The following question was assigned to my algebra class this week:
Let $V$ be a euclidean vector space (that is $\mathbb{F} = \mathbb{R}$) of dimension $n$ with some inner product $(,)$.
Let $\{v_1 \dots v_m \}$ be a set of vectors in $V$ that statisfy $(v_i,v_j) < 0$ for every $i \neq j$.
Show that $m \leq n + 1$. (Hint: project $\{v_1 \dots v_{m-1}\}$ over $\{v_m\}^\perp$)
I shall attempt to provide a rigorous proof below. The reader is welcome to point any errors and to provide an optimization suggestion to the proof as I believe it is not as polished as can be. (Btw, I saw in several meta posts that proof verifications are best posted as "answer your own question" type of post, so that's what I decided to go with)
Sincerest thanks for any comments or suggestions to the answer below. Have a great day!
Geometric Intuition
The way I see it, the question phrases a general statement to any dimension that describes a very simple concept in geometry.
Let us reduce the problem to a specific dimension ideally one where we can easily see an example that demonstrates the claim at hand. $\mathbb{R}^2$ is a good choice.
So now our given set consists of 4 vectors $\{v_1, v_2, v_3, v_4\}$. The provided information that $(v_i, v_j) < 0$ for any $i \neq j$ simply means that the angle between any vectors is an obtuse angle.
It's quite easy to understand (by drawing it for example) why that would be impossible for any 4 vectors in $V$ of dimension 2.
It is even more trivial when $V$ is of dimension 1 such that all vectors are linearly dependent (it is not to complicated to prove). The proof makes use of this idea.
Strategy of the Proof
The main idea of the proof is to reduce the problem into smaller and smaller dimensions until we reach a simple enough vector space where we can easily prove the claim.
We do so by creating a new set of vectors, this time consisting of $\{u_1 \dots u_{m - 1}\}$ that are all included in a subspace $U$ of $V$ where we have $\dim U = n-1$.
The trick is to show that the new set $\{u_1 \dots u_{m - 1}\}$ statisfies $(u_i,u_j) < 0$ for any $i \neq j$, or else we shall reach a contradiction to the given information. This can be done by employing the basic properties and axioms of an inner product and the orthogonal projection. Note that because our newly built set satisfies the same conditions as the original, we can repeat the process and get a smaller set of $m - 2$ vectors with a new subspace of dimension $n - 2$.
By induction, repeat the process again and again until reaching a subspace of dimension 1 and a set of at least 3 vectors. From there it is trivial to reach a contradiction.
Proof
Suppose, for the sake of contradition, that $m \geq n + 2$.
Let $\{v_m\}^\perp$ be the orthogonal complement of the set $\{v_m\}$.
Then, $V = \{v_m\}^\perp \oplus Sp\{v_m\}$.
Thus, we know that $\dim \{v_m\}^\perp = \dim V - \dim Sp\{v_m\} = n - 1$.
We shall proceed by projecting each of the remaining vectors in the given set $\{v_1 \dots v_{m - 1}\}$ onto $\{v_m\}^\perp$.
So let $\{u_1, \dots, u_{n-1}\}$ be an orthonormal basis to $\{v_m\}^\perp$. We can build such using some basis and the Gram-Schmidt process.
Denote the projection and it's corresponding perpendicular of each vector in $\{v_1, \dots, v_{m-1}\}$ by:
\begin{align*} v_1' = \sum_{i=1}^{n-1}(v_1, u_i)u_i,\;\;& \tilde{v_1} = v_1 - \sum_{i=1}^{n-1}(v_1, u_i)u_i \\ \vdots \\ v_{m-1}' = \sum_{i=1}^{n-1}(v_{m-1}, u_i)u_i,\;\;& \tilde{v_{m-1}} = v_{m-1} - \sum_{i=1}^{n-1}(v_{m-1}, u_i)u_i \end{align*}
(Remark (*): Recall that for every $i$, $v_i' \in \{v_m\}^\perp$, is a linear combination of $\{u_1, \dots, u_{n-1}\}$ and that
$\tilde{v_i} \in Sp\{v_m\}$ (for example because the product $(\tilde{v_i}, u_j)=0$ for any $j$))
Claim: $\{v_1', \dots, v_{m-1}'\}$ must satisfy: $(v_i', v_j') < 0 $ for all $i \neq j$.
So let $i \neq j$, \begin{align*} \boldsymbol{(1)}\;\;\;(v_i', v_j') = (v_i - \tilde{v_i}, v_j - \tilde{v_j}) = (v_i, v_j) - (v_i, \tilde{v_j}) - (\tilde{v_i}, v_j) + (\tilde{v_i}, \tilde{v_j}) \end{align*} We can also deduce that: \begin{align*} (v_i, \tilde{v_j}) = (v_i' + \tilde{v_i}, \tilde{v_j}) = 0 + (\tilde{v_i}, \tilde{v_j}) \end{align*} The same can be similiraly proved regarding $(\tilde{v_i}, v_j) = (\tilde{v_i}, \tilde{v_j})$.
Using (1) above: \begin{gather*} \boldsymbol{(2)} \;\; (v_i', v_j') = (v_i, v_j) - (\tilde{v_i}, \tilde{v_j}) \end{gather*} Now suppose that $(\tilde{v_i}, \tilde{v_j}) \leq 0$. Then, if $(v_i, v_m) = (\tilde{v_i}, v_m) < 0$, and (by Remark (*)) $\tilde{v_i} = \lambda \tilde{v_j}$.
\begin{gather*} (\tilde{v_i}, \tilde{v_j}) = \lambda(\tilde{v_j}, \tilde{v_j}) \leq 0 \\ \Downarrow \\ \lambda \leq \: 0\\ \Downarrow \\ (v_j, v_m) = (\tilde{v_j}, v_m) = \: \lambda(\tilde{v_i}, v_m) \geq 0 \end{gather*} In condradiction with the given information about $\{v_1 \dots v_m \}$. Thus, to statisfy the conditions of the question it must be true that $(\tilde{v_i}, \tilde{v_j}) > 0$.
By (2) $(v_i', v_j') < 0$. Note, that the set $\{v_1', \dots, v_{m-1}'\}$ also satisfies the question's conditions, lets denote it by $\Omega_1$. Also note that $\Omega_1 \subseteq \{v_m\}^\perp$.
Now, choose $v_{m-1}' \in \Omega_1$ and repeat the exact same process to reduce the number of vectors by 1 and the dimension of the corresponding subspace by 1. Denote the new set $\Omega_2$.
Repeat the process until the set $\Omega_{n-1}$. Note that there are at least $m - n + 1 = 3$ vectors in $\Omega_{n-1}$ and the dimension of the subspace $U$ s.t. $\Omega_{n-1} \subseteq U$ is 1. Choose any 3 vectors for example $w_1, w_2, w_3 \in \Omega_{n-1}$. These are linearly dependent as these exist in a subspace of dimension 1 and satisfy the conditions of the question regarding their negative inner product. So if: \begin{gather*} w_1 = \lambda_1 w_2, \; \; w_2 = \lambda_2 w_3 \end{gather*} By the condition of the question we have:
\begin{gather*} (w_1, w_2) = \lambda_1(w_2, w_2) < 0 \Longrightarrow \lambda_1 < 0 \\ (w_2, w_3) = \lambda_2(w_3, w_3) < 0 \Longrightarrow \lambda_2 < 0 \end{gather*} But then we also have:
\begin{gather*} (w1, w_3) = \lambda_1\lambda_2(w3, w3) > 0 \end{gather*} A contradiction.
In summary, the proof tries to show that if the conditions of the question are satisfied, it must also be true for a subspace of dimension 1. From there it is easy to show these conditions cannot hold.
I wasn't sure if the reduction process is redundant but currently I cannot see a different way to explain myself.
Regards to the reader! I'll be grateful for any improvement suggestions or comments pointing any mistakes or inaccuracies in the proof. Thank you!