Let $ V $ be an Euclidian space of dimension $n$ and let $v_1,...,v_m \in V$ s.t. $\langle v_i,v_j \rangle<0$ for all $i\neq j$. Show $m \leq n + 1 $

Theorem: Let $ V $ be an Euclidian space of dimension $ n $ and let $ v_1,...,v_m \in V $ s.t. $ \langle v_i,v_j \rangle < 0 $ for all $ i \neq j $.
Show that $ m \leq n + 1 $ ( Hint: project $ v_1,...,v_{m-1} $ on $ \{ v_m \}^{\perp} $ )

I have no idea what to do, I feel like I'm stuck because I don't fully understand how to apply the definition of projection in this problem, nevertheless, this is an exercise I must do.

I was given the following ( Suppose the vector spaces we are talking about are finitely created ):

(Definition) Let $ W \subseteq V $ be a subspace and let $ w_1,...,w_r $ be an orthonormal basis of $ W $. Then for all $ v \in V $ we'll define the projection of $ v $ on $ W$ as $ w = \langle v,w_1 \rangle w_1 + ... + \langle v,w_r \rangle w_r $ .

(Definition) Let $ W \subseteq V $ be a subspace, we'll define the orthogonal complement of $ W $ as $ W^{\perp} = \{ v \in V | \forall w > \in W. \langle v,w \rangle = 0 \} $ .

(Theorem) Let $ W \subseteq V $ be a subspace, then $ V = W \oplus W^{\perp} $.

I thought that I'd prove the theorem by induction on $ n $ and at base case and induction step I'd look at a basis $ ( \tau_1, ..., \tau_q ) $ of $ \{ v_m \}^{\perp} $ where $ \dim \{ v_m \}^{\perp} = q $ and then using the hint, the projection of $ v_1,...,v_{m-1} $ on $ \{ v_m \}^{\perp} $ will give me the vectors:
$ \delta_1 = \langle v_1,\tau_1 \rangle \tau_1 + ... + \langle v_1,\tau_q \rangle \tau_q $
.
.
.
$ \delta_{m-1} = \langle v_{m-1},\tau_1 \rangle \tau_1 + ... + \langle v_{m-1},\tau_q \rangle \tau_q $

But I don't know if what I've done will give me anything fruitful... Can you please help? I don't know how to prove the theorem.

Thanks in advance!

Solution 1:

Let $m > n + 1$ and $v_1, \ldots, v_m$ be any vectors in $\mathbb R^n$. Let $w_j$ equal $v_j$ extended with an additional $1$, so $w_j \in \mathbb R^{n+1}$. Then the $w_j$ are linearly dependent: there are coefficients $c_j$, not all zero, such that both

1. $\sum c_j v_j = 0$
2. $\sum c_j = 0$

The second equality implies that not all non-zero $c_j$ can have the same sign. Let $I$ be the set of indices for which $c_j >0$ and $J$ the set for which $c_j<0$. Then 1. can be written as $$v = \sum_I c_j v_j = -\sum_J c_j v_j.$$ It follows that $$0 \leq \langle v, v \rangle = \sum_{(j,k) \in I\times J} -c_j c_k \langle v_j, v_k \rangle.$$ Now all coeficients $-c_j c_k$ are positive and therefore not all $\langle v_j, v_k \rangle$ can be negative.

Solution 2:

Here's a way to visualize it, hopefully you can turn it into a proof.

First rotate coordinates so that $v_m$ is a positive multiple of ${\bf e}_n$, the $n$th unit coordinate vector. Then the vectors $v_1,...v_{m-1}$ all have negative $n$th entry. Then repeat on the vectors $v_1,...v_{m-1}$; rotate in the first $m-1$ variables so that the second to last entry of $v_{m-1}$ is positive and the previous entries are all zero. (You don't have to worry about $v_{m-1}$ being a negative multiple of ${\bf e}_n$ since if that were the case there would be no other vectors; you can't have negative inner products with both a positive and a negative multiple of ${\bf e}_n$.)

Then each of $v_1,...,v_{m-2}$ will have their last two entries negative. Now repeat the above procedure, each time rotating so that the $k$th to last vector has its $k$th to last entry positive, its later entries negative, and its earlier entries zero. You can do this at most $n$ times. The remaining $m - n$ vectors will have to have all negative entries. Since any two such vectors will have positive inner product with each other, there can be at most one of them. Hence a maximum of $n + 1$ total vectors.