Recipe

Apply the annihilator dimension formula

Use when: you know two of \dim V, \dim W, \dim(W^{\circ}) and need the third, or want to relate \dim(W^{\circ\circ}) back to \dim W

Goal: the missing dimension, or the double-annihilator identity, without ever constructing a basis of W or W^{\circ}

Taught in Annihilators and dimension

1. 1

   Write down \dim(W^{\circ}) = \dim V - \dim W and check which of the three symbols the problem never gave you.

   Why: the formula has exactly three symbols; knowing which one is missing tells you how to rearrange it before you touch any numbers.

2. 2

   Rearrange the same identity so the unknown sits alone on one side, then substitute the two known numbers.

   Example

   W is 2-dimensional inside a 5-dimensional V: \dim(W^{\circ}) = \dim V - \dim W = 5-2=3.

   \dim(W^{\circ})=4 inside a 7-dimensional space: \dim W = \dim V - \dim(W^{\circ}) = 7-4=3.

   Why: the identity is symmetric in what it lets you find -- there's no separate formula to memorize for "solve for \dim W" versus "solve for \dim(W^{\circ})."

3. 3

   Treat W^{\circ} itself as "the subspace being annihilated," now inside V^{*}, and apply the identical formula with V^{*} in place of V and W^{\circ} in place of W.

   Example

   \dim(W^{\circ\circ}) = \dim(V^{*}) - \dim(W^{\circ}) = \dim V - (\dim V-\dim W) = \dim W.

   Why: this is the same one-line identity run twice, not a second formula -- and it's what makes \dim(W^{\circ\circ})=\dim W provable before section 10 ever constructs the natural isomorphism W^{\circ\circ}\cong W directly.