Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
up vote 2 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 $(1)\Leftrightarrow(2)\Leftrightarrow(3)\Leftrightarrow(4)\Leftrightarrow(5)\Leftrightarrow(6)$.

share|cite|improve this answer
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 $Ay=0$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.