What Is an Example of a Degenerate Bilinear Map?

A bilinear map takes two vectors as inputs and returns a scalar, with the output linear in each input when the other is held fixed. It is degenerate if some nonzero vector pairs to zero with every vector in the other input. The simplest example is a form on $\mathbb{R}^2$ that ignores the second coordinate.

This example also shows how to recognize degeneracy from a matrix. A second example, using an alternating form on $\mathbb{R}^3$, illustrates why a degenerate alternating form is not a symplectic form.

A simple degenerate bilinear form

Define $B: \mathbb{R}^2\times\mathbb{R}^2\to\mathbb{R}$ by

$$B((a,b),(c,d))=ac.$$

This is bilinear: it is linear in $(a,b)$ when $(c,d)$ is fixed, and linear in $(c,d)$ when $(a,b)$ is fixed. It is not the zero map; for example, $B((1,0),(1,0))=1$.

But the nonzero vector $(0,1)$ pairs to zero with every vector. For any $(c,d)\in\mathbb{R}^2$,

$$B((0,1),(c,d))=0\cdot c=0.$$

Thus $B$ is degenerate. In fact, all vectors of the form $(0,b)$ have this property, so the radical is the line spanned by $(0,1)$.

For a numerical check, $B((2,3),(4,5))=2\cdot4=8$, while $B((0,1),(4,5))=0$. The second vector is not merely orthogonal to one particular vector; it pairs to zero with every vector, which is what makes it evidence of degeneracy.

The matrix test

A bilinear form on $\mathbb{R}^n$ can be written as

$$B(x,y)=x^{\mathsf T}Ay,$$

where $A$ is its matrix and $x,y$ are column vectors. For the example above,

$$A=\begin{pmatrix}1&0\\0&0\end{pmatrix}, \qquad B\big((a,b),(c,d)\big) =\begin{pmatrix}a&b\end{pmatrix} A\begin{pmatrix}c\\d\end{pmatrix}=ac.$$

The vector $x$ is in the left radical if $B(x,y)=0$ for every $y$. Since $B(x,y)=x^{\mathsf T}Ay$, this condition is $A^{\mathsf T}x=0$. The right radical consists of vectors $y$ for which $Ay=0$.

For a form on one finite-dimensional space, $A$ is square. It is degenerate exactly when $A$ is singular, or equivalently when its determinant is zero. Here $\det(A)=0$, and both radicals are $\operatorname{span}\{(0,1)\}$.

A degenerate alternating example

An alternating bilinear form satisfies $B(x,x)=0$ for every vector $x$. On $\mathbb{R}^3$, consider

$$B(x,y)=x_1y_2-x_2y_1.$$

It is alternating, but the third coordinate does not affect the result. In particular, $e_3=(0,0,1)$ is nonzero and

$$B(e_3,y)=0$$

for every $y\in\mathbb{R}^3$. So this alternating form is degenerate. Its matrix is

$$\begin{pmatrix} 0&1&0\\ -1&0&0\\ 0&0&0 \end{pmatrix},$$

which is singular. This is not a symplectic form: a symplectic form is, by definition, a nondegenerate alternating form. The standard symplectic form on $\mathbb{R}^2$, $B((x_1,x_2),(y_1,y_2))=x_1y_2-x_2y_1$, is nondegenerate instead.

Common mistakes

• Checking only one pair: Finding nonzero $x,y$ with $B(x,y)=0$ does not show degeneracy. The required vector must pair to zero with every vector in the other input.
• Thinking “degenerate” means “always zero”: A degenerate form can have many nonzero values, as the $\mathbb{R}^2$ example does.
• Ignoring left and right radicals: For a symmetric form, the left and right radicals agree. For a general nonsymmetric bilinear form, they need not. Nondegeneracy requires both radicals to be zero; for a square matrix in finite dimensions, this is equivalent to invertibility.
• Confusing alternating with symplectic: Alternating describes the condition $B(x,x)=0$. It does not guarantee nondegeneracy, as the $\mathbb{R}^3$ example shows.