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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 $(e_i, e_j) <0$ for all $i \neq j.$ Prove that any $n$ vectors of the system are linearly independent.

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    $\begingroup$ ... is $n$ the dimension of the space? $\endgroup$ – rschwieb Oct 26 '16 at 20:40
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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 $\lambda_1 > 0$.

    Then $\lambda_1 (e_1|e_1)+\lambda_2(e_1|e_2) =0$ and therefore $\lambda_2>0$.

    Then you can also write that $0=\lambda_1(e_1|e_3)+\lambda_2(e_2|e_3)<0$ which is absurd.

    You conclude that $e_1$ and $e_2$ are linearly independent.

  • Let $n>2$, 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 $i,j \neq n$). $$\begin{align}(f_i|f_j)&=(e_i-(e_i|e_n)e_n|e_j-(e_j|e_n)e_n)\\&=(e_i|e_j)-(e_i|(e_j|e_n)e_n)-(e_i|e_n)(e_n|e_j)+(e_i|e_n)(e_j|e_n)\|e_n\|^2\\&=(e_i|e_j)-(e_i|e_n)(e_j|e_n)<0\end{align}$$

    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.

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  • $\begingroup$ Great answer! Never thought of using induction in a problem of linear algebra. $\endgroup$ – bat_of_doom Jan 9 '17 at 6:15
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The result follows from the following Lemma, which is used in the theory of root systems to study bases of root systems.

Lemma: Let $(V, \langle\,{,}\,\rangle)$ be a Euclidean vector space. Let $\ell\in V^*$ a linear form and $A\subseteq V$ a subset such that:
(1) $\ell(x)>0$ for all $x\in A$.
(2) $\langle x,y\rangle \le 0$ for all $x,y\in A$, $x\neq y$.

Then $A$ is linearly independent.

Apply this Lemma to $\ell = \langle\,\cdot\,,-e_{i_0}\rangle$ and $A = \{e_1,\dotsc,\widehat{e_{i_0}},\dotsc,e_{n+1}\}$ (where $\widehat{e_{i_0}}$ denotes omission) for some fixed $i_0$.

Proof of the Lemma: Let $\sum_{x\in A}\lambda_xx = 0$ be a (finite) linear combination. Now, consider $$ v:= \sum_{x\in A_1}\lambda_xx = \sum_{x\in A_2}\mu_xx, $$ where $A = A_1\sqcup A_2$ and $\lambda_x\ge0$ for all $x\in A_1$, $\mu_x := -\lambda_x >0$ for all $x\in A_2$. Then $$ 0\le \langle v,v\rangle = \sum_{x\in A_1}\sum_{y\in A_2}\lambda_x \mu_x \langle x,y\rangle \le 0 $$ where we have used (2) for the last inequality. It follows that $v = 0$. Hence, $$ 0 = \ell(v) = \sum_{x\in A_1}\lambda_x \ell(x) = \sum_{x\in A_2} \mu_x \ell(x). $$ Since (1) says that $\ell(x)>0$ for all $x\in A$, it follows that $\lambda_x = 0$ for all $x\in A_1$ and $-\lambda_x = \mu_x = 0$ for all $x\in A_2$. Thus, $A$ is linearly independent.

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  • $\begingroup$ Very nice! .......... +1 $\endgroup$ – user1551 Jan 8 '17 at 14:25
  • $\begingroup$ How do you choose such a linear form? $\endgroup$ – bat_of_doom Jan 8 '17 at 16:23
  • $\begingroup$ @bat_of_doom I already answered how to apply this Lemma, directly below the Lemma! I only had the notation for the scalar product wrong, whence my edit. $\endgroup$ – Claudius Jan 9 '17 at 5:41
  • $\begingroup$ @user218931 Sorry, I didn't understand the notation earlier. $\ell(e_j) = <e_j|(-e_i)>$, isn't it? $\endgroup$ – bat_of_doom Jan 9 '17 at 6:14
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    $\begingroup$ @bat_of_doom yes, this is the linear form. The above Lemma came up in the study of root systems. I have never seen it being applied to other fields, so I wouldn't say that this Lemma is common. $\endgroup$ – Claudius Jan 9 '17 at 10:04
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Yet another proof. We shall prove the statement for $n\ge1$ only, because the case $n=0$ is a vacuous truth (an empty set is linearly independent by convention). Suppose the contrary that $e_1,\ldots,e_n$ are linearly dependent, so that some non-trivial linear combinations of them is zero. Among all such linear combinations, pick one with the minimal number of nonzero coefficients. Write it as $$ \sum_{i\in I} p_ie_i-\sum_{j\in J}q_je_j=0,\tag{1} $$ where $I,J$ are two non-overlapping subsets of $\{1,2,\ldots,n\}$ and every $p_i,\,q_j>0$.

The linear combination is non-trivial, so $I$ and $J$ cannot be both empty. We may assume that $I$ is non-empty. Then $J$ must be non-empty too, otherwise $(1)$ implies that $\sum_{i\in I} p_ie_i=0$ and contradiction arises by taking inner products on both sides with $e_{n+1}$.

Furthermore, $I$ cannot be singleton, otherwise, if $p_1e_1=\sum_{j\in J}q_je_j$, contradiction arises if we take inner products on both sides with $e_1$.

Thus $I$ has at least two indices. Pick any $i_0\in I$. Let $x=p_{i_0}e_{i_0}$. Then $y=\sum_{i\in I\setminus\{i_0\}} p_ie_i$ is a non-empty sum and $x+y=\sum_{j\in J}q_je_j$. Therefore \begin{align*} \langle x,y\rangle=\sum_{i\in I\setminus\{i_0\}}p_i\langle x,e_i\rangle&\lt0,\\ \langle x,x\rangle + \langle x,y\rangle=\langle x,x+y\rangle =\sum_{j\in J}q_j\langle x,e_j\rangle&\le0,\\ \langle y,x\rangle + \langle y,y\rangle=\langle y,x+y\rangle =\sum_{j\in J}q_j\langle y,e_j\rangle&\le0 \end{align*} and in turn $\langle x,x\rangle \langle y,y\rangle\le |\langle x,y\rangle|^2$. By Cauchy-Schwarz inequality, $x$ and $y$ must be linearly dependent. Consequently, we obtain a linearly dependent set $\{e_i: i\in I\}$ that is strictly smaller than $\{e_k: k\in I\cup J\}$ because $J$ is non-empty. This contradicts the minimality of $(1)$. Hence our initial assumption is wrong and $e_1,\ldots,e_n$ must be linearly independent.

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  • $\begingroup$ Thank you for the great answer. How did you come up with such an elegant solution? $\endgroup$ – bat_of_doom Jan 9 '17 at 6:43
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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 $t>0$, 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.

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