@@ -100,18+103,21 @@ If the inner product is nondegenerate, i.e.\ $\forall\vec{u}\, (\vec{v},\vec{u}
Since the dimension of $V$ and $V^{\dagger}$ is equal, it is additionally surjective.
Then, $V$ is isomorphic to $V^{\dagger}$.
Vector $\vec{v}^{\dagger}\equiv \varphi_{\vec{v}}\in V^{\dagger}$ is said to be the dual vector of $\vec{v}\in V$.
Since the dimension of $V$ and $V^{\dagger}$ is equal, it is additionally surjective.
Then, $V$ is isomorphic to $V^{\dagger}$.
Vector $\vec{v}^{\dagger}\equiv \varphi_{\vec{v}}\in V^{\dagger}$ is said to be the dual vector of $\vec{v}\in V$.
+The dual pairing $[\vec{v}^{\dagger},\vec{u}]=[\varphi_{\vec{v}},\vec{u}]=\varphi_{\vec{v}}(\vec{u})$ is associated with the inner product $(\vec{v},\vec{u})$.
-In physics and matrix algebra, the inner product is often defined with linearity in the second argument and conjugate linearity in the first argument.
-This allows to express the inner product $(\vec{u},\vec{v})$ as a product of vector $\vec{v}$ with a dual vector or linear functional of dual space $V^{\dagger}$
+Now, in physics and matrix algebra, the inner product is often defined with linearity in the second argument and conjugate linearity in the first argument.
+In this case, the antilinearity property is assigned to element $\varphi_{\vec{v}}=\vec{v}^{\dagger}$ of dual space indicating an isomorphism of $V$ to the conjugate complex of its dual space.