Annihilators from Matrix Equations

From Equations to Functionals

A system of linear equations can be viewed in two related ways. Each row of a matrix gives a linear functional on the space of unknown vectors; meanwhile, the right-hand side is tested by functionals that vanish on the matrix's column space. These viewpoints connect the solutions of matrix equations to annihilators in dual spaces.

Let \(\mathbb{F}\) be either \(\mathbb{R}\) or \(\mathbb{C}\), and let \(\mathbf{A}\in\mathbb{F}^{m\times n}\). A linear functional on \(\mathbb{F}^m\) is a linear map \(f:\mathbb{F}^m\to\mathbb{F}\). In standard coordinates, each such functional has a unique row \(\boldsymbol{\ell}\in\mathbb{F}^{1\times m}\) such that \(f(\mathbf{y})=\boldsymbol{\ell}\mathbf{y}\). We can also record its coordinates as the column \(\mathbf{y}_f=\boldsymbol{\ell}^{\top}\). The transpose here does not conjugate entries: this is the algebraic dual, whose functionals are linear over \(\mathbb{C}\), not conjugate-linear.

For a matrix \(\mathbf{A}\), its column space is \(\operatorname{Col}\mathbf{A}\subseteq\mathbb{F}^m\), and its null space is \(\operatorname{Nul}\mathbf{A}=\{\mathbf{x}\in\mathbb{F}^n:\mathbf{A}\mathbf{x}=\mathbf{0}\}\). The rows of \(\mathbf{A}\) are functionals on \(\mathbb{F}^n\): multiplying a row by \(\mathbf{x}\) gives the corresponding equation in \(\mathbf{A}\mathbf{x}=\mathbf{b}\). The following characterizations make the coordinate relationships precise.

Proof. First take a functional on \(\mathbb{F}^m\) with coordinate row \(\mathbf{y}^{\top}\). The columns of \(\mathbf{A}\) span \(\operatorname{Col}\mathbf{A}\), so the functional vanishes on that space exactly when it vanishes on every column. Its values on those columns, written together, form the row \(\mathbf{y}^{\top}\mathbf{A}\). Therefore it vanishes on \(\operatorname{Col}\mathbf{A}\) exactly when \(\mathbf{y}^{\top}\mathbf{A}=\mathbf{0}^{\top}\), equivalently \(\mathbf{A}^{\top}\mathbf{y}=\mathbf{0}\). This proves the first equality.

Each row of \(\mathbf{A}\) vanishes on \(\operatorname{Nul}\mathbf{A}\), since \(\mathbf{A}\mathbf{x}=\mathbf{0}\) makes every row's value at \(\mathbf{x}\) zero. Hence every linear combination of the rows lies in \((\operatorname{Nul}\mathbf{A})^\circ\). For the reverse inclusion, let \(f\) be a functional on \(\mathbb{F}^n\) that vanishes on \(\operatorname{Nul}\mathbf{A}\). Define \(g\) on \(\operatorname{Col}\mathbf{A}\) by \(g(\mathbf{A}\mathbf{x})=f(\mathbf{x})\). This is well-defined: if \(\mathbf{A}\mathbf{x}=\mathbf{A}\mathbf{x}'\), then \(\mathbf{x}-\mathbf{x}'\in\operatorname{Nul}\mathbf{A}\), so \(f(\mathbf{x})-f(\mathbf{x}')=0\). It is linear because both \(f\) and matrix multiplication are linear. Choose a basis of \(\operatorname{Col}\mathbf{A}\), extend it to a basis of \(\mathbb{F}^m\), and define a functional \(\widetilde g\) on \(\mathbb{F}^m\) by using \(g\) on the first basis vectors and assigning value zero to the added basis vectors. Then \(\widetilde g\) extends \(g\). Write \(\widetilde g(\mathbf{y})=\mathbf{r}\mathbf{y}\) for its coordinate row \(\mathbf{r}\). For every \(\mathbf{x}\in\mathbb{F}^n\), \[ f(\mathbf{x})=g(\mathbf{A}\mathbf{x})=\widetilde g(\mathbf{A}\mathbf{x})=\mathbf{r}\mathbf{A}\mathbf{x}. \] Since this holds for every \(\mathbf{x}\), the coordinate row of \(f\) is \(\mathbf{r}\mathbf{A}\), a linear combination of the rows of \(\mathbf{A}\). This proves the reverse inclusion and the theorem. \(\square\)

Annihilating the Column Space

The first equality gives a direct coordinate method. A functional on \(\mathbb{F}^m\), recorded by a row \(\mathbf{y}^{\top}\), annihilates the columns of \(\mathbf{A}\) exactly when \(\mathbf{A}^{\top}\mathbf{y}=\mathbf{0}\). In other words, solve the homogeneous system with coefficient matrix \(\mathbf{A}^{\top}\). The resulting column vectors are the coordinate columns of the annihilating functionals.

Annihilating the Null Space

The second equality has a useful interpretation: every linear functional that vanishes on all solutions of \(\mathbf{A}\mathbf{x}=\mathbf{0}\) is a linear combination of the equations in that system. Conversely, every linear combination of the equations vanishes on every solution. A functional written as a row is therefore in the annihilator exactly when it belongs to the row space.

Testing Whether a System Is Consistent

A system \(\mathbf{A}\mathbf{x}=\mathbf{b}\) is consistent precisely when \(\mathbf{b}\in\operatorname{Col}\mathbf{A}\), because \(\mathbf{A}\mathbf{x}\) is a linear combination of the columns of \(\mathbf{A}\). The annihilator of that column space supplies a test on \(\mathbf{b}\): every functional that vanishes on the columns must also vanish on \(\mathbf{b}\). The converse holds too; if \(\mathbf{b}\) lies outside the column space, a functional can be constructed to vanish on the column space but not on \(\mathbf{b}\).

Proof. If the system is consistent, then \(\mathbf{b}=\mathbf{A}\mathbf{x}\) for some \(\mathbf{x}\). For every \(\mathbf{y}\in\operatorname{Nul}\mathbf{A}^{\top}\), \[ \mathbf{y}^{\top}\mathbf{b}=\mathbf{y}^{\top}\mathbf{A}\mathbf{x} =(\mathbf{A}^{\top}\mathbf{y})^{\top}\mathbf{x} =\mathbf{0}^{\top}\mathbf{x}=0. \] For the converse, suppose \(\mathbf{b}\notin\operatorname{Col}\mathbf{A}\). Choose a basis of \(\operatorname{Col}\mathbf{A}\). Appending \(\mathbf{b}\) gives a linearly independent list, since \(\mathbf{b}\) is not in that column space. Extend this list to a basis of \(\mathbb{F}^m\), and define a functional to be zero on the chosen basis of \(\operatorname{Col}\mathbf{A}\), equal to \(1\) on \(\mathbf{b}\), and zero on any other added basis vectors. This defines a linear functional on all of \(\mathbb{F}^m\). Write its coordinate row as \(\mathbf{y}^{\top}\). It vanishes on \(\operatorname{Col}\mathbf{A}\), so the first equality of the theorem gives \(\mathbf{A}^{\top}\mathbf{y}=\mathbf{0}\); but \(\mathbf{y}^{\top}\mathbf{b}=1\). Thus the condition fails whenever the system is inconsistent, proving the converse. \(\square\)

Common Pitfall: Transpose Is Not Conjugate Transpose

Over \(\mathbb{C}\), the dual space consists of complex-linear functionals. In standard coordinates these are still rows acting by ordinary matrix multiplication, without conjugating the input. Therefore the column-coordinate equation for the annihilator of a column space uses \(\mathbf{A}^{\top}\), not \(\mathbf{A}^{*}\). The conjugate transpose belongs to inner-product calculations and does not replace the algebraic dual in these identities.

For a concrete check, take \(\mathbf{A}=\begin{bmatrix}1\\i\end{bmatrix}\in\mathbb{C}^{2\times1}\). The row \(\begin{bmatrix}-i&1\end{bmatrix}\) annihilates its column, since \[ \begin{bmatrix}-i&1\end{bmatrix}\begin{bmatrix}1\\i\end{bmatrix} =-i+i=0. \] Its coordinate column is \(\mathbf{y}=(-i,1)^{\top}\), and indeed \[ \mathbf{A}^{\top}\mathbf{y} =\begin{bmatrix}1&i\end{bmatrix}\begin{bmatrix}-i\\1\end{bmatrix} =-i+i=0. \] A mistaken conjugate-transpose test gives \[ \mathbf{A}^{*}\mathbf{y} =\begin{bmatrix}1&-i\end{bmatrix}\begin{bmatrix}-i\\1\end{bmatrix} =-i-i=-2i\ne0. \] So using \(\mathbf{A}^{*}\) here incorrectly rejects a functional that does vanish on the column space.

The row and column descriptions are two coordinate recordings of the same functional spaces. The annihilator of \(\operatorname{Col}\mathbf{A}\) is found by solving \(\mathbf{A}^{\top}\mathbf{y}=\mathbf{0}\); the annihilator of \(\operatorname{Nul}\mathbf{A}\) consists of the rows of \(\mathbf{A}\) and their linear combinations. In applications, the first description also detects inconsistent right-hand sides, while the second identifies which linear combinations of the equations vanish on every homogeneous solution.