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Let $V$ be a vector space of dimension $n$, $B=\{u_1,....,u_n\}$ be a basis of it, and $\langle\, ,\,\rangle$ be an inner product on $V$.

Now, my question is that if the set $$A=\bigg\{\begin{pmatrix}\langle u_1,u_1\rangle\\ \vdots\\ \langle u_1,u_n\rangle\end{pmatrix},\begin{pmatrix}\langle u_2,u_1\rangle\\ \vdots\\ \langle u_2,u_n\rangle\end{pmatrix},\ldots,\begin{pmatrix}\langle u_n,u_1\rangle\\ \vdots \\ \langle u_n,u_n\rangle \end{pmatrix}\bigg\}$$ is linearly independent?

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  • $\begingroup$ That is not a question. $\endgroup$ – José Carlos Santos Jul 22 '17 at 13:58
  • $\begingroup$ @JoséCarlosSantos why? $\endgroup$ – Majid Jul 22 '17 at 13:59
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    $\begingroup$ Yes. This is the the Gramian matrix of the vectors $u_1,\ldots,u_n$. $\endgroup$ – José Carlos Santos Jul 22 '17 at 14:18
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    $\begingroup$ Inner products need to be typeset with \langle and \rangle, not < and >, to look correct. I've fixed the post. $\endgroup$ – Greg Martin Jul 22 '17 at 14:28
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    $\begingroup$ Following up on José Carlos Santos's comment: the Gramian matrix Wikipedia page says that this matrix has nonzero determinant if and only if the original set of vectors is linearly independent; that implies a positive answer to your question. $\endgroup$ – Greg Martin Jul 22 '17 at 14:35
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Let $\alpha_1,\ldots,\alpha_n$ some scalars satisfying $$\sum_{j=1}^n \alpha_j\begin{pmatrix}\langle u_j,u_1\rangle \\ \vdots \\ \langle u_j,u_n\rangle \end{pmatrix}=0$$ Then we have $$\langle \sum_{j=1}^n \alpha_j u_j, u_k \rangle =0 \qquad \forall k=1,\ldots,n.$$ It follows that $$\|\sum_{j=1}^n \alpha_j u_j\|^2=\langle \sum_{j=1}^n \alpha_j u_j, \sum_{k=1}^n \alpha_k u_k \rangle=\sum_{k=1}^n \alpha_k\underbrace{\langle \sum_{j=1}^n \alpha_j u_j, u_k \rangle}_{=0} =0.$$ Hence, we have $\sum_{j=1}^n \alpha_j u_j=0$ and since $u_1,\ldots,u_n$ are linearly independent, we must have $\alpha_1=\ldots=\alpha_n=0$ implying that the vectors in $A$ are linearly independent.

The converse is also true, indeed, if $B$ is not linearly independent, there exists $\alpha_1,\ldots,\alpha_n$ not all $0$ such that $\sum_{j=1}^n \alpha_j u_j=0$. Then, we have $$\sum_{j=1}^n \alpha_j\begin{pmatrix}\langle u_j,u_1\rangle \\ \vdots \\ \langle u_j,u_n\rangle \end{pmatrix}=\begin{pmatrix}\langle \sum_{j=1}^n \alpha_j u_j,u_1\rangle \\ \vdots \\ \langle \sum_{j=1}^n \alpha_ju_j,u_n\rangle \end{pmatrix}=\begin{pmatrix}\langle 0,u_1\rangle \\ \vdots \\ \langle 0,u_n\rangle \end{pmatrix}=0$$ and so the vectors in $A$ are linearly dependent.

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    $\begingroup$ @Surd Thanks a lot! $\endgroup$ – Majid Jul 22 '17 at 14:48

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