# Definition A [[Quadratic Forms on Vector Spaces|quadratic form]] on a [[vector space]] $V$ is non-degenerate if for each $v \neq 0 \in V$, there exists $w \in V$ such that $\braket{\pmb{x}|\pmb{y}}$ = 0, where $\braket{\pmb{x}|\pmb{y}}$ is the [[Hermitian Form]]. In other "words", $ \forall \pmb{x} \in V \exists \pmb{y} \in V | \braket{\pmb{x}|\pmb{y}} \neq 0 $ If the condition is violated such that $\pmb{x}^* \in V \exists$ and $\braket{\pmb{x}^*|\pmb{y}}$, then $\pmb{x}$ and $\pmb{x}+\pmb{x}^*$ are degenerate and cannot be distinguished by the [[Hermitian Form]]. This also means that if $\braket{\pmb{x}|\pmb{y}} = \braket{\pmb{x}^*|\pmb{y}} \forall \pmb{y} \in V$ then the condition is violated. # Weak and Strong Non-Degeneracy ## Weak Non-Degeneracy Let $Z$ be a real [[Banach Space]], possibly infinite-[[dimension|dimensional]], and $\Omega$ a [[bilinear form]]. We say $\Omega$ is *weakly non-degenerate* if $\Omega(z_1, z_2) = 0\, \forall\, z_2 \in Z$ implies $z_1 = 0$. This is the same as the definition above. Recall that we have an [[Bilinear Form#Associated Linear Map|induced continuous linear mapping]] $\Omega^\flat: Z \rightarrow Z^*$ such that $ \Omega^\flat(z_1)(z_2) = \Omega(z_1, z_2) $ and weak non-degeneracy is equivalent to $\Omega^\flat$ being [[Types of Maps|injective]], i.e., $\Omega^\flat(z) = 0 \implies z = 0$ (since, again, $\Omega^\flat$ is [[Continuous Map|continuous]] and [[Linear Map|linear]].) ## Strong Non-Degeneracy If, in addition to being an injection, $\Omega^\flat$ is also a [[Types of Maps|surjection]], i.e., onto, then $\Omega^\flat$ is a [[Types of Maps|bijection]] and thus an [[isomorphism]] (see [[Linear Map#Homomorphisms and Isomoprhisms|here]]). If this condition is obeyed, then $\Omega$ is said to be *strongly non-degenerate*. ## Finite-Dimensional Spaces Note that a linear map between finite-dimensional spaces of the same dimension is one-to-one if and only if it is also onto, thus, weak and strong non-degeneracy are equivalent for finite-dimensional spaces (since $Z$ and $Z^*$ have the same dimension in this case). Let $\{e^I\}$ be a [[basis]] for $Z$, then the [[matrix]] elements of $\Omega$ in this basis are defined by: $ \Omega(e_I, e_J) = \Omega_{IJ} $ Now if $\{e^J\}$ is the [[dual basis]] of $Z^*$ dual to $\{e_I\}$, then we can write: $ \Omega(z, w) = z^I \Omega_{IJ} w^J \implies [z]^T [\Omega][w] $ where we employ the [[Einstein summation convention]] and $[w] = w^I e_I$, $[z] = z^I e_I$. [[Bilinear Form#Associated Linear Map|Recall that]] the matrix of $\Omega^\flat$ relative to the bases $\{e_I\}$ and $\{e^J\}$ is the transpose of the matrix of $\Omega$ in the $\{e_I\}$ basis, i.e., $\Omega^\flat_{IJ} = \Omega_{JI}$ Let's write this out the degeneracy condition $ ([z]^T [\Omega] [w] = 0 \, \forall\, z) \implies [\Omega][w] = 0 $ If the $\Omega$ is non-degenerate then: $ [\Omega][w] = 0 \iff [w] = 0 $ This means that $[\Omega]$ is invertible, i.e., $\det([\Omega]) \neq 0$. Thus, a bilinear form $\Omega$ is non-degenerate if and only if $\det([\Omega]) = \det([\Omega^\flat]) \neq 0$. If the form is skew-symmetric, then the space $Z$ must be even dimensional, since the [[determinant]] of a [[skew-symmetric matrix|antisymmetric]] with an odd number of rows vanishes.