Is $A=\{(\langle u_1,u_1\rangle,...,\langle u_1,u_n\rangle),...,(\langle u_n,u_1\rangle,...,\langle u_n,u_n\rangle)\}$ linearly independent? 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?
 A: 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.
