Dual Spaces and Bilinear Forms · Tutorial 894 of 1000

Radicals and Orthogonality for Bilinear Forms

Learn to translate radicals and orthogonality into matrix equations, and determine whether a bilinear form is degenerate.

Advanced 10 min read

What You'll Learn

Define left and right radicals of a bilinear form
Distinguish symmetric and nonsymmetric orthogonality
Express radicals as null spaces of matrices
Find radicals by exact row reduction
Decide whether a finite-dimensional bilinear form is nondegenerate

Orthogonality Depends on the Form

For the usual dot product, the statement that two vectors are orthogonal does not depend on their order. A general bilinear form need not have that property. Its value on $(\mathbf{x},\mathbf{y})$ may differ from its value on $(\mathbf{y},\mathbf{x})$, so it is useful to distinguish which argument is being varied. The vectors that pair to zero with every vector form the radical, or, for a nonsymmetric form, the left and right radicals. These ideas can be studied by solving homogeneous matrix equations.

Throughout this tutorial, $V=\mathbb{R}^n$ is finite-dimensional, and $B:V\times V\to\mathbb{R}$ is a bilinear form: it is linear in each argument when the other is held fixed. In the standard basis, there is a matrix $\mathbf{A}\in\mathbb{R}^{n\times n}$ such that $B(\mathbf{x},\mathbf{y})=\mathbf{x}^{\top}\mathbf{A}\mathbf{y}$. The matrix need not be symmetric. The form is symmetric when $B(\mathbf{x},\mathbf{y})=B(\mathbf{y},\mathbf{x})$ for all $\mathbf{x},\mathbf{y}$, equivalently when $\mathbf{A}^{\top}=\mathbf{A}$.

Definition: Vectors $\mathbf{x},\mathbf{y}\in V$ are orthogonal with respect to $B$ if $B(\mathbf{x},\mathbf{y})=0$. The left radical and right radical are, respectively, $$ \operatorname{Rad}_L(B)=\{\mathbf{x}\in V:B(\mathbf{x},\mathbf{y})=0\text{ for every }\mathbf{y}\in V\},\qquad \operatorname{Rad}_R(B)=\{\mathbf{y}\in V:B(\mathbf{x},\mathbf{y})=0\text{ for every }\mathbf{x}\in V\}. $$ The form is nondegenerate if both radicals are $\{\mathbf{0}\}$; otherwise it is degenerate.

The labels record which argument is fixed as the vector in the radical: the left radical occupies the first argument, and the right radical occupies the second. For a symmetric form the two radicals coincide, but that is not true for an arbitrary bilinear form. Likewise, for a nonsymmetric form, $B(\mathbf{x},\mathbf{y})=0$ does not in general imply $B(\mathbf{y},\mathbf{x})=0$.

Radicals as Matrix Equations

Theorem: If $B(\mathbf{x},\mathbf{y})=\mathbf{x}^{\top}\mathbf{A}\mathbf{y}$, then $$ \operatorname{Rad}_L(B)=\operatorname{Nul}(\mathbf{A}^{\top}),\qquad \operatorname{Rad}_R(B)=\operatorname{Nul}(\mathbf{A}). $$ In particular, both radicals are subspaces of $V$.

Proof. Fix $\mathbf{x}\in V$. It belongs to the left radical exactly when $\mathbf{x}^{\top}\mathbf{A}\mathbf{y}=0$ for every $\mathbf{y}\in V$. Since $\mathbf{x}^{\top}\mathbf{A}\mathbf{y}=(\mathbf{A}^{\top}\mathbf{x})^{\top}\mathbf{y}$, this holds exactly when $\mathbf{A}^{\top}\mathbf{x}=\mathbf{0}$: if this vector is zero, every product is zero; if it is nonzero, choosing $\mathbf{y}=\mathbf{A}^{\top}\mathbf{x}$ gives the positive number $\|\mathbf{A}^{\top}\mathbf{x}\|^2$, not zero. Thus $\operatorname{Rad}_L(B)=\operatorname{Nul}(\mathbf{A}^{\top})$. Similarly, $\mathbf{y}$ belongs to the right radical exactly when $\mathbf{x}^{\top}\mathbf{A}\mathbf{y}=0$ for every $\mathbf{x}$. This holds exactly when $\mathbf{A}\mathbf{y}=\mathbf{0}$, by choosing $\mathbf{x}=\mathbf{A}\mathbf{y}$ if that vector is nonzero. Hence $\operatorname{Rad}_R(B)=\operatorname{Nul}(\mathbf{A})$. Each null space is a subspace, so both radicals are subspaces. $\square$

This result gives a practical distinction: use $\mathbf{A}^{\top}\mathbf{x}=\mathbf{0}$ for the left radical and $\mathbf{A}\mathbf{y}=\mathbf{0}$ for the right radical. Row-reducing $\mathbf{A}$ answers the right-radical question; row-reducing $\mathbf{A}^{\top}$ answers the left-radical question. If $B$ is symmetric, these equations are the same.

Theorem: For a bilinear form on finite-dimensional $V=\mathbb{R}^n$, the following are equivalent: $B$ is nondegenerate; $\operatorname{Rad}_R(B)=\{\mathbf{0}\}$; $\operatorname{Rad}_L(B)=\{\mathbf{0}\}$; and its matrix $\mathbf{A}$ is invertible.

Proof. The matrix equations show that the right radical is $\operatorname{Nul}(\mathbf{A})$ and the left radical is $\operatorname{Nul}(\mathbf{A}^{\top})$. By the Invertible Matrix Theorem for square matrices, $\operatorname{Nul}(\mathbf{A})=\{\mathbf{0}\}$ exactly when $\mathbf{A}$ is invertible. The transpose of an invertible matrix is invertible, since $(\mathbf{A}^{-1})^{\top}\mathbf{A}^{\top}=(\mathbf{A}\mathbf{A}^{-1})^{\top}=\mathbf{I}_n$; the reverse implication follows by applying the same fact to $\mathbf{A}^{\top}$. Thus $\mathbf{A}$ is invertible exactly when both null spaces are trivial. This proves all the stated equivalences. $\square$

Finding Radicals in Examples

Worked Example: A Nondegenerate Symmetric Form

Let $B(\mathbf{x},\mathbf{y})=\mathbf{x}^{\top}\mathbf{A}\mathbf{y}$ on $\mathbb{R}^2$, where $$ \mathbf{A}=\begin{bmatrix}2&1\\1&1\end{bmatrix}. $$ Find both radicals and determine whether $B$ is nondegenerate.

The matrix is symmetric, so the left- and right-radical equations are the same. Row-reduce $\mathbf{A}$:

\begin{bmatrix}2&1\\1&1\end{bmatrix} \xrightarrow{R_1\leftrightarrow R_2} \begin{bmatrix}1&1\\2&1\end{bmatrix} \xrightarrow{R_2\to R_2-2R_1} \begin{bmatrix}1&1\\0&-1\end{bmatrix} \xrightarrow{R_2\to -R_2} \begin{bmatrix}1&1\\0&1\end{bmatrix} \xrightarrow{R_1\to R_1-R_2} \begin{bmatrix}1&0\\0&1\end{bmatrix}.

The equation $\mathbf{A}\mathbf{y}=\mathbf{0}$ therefore has only $\mathbf{y}=\mathbf{0}$. Because $\mathbf{A}^{\top}=\mathbf{A}$, the left-radical equation has only $\mathbf{x}=\mathbf{0}$ as well. Both radicals are trivial, so the form is nondegenerate.

Check: For the only proposed radical vector, $\mathbf{A}\mathbf{0}=\mathbf{0}$ and $\mathbf{A}^{\top}\mathbf{0}=\mathbf{0}$. The row reduction also shows there can be no nonzero solution to either homogeneous system.

Worked Example: A Degenerate Symmetric Form

Let $$ \mathbf{A}=\begin{bmatrix}1&2&0\\2&4&0\\0&0&0\end{bmatrix}. $$ Find the left and right radicals of the associated form on $\mathbb{R}^3$.

Since $\mathbf{A}^{\top}=\mathbf{A}$, the two radical equations agree. Reduce the matrix:

\begin{bmatrix}1&2&0\\2&4&0\\0&0&0\end{bmatrix} \xrightarrow{R_2\to R_2-2R_1} \begin{bmatrix}1&2&0\\0&0&0\\0&0&0\end{bmatrix}.

Writing $\mathbf{y}=(y_1,y_2,y_3)^{\top}$, the system $\mathbf{A}\mathbf{y}=\mathbf{0}$ reduces to $y_1+2y_2=0$. Thus $y_1=-2y_2$, while $y_2$ and $y_3$ are free. Every solution has the form $$ \mathbf{y}=y_2\begin{bmatrix}-2\\1\\0\end{bmatrix} +y_3\begin{bmatrix}0\\0\\1\end{bmatrix}. $$

Therefore $\operatorname{Rad}_R(B)=\operatorname{Rad}_L(B)=\operatorname{span}\{(-2,1,0)^{\top},(0,0,1)^{\top}\}$, and the form is degenerate.

Check: Multiplying each displayed spanning vector by $\mathbf{A}$ gives $$ \begin{bmatrix}1&2&0\\2&4&0\\0&0&0\end{bmatrix} \begin{bmatrix}-2\\1\\0\end{bmatrix} =\begin{bmatrix}0\\0\\0\end{bmatrix},\qquad \begin{bmatrix}1&2&0\\2&4&0\\0&0&0\end{bmatrix} \begin{bmatrix}0\\0\\1\end{bmatrix} =\begin{bmatrix}0\\0\\0\end{bmatrix}. $$ The matrix is symmetric, so these checks also verify the left-radical equation.

Worked Example: Different Left and Right Radicals

Consider the nonsymmetric matrix $$ \mathbf{A}=\begin{bmatrix}1&1\\0&0\end{bmatrix}. $$ Find both radicals and compare the two orders of orthogonality.

For the right radical, solve $\mathbf{A}\mathbf{y}=\mathbf{0}$:

\begin{bmatrix}1&1\\0&0\end{bmatrix} \xrightarrow{\text{already in RREF}} \begin{bmatrix}1&1\\0&0\end{bmatrix}.

The equation is $y_1+y_2=0$, so $y_1=-y_2$ and $\operatorname{Rad}_R(B)=\operatorname{span}\{(-1,1)^{\top}\}$. For the left radical, solve $\mathbf{A}^{\top}\mathbf{x}=\mathbf{0}$:

\begin{bmatrix}1&0\\1&0\end{bmatrix} \xrightarrow{R_2\to R_2-R_1} \begin{bmatrix}1&0\\0&0\end{bmatrix}.

This gives $x_1=0$, with $x_2$ free, so $\operatorname{Rad}_L(B)=\operatorname{span}\{(0,1)^{\top}\}$. The form is degenerate, and the two radicals differ.

Check: Multiplication verifies the right-radical generator: $$ \mathbf{A}\begin{bmatrix}-1\\1\end{bmatrix} =\begin{bmatrix}0\\0\end{bmatrix}, $$ and verifies the left-radical generator: $$ \mathbf{A}^{\top}\begin{bmatrix}0\\1\end{bmatrix} =\begin{bmatrix}0\\0\end{bmatrix}. $$ The order of the arguments matters: $B(\mathbf{e}_1,\mathbf{e}_2)=\mathbf{e}_1^{\top}\mathbf{A}\mathbf{e}_2=1$, whereas $B(\mathbf{e}_2,\mathbf{e}_1)=\mathbf{e}_2^{\top}\mathbf{A}\mathbf{e}_1=0$.

Orthogonal Complements and a Common Error

For a subspace $U\subseteq V$, define its left and right orthogonal complements by $$ U^{\perp_L}=\{\mathbf{x}\in V:B(\mathbf{x},\mathbf{u})=0\text{ for all }\mathbf{u}\in U\},\qquad U^{\perp_R}=\{\mathbf{x}\in V:B(\mathbf{u},\mathbf{x})=0\text{ for all }\mathbf{u}\in U\}. $$ Each is a subspace: if two vectors satisfy the relevant zero equations, every linear combination does too, by bilinearity. Taking $U=V$ gives $V^{\perp_L}=\operatorname{Rad}_L(B)$ and $V^{\perp_R}=\operatorname{Rad}_R(B)$. For a symmetric form, the left and right complements agree; without symmetry, they need not.

If the columns of $\mathbf{C}$ are a basis for $U$, then every $\mathbf{u}\in U$ has the form $\mathbf{C}\mathbf{c}$. The right-complement condition is $B(\mathbf{C}\mathbf{c},\mathbf{x})=\mathbf{c}^{\top}\mathbf{C}^{\top}\mathbf{A}\mathbf{x}=0$ for every $\mathbf{c}$, which is equivalent to $\mathbf{C}^{\top}\mathbf{A}\mathbf{x}=\mathbf{0}$. The left-complement condition is $B(\mathbf{x},\mathbf{C}\mathbf{c})=0$ for every $\mathbf{c}$, equivalent to $\mathbf{C}^{\top}\mathbf{A}^{\top}\mathbf{x}=\mathbf{0}$. These equations extend the radical method to orthogonality with respect to a subspace.

Common mistake: using the same matrix equation for both radicals. For the matrix in the third example, using $\operatorname{Nul}(\mathbf{A})$ for the left radical would incorrectly give $\operatorname{span}\{(-1,1)^{\top}\}$. That is the right radical, not the left. The correction is to use $\operatorname{Nul}(\mathbf{A}^{\top})$ for the left radical; it gives $\operatorname{span}\{(0,1)^{\top}\}$. The distinction follows directly from the positions of the vectors in $\mathbf{x}^{\top}\mathbf{A}\mathbf{y}$.

For symmetric bilinear forms, radicals identify precisely the vectors that pair to zero with every vector, and they can be found by one homogeneous system. For nonsymmetric forms, checking both sides matters: the two null spaces can describe different subspaces. In either case, row reduction gives exact equations and makes degeneracy a concrete question about whether the relevant null spaces contain nonzero vectors.

Check Your Understanding

For each question, write down the relevant matrix equation before solving it.

For a matrix $\mathbf{A}$ representing a bilinear form, which homogeneous system determines the left radical, and which determines the right radical?
Find both radicals for the form with matrix $\begin{bmatrix}1&0\\0&0\end{bmatrix}$. Is it degenerate?
Explain why the left and right radicals coincide when the matrix of the form is symmetric.
For the matrix $\begin{bmatrix}1&1\\0&0\end{bmatrix}$, compute $B(\mathbf{e}_1,\mathbf{e}_2)$ and $B(\mathbf{e}_2,\mathbf{e}_1)$. What do the results show?
If the columns of $\mathbf{C}$ span $U$, write a matrix equation whose solutions are $U^{\perp_R}$.

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