Showing existence of $L: \mathbb{R}^d \to \mathbb{R}^d$ s.t $\langle x,y\rangle^{'} = \langle L(x), y\rangle$ 
Let $\langle \,\cdot\,,\,\cdot\, \rangle'$ be some inner product in $\mathbb{R}^d$. Show that there is a positive linear operator $L: \mathbb{R}^d \to \mathbb{R}^d$ s.t $\langle x,y\rangle^{'} = \langle L(x), y\rangle$ where $\langle\,\cdot\,,\,\cdot\,\rangle$ is the standard inner product.

I'm not even sure what a positive linear operator means, I know what is a positive definite matrix but from some reading online it seems those are different definitions(?). Help appreciated.
 A: $\langle \cdot, \cdot \rangle$ (respectively $\langle \cdot, \cdot \rangle^\prime$) induces a non degenerate form $B : \mathbb  R^d \times \mathbb R^d \to \mathbb R$ (respectively $B^\prime : \mathbb  R^d \times \mathbb R^d \to \mathbb R$).
$B$ (respectively $B^\prime$) induces a linear isomorphism
$$\begin{array}{l|rcl}
\tilde{B} : & \mathbb R^d & \longrightarrow & \left(\mathbb R^d\right)^\prime \\
    & x & \longmapsto & y \mapsto B(x,y)
\end{array}$$ where $\left(\mathbb R^d\right)^\prime$ stands for the dual of $\mathbb R^d$.
Hence $L = \tilde{B}^{-1} \circ \tilde{B}^\prime$ is a linear automorphism of $\mathbb R^d$. We have $\tilde{B} \circ L = \tilde{B}^\prime$ and therefore
$$\langle x, y \rangle^\prime = B^\prime(x,y) = (\tilde{B}^\prime(x))(y)= ((\tilde{B}\circ L)(x))(y) = B(L(x),y) = \langle L(x), y \rangle$$ for all $x,y \in \mathbb R^d$ as desired.
Moreover
$$\langle L(x), x \rangle = \langle x, x \rangle^\prime >0$$ for all $0 \neq x \in \mathbb R^d$ meaning that $L$ is positive definite.
