Prove that vectors in Euclidean space are linearly independent Could you help me with an idea of solving the following problem? I think that proof involves the positive definiteness of Gram matrix, but I don't know how.

Consider a system of vectors $e_1, e_2, ..., e_n, e_{n+1}$ in some Euclidean space such that dot product $(e_i, e_j) &lt;0$ for all $i \neq j.$ Prove that any $n$ vectors of the system are linearly independent.
Let us try by recurrence and suppose that $n$ is the dimension of the space.

for $n=2$ we have $(e_1, e_2, e_3)$ in an Euclidian space of dimension 2, such that the scalar products are non positive.
Let us suppose that $e_1$ and $e_2$ (without loss of generality) are linearly dependent, i.e that there exist $(\lambda_1, \lambda_2) \neq (0,0) $ such that $$\lambda_1 e_1 + \lambda_2 e_2 = 0$$ If $\lambda_1 = 0$, then $\lambda_2 = 0$ therefore we can suppose that $\lambda_1 &gt; 0$.
Then $\lambda_1 (e_1|e_1)+\lambda_2(e_1|e_2) =0$ and therefore $\lambda_2&gt;0$.
Then you can also write that $0=\lambda_1(e_1|e_3)+\lambda_2(e_2|e_3)&lt;0$ which is absurd.
You conclude that $e_1$ and $e_2$ are linearly independent.

Let $n&gt;2$, and suppose that the property we want to prove is right at the rank $n-1$. Let $e_1, \dots e_{n+1}$ be vectors verifying the hypothesis from the question. We can suppose that each of these vectors have unit norm (by dividing them by their norm).
Let us take $n$ of these vectors, let us say $e_1, \dots, e_n$ and suppose they are linearly dependent, i.e. that there exist $(\lambda_1, \dots, \lambda_n) \neq (0, \dots, 0)$ such that $$\sum_{i=1}^n \lambda_i e_i =0$$ Then $$\sum_{i=1}^n \lambda_i (e_i|e_n)e_n =0$$ If we substract the two previous equalities we obtain $$\sum_{i=1}^n \lambda_i f_i =0$$ where we note $f_i=e_i-(e_i|e_n)e_n$.
We notice that $f_n=0$ (since $\|e_n\|^2=1$). This shows that $$\sum_{i=1}^{n-1} \lambda_i f_i =0$$
Let us know compute $(f_i|f_j)$ for $i \neq j$ (and $i,j \neq n$). $$\begin{align}(f_i|f_j)&amp;=(e_i-(e_i|e_n)e_n|e_j-(e_j|e_n)e_n)\\&amp;=(e_i|e_j)-(e_i|(e_j|e_n)e_n)-(e_i|e_n)(e_n|e_j)+(e_i|e_n)(e_j|e_n)\|e_n\|^2\\&amp;=(e_i|e_j)-(e_i|e_n)(e_j|e_n)&lt;0\end{align}$$
Moreover all the $f_i$ ($i \neq n$) are non-zero vectors located in $(e_n)^{\perp}$ space, which is of dimension $n-1$.
We have now $n$ vectors verifying the hypothesis at rank $n-1$, therefore any $n-1$ combination of the vectors $(f_i)_{i \in \{1, \dots, n-1, n+1 \}}$ is linearly independant. In particular $f_1, \dots, f_{n-1}$ are independant and consequently $\lambda_1, \dots, \lambda_{n-1} = 0$.
This proves that $(e_1, \dots, e_n)$ are independant. And the property is true at rank n.
Here is an overkill, but it has an interesting connection to another area of matrix theory.

Let $E=[e_1|\cdots|e_n|e_{n+1}]$. By the given conditions, $P=E^\top E\in M_{n+1}(\mathbb R)$ is a positive semidefinite irreducible $Z$-matrix. Therefore $P+tI$ is an $M$-matrix for every $t&gt;0$, so that $(P+tI)^{-1}\ge0$ entrywise. As $P+tI$ is irreducible, so is $(P+tI)^{-1}$. By Perron-Frobenius theorem, the largest eigenvalue of $(P+tI)^{-1}$ is simple. Thus the smallest eigenvalue of $P$ is simple too. Hence the rank deficiency of $P$ is at most $1$ because $P$ is positive semidefinite.

It follows that $\operatorname{rank}(E)=\operatorname{rank}(P)\ge n$. That is, some $n$ vectors of $\{e_1,e_2,\ldots,e_n,e_{n+1}\}$ are linearly independent. Without loss of generality, suppose the subset $\{e_2,\ldots,e_{n+1}\}$ is linearly independent. We want to prove that $\{e_1,e_2,\ldots,e_n\}$ (and similarly for other subsets of size $n$ containing $e_1$) is also linearly independent. Suppose
$$
\sum_{i=1}^k p_ie_i = \underbrace{\sum_{i=k+1}^n p_ie_i}_{:=\, x},
$$
where $k\ge1$ and each $p_i$ is nonnegative. By taking inner products on both sides with $x$, we see that $x$ must be zero. As $e_{k+1},\ldots,e_n$ are linearly independent, $p_i$ must be zero for every $i\ge k+1$. In turn, $\sum_{i=1}^k p_ie_i = 0$. Take inner products again on both sides with $e_{n+1}$, we see that $p_i$ must also be zero for each $i\le k$. Hence all coefficients are zero and $e_1,\ldots,e_n$ are linearly independent. QED

Our conclusion also translates to the following


  Proposition. The rank deficiency of a positive semidefinite irreducible $Z$-matrix is at most $1$. Furthermore, if the matrix has strictly negative off-diagonal entries, then its proper principal submatrices are all nonsingular.