Can all the above sums be negative? So, I have an exercise for my school and I am a little lost.
Every answer will be helpful!
So about the exercise: Let $a,b,c,d,e,f,g,h$ be real numbers. Can all the following sums $$ac+bd\\ ae+fb\\ ag+bh\\ ce+df\\ cg+dh\\ eg+fh$$ be negative numbers?.
I assumed that all of them are negative, I then assumed that every sum is the det of a matrix but it didn't help.
Is there any way I can prove that it's not possible?
Thank you all for your time!
 A: The six inequalities can be written, using the standard scalar product $(x,y)\bullet(z,w):=xz+yw$, as:
$(a,b)\bullet(c,d)<0$; $(a,b)\bullet(e,f)<0$; $(a,b)\bullet(g,h)<0$;
$(c,d)\bullet(e,f)<0$; $(c,d)\bullet(g,h)<0$; $(e,f)\bullet(g,h)<0$.
Now, this is impossible since you get that the four vectors $(a,b)$, $(c,d)$, $(e,f)$ and $(g,h)$ in the cartesian plane $\mathbb R^2$ have all angles of more than $\pi/2$ between them: but the sum becomes greater than $4\cdot \pi/2>2\pi$ that is impossible on a plane
A: This is the special case $n=2$ of the following proposition.
Proposition. When $n\ge1$, the maximum number of vectors in an $n$-dimensional real inner product space such that their pairwise inner products are negative is $n+1$.
We give two proofs, one based on simple geometry and the other based on Perron-Frobenius theorem.
Proof 1. We use mathematical induction on $n$. The base case $n=1$ is trivial. In the inductive case, suppose $v_1,v_2,\ldots,v_k\in\mathbb R^n$ are such that $\langle v_i,v_j\rangle<0$ for all $i\ne j$. Then each $v_i$ is nonzero. For each $i<k$, let
$$
w_i=v_i-\frac{\langle v_i,v_k\rangle}{\|v_k\|^2} v_k.
$$
That is, $w_i$ is the component of $v_i$ orthogonal to $v_k$. Then
$$
\langle w_i,w_j\rangle
=\left\langle v_i,v_j\right\rangle
-\frac{\langle v_i,v_k\rangle\langle v_j,v_k\rangle}{\|v_k\|^2}
<\left\langle v_i,v_j\right\rangle<0
$$
whenever $i\ne j$. Hence $w_1,w_2,\ldots,w_{k-1}$ are vectors in the $(n-1)$-dimensional inner product space $\{v_k\}^\perp$ such that their pairwise inner products are negative. So, by induction assumption, we have $k-1\le(n-1)+1$. In turn, $k\le n+1$.
Proof 2. We may assume that the inner product space is $\mathbb R^n$. Suppose the contrary that there are some $v_1,v_2,\ldots,v_{n+2}\in\mathbb R^n$ with $\langle v_i,v_j\rangle<0$ for all $i\ne j$. Normalise these vectors and put them into the columns of a matrix $V\in\mathbb R^{n\times(n+2)}$. Then $V^TV$ is a positive semidefinite matrix with a diagonal of ones, negative off-diagonal entries and nullity $\ge2$.
Let $F=I-V^TV$. Then $V^TV=(c+1)I-(F+cI)$ for every real number $c$. When $c>0$ is sufficiently large, we obtain $(c+1)I\succeq F+cI\succ0$. Since $V^TV$ is singular, we must also have $\rho(F+cI)=c+1$. As $F+cI$ is entrywise positive, by Perron-Frobenius theorem $c+1$ is a simple eigenvalue of $F+cI$. Hence $0$ is a simple eigenvalue of $V^TV=(c+1)I-(F+cI)$, which is a contradiction because the nullity of $V^TV$ is at least $2$.
