\in\mathbb{R}^{d}\setminus\left\{ 0\right\} $
and $q\in\mathbb{R}^{d}$ are two vectors such that every $u\in\mathbb{R}^{d}$
satisfying $p^{T}u>0$ satisfies $q^{T}u\geq0$, then
(1) there exists a nonnegative real $\lambda$ such that $q=\lambda p$.
Proof of (1). Let $p\in\mathbb{R}^{d}\setminus\left\{ 0\right\} $ and
$q\in\mathbb{R}^{d}$ be two vectors such that every $u\in\mathbb{R}^{d}$
satisfying $p^{T}u>0$ satisfies $q^{T}u\geq0$. We have $p\neq0$ and thus
$p^{T}p>0$.
If $p$ and $q$ were linearly independent, then there would be a vector
$v\in\mathbb{R}^{d}$ satisfying $p^{T}v=1$ and $q^{T}v=-1$; but this would
contradict the assumption that every $u\in\mathbb{R}^{d}$ satisfying
$p^{T}u>0$ satisfies $q^{T}u\geq0$. Hence, $p$ and $q$ must be linearly
dependent. Since $p\neq0$, this shows that there exists a $\lambda
\in\mathbb{R}$ such that $q=\lambda p$. It remains to prove that this
$\lambda$ is nonnegative. Indeed, recall that every $u\in\mathbb{R}^{d}$
satisfying $p^{T}u>0$ satisfies $q^{T}u\geq0$. Applying this to $u=p$, we get
$q^{T}p\geq0$ (since $p^{T}p>0$). Since $q=\lambda p$, this rewrites as
$\lambda p^{T}p\geq0$, and thus $\lambda\geq0$ (since $p^{T}p>0$). This
finishes the proof of (1).
Another lemma, which is really well-known: If $V$ is a finite-dimensional
vector space, and if $\phi$ is a linear endomorphism of $V$ such that $\phi$
sends every vector in $V$ to a scalar multiple of this vector, then
(2) the endomorphism $\phi$ is a scalar multiple of the identity.
(In order to prove (2), fix a basis of $V$ and see what $\phi$ does to the
basis vectors and their pairwise sums.)
Now, let $M\in\mathbb{R}^{d\times d}$ be any transitive matrix. We need to
prove that $M$ is symmetric.
If $M=0$, then this is obvious. Thus, WLOG assume that $M\neq0$.
Let $w\in\mathbb{R}^{d}$ be any vector such that $M^{T}w\neq0$. Then, $w\neq0$
(since $M^{T}w\neq0$).
Let $u\in\mathbb{R}^{d}$ be any vector such that $\left( M^{T}w\right)
^{T}u>0$. Let us now show that $w^{T}M\left( M^{T}u\right) \geq0$. Indeed,
if $M^{T}u=0$, then this is clear; otherwise it follows from the transitivity
of $M$ (in fact, from $w^{T}Mu=\left( M^{T}w\right) ^{T}u>0$ and
$u^{T}M\left( M^{T}u\right) =\left( M^{T}u\right) ^{T}\left(
M^{T}u\right) >0$ (since $M^{T}u\neq0$), we obtain $w^{T}M\left(
M^{T}u\right) >0$ (since $M$ is transitive)). Thus, we have proven that
$w^{T}M\left( M^{T}u\right) \geq0$. Hence, $\left( MM^{T}w\right)
^{T}u=w^{T}MM^{T}u=w^{T}M\left( M^{T}u\right) \geq0$.
Let us now forget that we fixed $u$. We thus have shown that every
$u\in\mathbb{R}^{d}$ satisfying $\left( M^{T}w\right) ^{T}u>0$ satisfies
$\left( MM^{T}w\right) ^{T}u\geq0$. Thus, (1) (applied to $p=M^{T}w$ and
$q=MM^{T}w$) yields that
(3) there exists a nonnegative real $\lambda$ such that $MM^{T}w=\lambda
M^{T}w$.
Now, let us forget that we fixed $w$. We thus have proven that for every
$w\in\mathbb{R}^{d}$ satisfying $M^{T}w\neq0$, we have (3). But (3)
also holds for every $w\in\mathbb{R}^{d}$ satisfying $M^{T}w=0$ (since we can
use $\lambda=0$). Hence, (3) holds for every $w\in\mathbb{R}^{d}$.
Let $V=M^{T}\left( \mathbb{R}^{d}\right) $ be the image of $M^T$. Then,
$M\left( V\right) \subseteq V$ (because (3) holds for every
$w\in\mathbb{R}^{d}$). Thus, $M$ restricts to an $\mathbb{R}$-linear
endomorphism $\phi$ of $V$. This endomorphism $\phi$ sends every vector in $M$
to a scalar multiple of this vector (because (3) holds for every
$w\in\mathbb{R}^{d}$). Thus, $\phi$ must be a scalar multiple of the identity
(due to (2)). In other words, there exists some $\mu\in\mathbb{R}$ such
that every $v\in V$ satisfies $\phi\left( v\right) =\mu v$. In other words,
there exists some $\mu\in\mathbb{R}$ such that every $w\in\mathbb{R}^{d}$
satisfies $MM^{T}v=\mu M^{T}v$ (because of what $V$ is and what $\phi$ is). In
other words, there exists some $\mu\in\mathbb{R}$ such that $MM^{T}=\mu M^{T}
$. Consider this $\mu$.
We are working over $\mathbb{R}$. Hence, a well-known fact says that
$\operatorname*{Ker}\left( MM^{T}\right) =\operatorname*{Ker}\left(
M^{T}\right) $. Thus, from $M^{T}\neq0$, we obtain $MM^{T}\neq0$, so that
$\mu M^{T}=MM^{T}\neq0$ and therefore $\mu\neq0$. Thus, we can transform
$MM^{T}=\mu M^{T}$ into $M^{T}=\dfrac{1}{\mu}MM^{T}$. The matrix $M^{T}$ is
thus symmetric (since $MM^{T}$ is symmetric). In other words, the matrix $M$
is symmetric.
