Theorem 6.8
Let \(V\) be a finite-dimensional inner product space over \(\mathbf{F}\), and let \(g: V \rightarrow \mathbf{F}\) be a linear transformation. Then, there exists a unique vector \(y \in V\) such that \(g(x) = \langle x, y \rangle\) such that \(g(x) = \langle x, y \rangle\) for all \(x \in V\)


Proof:

Let \(\beta = \{v_1,...,v_n\}\) be an orthonormal basis for \(V\), and let

$$ \begin{align*} y = \sum_{i=1}^{n} \overline{g(v_i)}v_i \end{align*} $$

Define \(h: V \rightarrow \mathbf{F}\) by \(h(x) = \langle x, y \rangle\) which is linear. Furthermore, for \(i \leq j \leq n\), we have

$$ \begin{align*} h(v_j) = \langle v_j \rangle &= \left\langle v_j, \sum_{i=1}^{n} \overline{g(v_i)}v_i \right\rangle \\ &= \sum_{i=1}^{n} g(v_i) \langle v_j, v_i \rangle \\ &= \sum_{i=1}^{n} g(v_i) \delta_{ij} \\ &= g(v_j) \end{align*} $$

But \(h\) and \(g\) agree on a basis so by the corollary from 2.6, \(g = h\).

To show that \(y\) is unique. Suppose it is not and let \(g(x) = \langle x, y' \rangle\). for all \(x\). Then, \(\langle x, y' \rangle = \langle x, y \rangle\) but by Theorem 6.1(e), this means that \(y = y'\) as we wanted to show. \(\ \blacksquare\)


Notes: So my understanding is that we can turn any linear transformation from \(V\) to \(\mathbf{F}\) into an inner product with some unique vector \(y \in V\). For example if \(V = \mathbf{R}^2\) and we have some \(g(x): V \rightarrow V\). where \(g(a,b) = 3a + 2b\). Then,

$$ \begin{align*} y = (3, 2) \end{align*} $$

is a unique vector in \(V\) such that

$$ \begin{align*} g(x) &= \langle x, y \rangle \\ &= \langle x, (3, 2) \rangle \\ &= 2a + 3b \end{align*} $$

At least this is what I understood!


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