Hope you're having a great day.
Well I've been learning about Analytic Geometry in the mml book, but I've found myself stuck the question statement presented in the photo. Why does A(i,j) := <b(i),b(j)>? I'm really confused.
The quote in the text book goes as follows:
"Consider an n-dimensional vector space V with an inner product h·, ·i : V × V → R (see Definition 3.3) and an ordered basis B = (b1, . . . , bn) of V . Recall from Section 2.6.1 that any vectors x, y ∈ V can be written as linear combinations of the basis vectors so that $ \boldsymbol{x}=\sum_{i=1}^{n} \psi_{i} \boldsymbol{b}_{i} \in V$ and $y=\sum_{j=1}^{n} \lambda_{j} b_{j} \in V$ for suitable ψi , λj ∈ R. Due to the bilinearity of the inner product, it holds for all x, y ∈ V that
$\langle\boldsymbol{x}, \boldsymbol{y}\rangle=\left\langle\sum_{i=1}^{n} \psi_{i} \boldsymbol{b}_{i}, \sum_{j=1}^{n} \lambda_{j} \boldsymbol{b}_{j}\right\rangle=\sum_{i=1}^{n} \sum_{j=1}^{n} \psi_{i}\left\langle\boldsymbol{b}_{i}, \boldsymbol{b}_{j}\right\rangle \lambda_{j}=\hat{\boldsymbol{x}}^{\top} \boldsymbol{A} \hat{y}$
where $ A_{i j}:=\left\langle\boldsymbol{b}_{i}, \boldsymbol{b}_{j}\right\rangle $ and $ \hat{\boldsymbol{x}}, \hat{\boldsymbol{y}} $ are the coordinates of x and y with respect to the basis B."
Thank you in advance for your responses.
It is just how matrix and vector multiplication works. If you set $$ \hat x =\begin{pmatrix} \psi_1 \\ \vdots \\ \psi_n \end{pmatrix} \;\;\;\mathrm{and} \;\;\; \hat y =\begin{pmatrix} \lambda_1 \\ \vdots \\ \lambda_n \end{pmatrix} $$ you will note that $$ \sum\limits_{i=1}^n \sum\limits_{j=1}^n \psi_i A_{ij} \lambda_j = \hat x^T A \hat y\; . $$ So you get $$ \sum\limits_{i=1}^n \sum\limits_{j=1}^n \psi_i \,\langle \mathbf{b}_i , \mathbf{b}_j\rangle \,\lambda_j = \hat x^T A \hat y $$ if you set $$ A_{ij} = \langle \mathbf{b}_i , \mathbf{b}_j\rangle $$