Hope you're doing good!
Well I have a problem. The thing is when wanting to prove that a matrix A is a symmetric positive definite matrix then all you have to do is prove that A is symmetric and that $\forall \boldsymbol{x} \in V \backslash\{\mathbf{0}\}: \boldsymbol{x}^{\top} \boldsymbol{A} \boldsymbol{x}>0$. But assuming the following:
Consider an $n$ -dimensional vector space $V$ with an inner product $\langle\cdot, \cdot\rangle$ : $V \times V \rightarrow \mathbb{R}$ (see Definition 3.3 ) and an ordered basis $B=\left(b_{1}, \ldots, b_{n}\right)$ of $V$. Recall from Section 2.6 .1 that any vectors $x, y \in V$ can be written as linear combinations of the basis vectors so that $x=\sum_{i=1}^{n} \psi_{i} b_{i} \in V$ and $\boldsymbol{y}=\sum_{j=1}^{n} \lambda_{j} \boldsymbol{b}_{j} \in V$ for suitable $\psi_{i}, \lambda_{j} \in \mathbb{R}$. Due to the bilinearity of the inner product, it holds for all $x, y \in V$ that $$ \langle\boldsymbol{x}, \boldsymbol{y}\rangle=\left\langle\sum_{i=1}^{n} \psi_{i} \boldsymbol{b}_{i}, \sum_{j=1}^{n} \lambda_{j} \boldsymbol{b}_{j}\right\rangle=\sum_{i=1}^{n} \sum_{j=1}^{n} \psi_{i}\left\langle\boldsymbol{b}_{i}, \boldsymbol{b}_{j}\right\rangle \lambda_{j}=\hat{\boldsymbol{x}}^{\top} \boldsymbol{A} \hat{\boldsymbol{y}} $$
Then why don't we have to prove that $A_{i j}:=\left\langle\boldsymbol{b}_{i}, \boldsymbol{b}_{j}\right\rangle$ to know if the matrix represented an inner product of V?
Example:
$$ \boldsymbol{A}_{1}=\left[\begin{array}{ll} 9 & 6 \\ 6 & 5 \end{array}\right] $$ $A_{1}$ is positive definite because it is symmetric and $$ \begin{aligned} \boldsymbol{x}^{\top} \boldsymbol{A}_{1} \boldsymbol{x} &=\left[\begin{array}{ll} x_{1} & x_{2} \end{array}\right]\left[\begin{array}{ll} 9 & 6 \\ 6 & 5 \end{array}\right]\left[\begin{array}{l} x_{1} \\ x_{2} \end{array}\right] \\ &=9 x_{1}^{2}+12 x_{1} x_{2}+5 x_{2}^{2}=\left(3 x_{1}+2 x_{2}\right)^{2}+x_{2}^{2}>0 \end{aligned} $$ Why don't we have to prove that Aij = ⟨bi,bj⟩?
Thanks in advanced for your reponses.