Dot product is "just" a specific example of an inner product. Is this right?
I am asked to find a scalar product (I think they mean the inner product) on $$ V = \mathbb{R}^3 s.t. \left( \begin{array}{c} 1 \\ 1 \\ 0 \\ \end{array} \right), \left( \begin{array}{c} 1 \\ 0 \\ 1 \\ \end{array} \right), \left( \begin{array}{c} 0 \\ 1 \\ 1 \\ \end{array} \right)$$ is an orthonormal basis.
There is solution for this kind of task in this forum, but I have a few doubts.
Solution:
Call the $i$th such vector $v_i$. Let $V$ denote the matrix satisfying $V_{ik}=(v_i)_k$. We seek an inner product $a,\,b\mapsto a_kM_{kl}b_l$, with implicit summation over repeated indices, so that $\delta_{ij}=V_{ik}M_{kl}V_{jl}$ i.e. $VMV^T=I$. So take $M=V^{-1}(V^T)^{-1}=(V^TV)^{-1}$. In this case $$V=\left(\begin{array}{ccc} 1 & 1 & 0\\ 1 & 0 & 1\\ 0 & 1 & 1 \end{array}\right),\,V^{T}V=\left(\begin{array}{ccc} 2 & 1 & 1\\ 1 & 2 & 1\\ 1 & 1 & 2 \end{array}\right),\,M=\frac{1}{4}\left(\begin{array}{ccc} 3 & -1 & -1\\ -1 & 3 & -1\\ -1 & -1 & 3 \end{array}\right).$$
Questions:
Can someone explain the part "we seek an inner product..". Is $M$ a matrix representation of the inner product? Why do a and b have different indices? Shouldn't it be like: $a,\,b\mapsto a_kM_{kl}b_k$ So hat a and b have the same indices.
Is this a general formula for calculating the inner product? $VMV^T=I$? Why does this hold?
Thank you very much for the explanations.
The idea here is that an inner product is just a map \begin{align*} \langle\cdot, \cdot \rangle: \ V^2 \rightarrow \mathbb{R} \end{align*} that maps two vectors $a,b \in V$ onto a real number $\langle a,b \rangle \in \mathbb{R}$. Now, we seek a specific map such that the given vectors $v_i$ are actually orthonormal, i.e. $\langle v_i, v_j \rangle = \delta_{ij}$.
It is best to explain the following with the simplest possible example, the standard Euclidean inner product, at hand. It is given by \begin{align*} v \cdot w = \sum_i v_i w_i = v^T w = v^T I w = \sum_{ij} v_i I_{ij} w_j \,. \end{align*} Thus, the identity matrix $I$ describes the Euclidean inner product.
Yes. The same way as the Euclidean inner product is described by the identity matrix.
Because they are multiplied from different sides onto the matrix $M$. Refer to the definition of multiplication of matrices and then write down the product $a^TMb$ of row vector $a^T$, matrix $M$, and column vector $b$.
No. See explanation above.
The inner product is orthonormal w.r.t. the basis vectors $v_i$, i.e. $$ \langle v_i, v_j \rangle = v_i^T M v_j = \sum_{kl} \left(v_i\right)_k M_{kl} \left(v_j\right)_l \overset{!}{=} \delta_{ij} \quad \forall i,j=1,2,3 \,. $$ Note that there are actually 9 different scalar equations above. One can summarize them in matrix notation by defining a matrix $V$ who's columns are given by the basis vectors $v_i$. The matrix equation is then given by $$ V M V^T = I \,. $$