19 - Dimension of Annihilators

Dimension Of Annihilators

Let $V$ be finite dimensional and $U\subseteq V$ be a subspace then

$$\dim (U^0)=\dim (V)-\dim (U)$$

Proof

Let $\{e_1,\cdots ,e_m\}$ be a basis for $U$ and extend it to $\{e_1,\cdots ,e_n\}$ for $V$
Let $\{e_1^{\prime },\cdots ,e_n^{\prime }\}$ be the dual basis

As $e_j^{\prime }(e_i)=0$ for $j=m+1,\cdots ,n$ and $i=1,\cdots ,m$ then

$$e_j^{\prime }\in U^0\text{ for }j=m+1,\cdots ,n$$

Hence

$$⟨e_{m+1}^{\prime },\cdot ,e_n^{\prime }⟩\subseteq U^0$$

Conversely for $f\in U^0$ then there exists $a_i\in \mathbf{F}$ such that $f={\sum }_{i=1}^n{a}_i{e}_i^{\prime }$
As $e_i\in U$ for $i=1,\cdots ,m$ then

$$f(e_i)=0\text{ and hence }{a}_i=0$$

Hence

$$f\in ⟨e_{m+1}^{\prime },\cdots ,e_n^{\prime }⟩$$

Thus

$$U^0=⟨e_{m+1}^{\prime },\cdots ,e_n^{\prime }⟩$$

where the set of vectors is a subset of the dual basis, spanning and is linearly independent so a basis of $U^0$

$$\dim (U^0)=n-m=\dim (V)-\dim (U)$$