What is Annihilator?

Example 2.7 (An annihilator, both directions)

Let $F = \operatorname{Vect}\bigl((1, 2, 1),\ (1, 0, -1)\bigr) \subseteq \R^3$. A form $\varphi = a\,e_1^* + b\,e_2^* + c\,e_3^*$ annihilates $F$ iff

i.e. $c = a$ and $b = -a$: $F^\circ = \R\,(e_1^* - e_2^* + e_3^*)$, of dimension $3 - 2 = 1$ as Theorem 2.6 requires. Reading it backwards: $F = \{(x, y, z) : x - y + z = 0\}$ — the plane recovered as the kernel of the single form spanning $F^\circ$. Going from a spanning family to equations is computing an annihilator; going from equations to a parametrization is computing a pre-annihilator. (Check: both spanning vectors satisfy $x - y + z = 0$.)

Example 2.11 (Rank read on both sides)

Let

Column rank: the third row is the sum of the first two, so $\operatorname{rk} A \leq 2$; columns $1$ and $2$ are free: $\operatorname{rk} A = 2$. The transpose’s kernel: solving $A^{\mathsf T}y = 0$ gives $y \in \R\,(1, 1, -1)$, so $\ker A^{\mathsf T}$ has dimension $1 = 3 - 2$: exactly $(\operatorname{im} A)^\circ$ under the identification of $(\R^3)^*$ with row vectors, as Proposition 2.10 asserts — the single relation “row$_3$ = row$_1$ + row$_2$” is the annihilator of the column space. Row rank ($2$ free rows) and column rank agree not by accident but because both equal $\operatorname{rk} A = \operatorname{rk} A^{\mathsf T}$.

Example 2.24 (The trace pairing splits the matrix space)

On $\mathcal{M}_2(\R)$ with the pairing $\langle A, B\rangle = \operatorname{tr}(AB)$ of Exercise 2.9: decompose $M = \left(\begin{smallmatrix}1 & 4\\ 2 & 3\end{smallmatrix}\right)$ into symmetric and antisymmetric parts,

Then $\operatorname{tr}(SA) = \operatorname{tr} \left(\begin{smallmatrix}-3 & 1\\ -3 & 3\end{smallmatrix}\right) = 0$: the two parts are “orthogonal” for the trace pairing — an instance of the general fact (proved in the weekend problem of this chapter) that antisymmetric matrices form exactly the annihilator of the symmetric ones. Duality sees the decomposition $\mathcal{M}_n = \mathcal{S}_n \oplus \mathcal{A}_n$ before any inner product is chosen.