A solution without using Gram schmidt required.

Question

The question is given below:

I found a solution online using Gram Schmidt procedure but I need a solution without it, could anyone help me in this, please?

EDIT:

I was given a hint that I can show that there is a dual basis and that the inner product establishes an isomorphism $\delta$ between the space and its dual and to let $w_{j} = \delta^{-1} (v_{j}^*)$ where $v_{j}^*$ are the dual basis of $v_{j}$ and that $<v_{i}, w_{j}> \delta_{w_{j} (v_{i})}$ but still I do not understand the intuition of the hint. could anyone explain for me what is this hint doing?

Answer 1

Let us focus on $w_1$ first. Since $\langle v_i, w_1 \rangle = 0$ for $i = 2, \ldots, n$, we know $w_1$ must lie in the orthogonal complement of $\text{span}\{v_2, \ldots, v_n\}$. This orthogonal complement is nontrivial (since $\text{span}\{v_2, \ldots, v_n\}$ has dimension $<n$), so there exists some vector $u$ in it. We must have $\langle u, v_1 \rangle \ne 0$, else $v_1$ lies in $\text{span}\{v_2, \ldots, v_n\}$ which contradicts the fact that $\{v_1,\ldots, v_n\}$ is a basis. Thus, we can choose $w_1 := \frac{1}{\langle u, v_1 \rangle} u$ to obtain $\langle w_1, v_1 \rangle = 1$.

The other $w_2, \ldots, w_n$ can be determined similarly.

Answer 2

Write $$ w_j= \sum_k x_{jk} v_k.$$ Then the condition $\langle v_i, w_j\rangle = \delta_{ij}$ is equivalent to $$ \sum_k x_{jk} \langle v_i, v_k\rangle = \delta_{ij}, $$ which is the system of $n^2$ linear equations given in matrix form as $AX=I$, where $$ A=\begin{bmatrix} \langle v_i, v_j\rangle\end{bmatrix}, \quad X=\begin{bmatrix} x_{ij}\end{bmatrix}, $$ and $I$ is the $n\times n$ identity matrix.

Therefore, what really needs to be done is proving that $A$ is invertible.

To do that, try exploring the concept of Gramian matrix, for example on Wikipedia https://en.wikipedia.org/wiki/Gramian_matrix