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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.

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    $\begingroup$ b) isn't quite right for $d=1$. Actually for $d=1$ any matrix with a non-negative entry is an example. $\endgroup$ – Peter Mueller Jul 31 '15 at 19:42
  • $\begingroup$ Actually now I think that the rank of such matrices must be 1. Because every positive semidefinite matrix also has a eigen decomposition so I can again change my basis and reduce the problem to one where $M$ is diagonal with $1 \ge k < d$ positive values. If $k=1$ then fine otherwise if $k>1$ then I can use the counterexample in the $k$ dimensional space (used for b)) to create a counterexample here. $\endgroup$ – Pushpendre Jul 31 '15 at 21:45
  • $\begingroup$ @Pushpendre: Be careful; diagonalization might mess with the scalar product. $\endgroup$ – darij grinberg Jul 31 '15 at 23:09
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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 $1$.

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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.

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