An inner product space question

Question

Let $V$ be an inner product space and $u_1,\dots,u_k$ vectors of $V$. Let $G$ be the Gram matrix $$G_{ij}= (u_i| u_j)$$ I need to prove that if $u_1,.,u_k$ is a base of $V$ then product: $$ y^t*G*\overline{y}>0$$ for $y$ that belongs to $F^n$.

I've no clue about that, and I would much appreciate any help! thx

Answer 1

If the inner product $\langle u_i, u_j \rangle$ of two column vectors in $F^n$ (where the field $F$ is $\mathbb{R}$ or $\mathbb{C}$) is defined to be the usual scalar product $(u_i)^T (u_j)$, then we can form a matrix $U = [u_1 ~u_2 \cdots u_n]$ whose columns are the $u_i$'s. Define the so-called Gram matrix $G=U^T U$. Observe that $G_{ij} = (u_i)^T (u_j)$. Hence, $x^T Gx = x^T (U^T U)x = (x^T U^T)(Ux) = (Ux)^T (Ux)$. This last quantity is the sum of the squares of the components of the vector $(Ux)$ (it can be rewritten as $|Ux|^2$ if the absolute values denote the usual length of a vector). If the columns of $U$ form a basis for the space, then $U$ is nonsingular, and so $U$ takes every nonzero vector to a nonzero vector. So, $x \ne 0$ implies $Ux \ne 0$ and that the length of $|Ux|$ is strictly positive. This proves that if $x$ is nonzero and the $u_i$'s form a basis then $x^T G x > 0$.

Answer 2

A Gram matrix is one of the form $G=X^T X$ for some matrix $X$. Then the quadratic form corresponding to $G$ is $Q(y)=y^T G y = y^T X^T X y=(Xy)^T(Xy)=\| Xy \|^2$. Can you conclude?