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Let's consider a finite-dimensional vector space $E$ on the field $\mathbb{K}$ (where $\mathbb{K}=\mathbb{C} \ \text{or}\ \mathbb{R}$) and a sesquilinear (or bilinear if $\mathbb{K}=\mathbb{R}$) form $q:E\times E \rightarrow \mathbb{K}$.

The definition for a non-degenerate form is that $q(x,y)=0\ \forall y\in E$ implies $x=0$.

Now if we represent $q(x,y)$ with a matrix, so $q(x,y) =x^HAy$, why does the condition that the form be non-degenerate impose that $A$ is non-singular?

I tried to see it using the dual space as $M(x,A)=x^HA\in E^*$, so that $M:E\times L(E,E)\rightarrow E^*$, where $L(E,E)$ is the vector space of all linear transformations from $E$ to $E$ and playing with the nullspace of $A$, but I just can't see it

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2 Answers 2

up vote 1 down vote accepted

Let $q$ be a sesquilinear form on a vector space $E$, given by a matrix $A$. The following statements are equivalent:

  1. $q$ is degenerate.

  2. There exists a nonzero vector $x\in E$ so that $q(x,y)=0$ for all $y\in E$.

  3. There exists a nonzero vector $x\in E$ so that $x^H A y = 0$ for all $y\in E$.

  4. There exists a nonzero vector $x\in E$ so that $x^H A$ is the zero (row) vector.

  5. The left nullspace of $A$ is non-trivial.

  6. The matrix $A$ is singular.

It should be clear that $(1)\Leftrightarrow(2)\Leftrightarrow(3)\Leftrightarrow(4)\Leftrightarrow(5)\Leftrightarrow(6)$.

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Thanks! the word I needed was left nullspace –  Ralph Jun 20 '12 at 18:21

$A$ is singular if and only if there exist $0 \ne y\in E$ such that $Ay=0$. If this is the case, then clearly $q(x, y)=0$ for all $x$ (modulo identification of $E$ with $\mathbb{K}^n$). Viceversa, if $q(x, y)=0$ for all $x\in E$ then the vector $Ay$ is orthogonal to the whole $\mathbb{K}^n$. In particular $\lVert Ay\rVert^2=(Ay)^HAy=0,$ so $Ay=0$.

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