Let $\{A_1,A_2...A_K\}$ be a set of linearly independent matrices in $M^\mathbb R_{n \times n}$.
Prove that the real $k\times\ k$ matrix $U$, given by: $U_{ij} = (tr(A^t_iA_j))$ where $1 \leq i,j \leq k$, is invertible.
Not sure how to tackle this at all. My approach was to think of $U$ as a Gramian matrix in respect to the standard inner product, because obviously $U_{ij} = <A_i,A_j>$. I know that this should imply that $det(U) \neq 0$, but not sure why. Any help on this matter, and insight on general Gramiam matrices would be very helpful. Thanks a lot.
edit: I know that a Gramiam matrix is generally referring to inner products between vectors, but I'm hoping there's a general notion of it as just inner products.
Assume that $U$ is noninvertible, then its columns form a linearly dependent set of $\mathbb{R}^n$, namely there exists a nonzero vector $(\lambda_1,\ldots,\lambda_n)$ such that for all $i$, one has: $$\sum_{j=1}^n\lambda_j\langle A_i,A_j\rangle=0.$$ Hence, for all $i$, one gets: $$\left\langle\sum_{j=1}^n\lambda_jA_j,A_i\right\rangle=0.$$ This means that $\displaystyle\sum_{j=1}^n\lambda_jA_j\in\textrm{Span}(A_1,\ldots,A_n)^{\perp}$, but $\displaystyle\sum_{j=1}^n\lambda_jA_j\in\textrm{Span}(A_1,\ldots,A_n)$, therefore: $$\sum_{j=1}^n\lambda_jA_j\perp\sum_{j=1}^n\lambda_jA_j,$$ which implies that: $$\sum_{j=1}^n\lambda_jA_j=0.$$ Finally, the $A_j$ are linearly dependent, a contradiction.
Remark 1. We even proved better, $\textrm{rank}(U)=\dim\left(\textrm{Vect}(A_1,\ldots,A_n)\right)$.
Remark 2. The above works for any real prehilbertian space.