Existence and characterization of transitive matrices? We call a matrix $M \in \mathbb{R}^{d \times d}$ transitive if it satisfies the following: 

For any three vectors $u, v, w$ in $\mathbb{R}^d$. If $u^T M v > 0$ and $v^T M w > 0$ then $u^TMw > 0$. 

EDIT: The case when $d = 1$ is simple. For $d > 1$ the following is true: 
a) M must be Positive Semi Definite, because if $M$ is transitive and $\exists u$ s.t. $u^TMu < 0$ then we can let $v = -u, w = u$ and arrive at the contradiction that $u^t M u < 0 \implies u^tMv > 0, v^tMw > 0 \implies u^TMw > 0 \implies u^TMu > 0$
b) $M \ne I$ because if $M = I$ then we just need to construct three vectors such that $u^Tv > 0, v^Tw > 0$ but $u^Tw < 0$. Intuitively, I know that sum of two acute angles can be obtuse.
c) More generally, $M$ can not be full rank positive definite because then the factors in the Cholesky decomposition $LL^T$ of $M$ would also be full rank and the problem could be reduced to case b) by change of basis using $L$.
Now I am stuck in the space of low rank positive semidefinite matrices. Is there a way to characterize such matrices further? Can such a matrix even exist, apart from zero matrix?

EDIT2: The deduction that $M$ is PSD in step a) implicitly assumed that $M$ is symmetric. Keith Kearnes' answer explicitly shows that $M$ must be rank 1 in that case. The case when $M$ is anti-symmetric is open but I doubt it would be possible to prove anything in that case. I will wait a few days to get more answers. Thanks.
EDIT3: Darij Grinberg's answer proved that every transitive matrix must be symmetric. This completed the characterization.

 A: I assume from the wording of the question (positive semidefinite, Cholesky decomposition) that you intend $M$ to be symmetric. Write $M$ as $N^TN$ and let $V\leq \mathbb R^d$ be the range of $N$. For $u, v, w\in \mathbb R^d$ let $x=Nu, y=Nv, z= Nw$. Then your condition reduces to 
$$
x^Ty>0 \;\&\; y^Tz>0 \Rightarrow x^Tz>0$$ 
on $V$. If the dimension of $V$ is 2 or more, then (as you observed in your last sentence under your item (b)) it is possible to contradict this. That is, it is possible to find $x, y, z$ in the same plane in $V$ such that the angles between $x$ and $y$ and between $y$ and $z$ are acute, while the angle between $x$ and $z$ is obtuse. Once they are known, you can solve for $u, v, w$ which contradict the original condition.
On the other hand, if the dimension of $V$ is 1, there is no contradiction. In this case, $N$ is $1\times d$, so $u^TN^T = Nu$ is a real number, $u^TN^TNv = (Nu)(Nv)$, and you can argue that if 
$(Nu)(Nv) > 0$ and $(Nv)(Nw) > 0$,
then the number $(Nu)(Nw)$ has the same sign as $$(Nu)(Nv)^2(Nw) = [(Nu)(Nv)][(Nv)(Nw)] > 0.$$
The condition that the dimension of $V$ is 1 is that $N$ (and $M$) have rank $1$.
So, the answer for symmetric $M$ is: $M$ is positive semidefinite of rank at most $1$.
A: 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.
