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In finite dimension, $W^{⊥ ⊥} = W$ and $\dim W + \dim W^{⊥} = \dim V$

Statement

For every subspace $W$ of a finite-dimensional inner product space $V$ ,

$W^{⊥ ⊥} = W, \dim W + \dim W^{⊥} = \dim V .$

These formulas include $W = 0$ and $W = V$ .

Facts & Assumptions

Given: A subspace $W$ of a finite-dimensional inner product space $V$ .

[L1]

Orthogonal decomposition gives $V = W \oplus W^{⊥}$ (For a subspace $W$ of a finite-dimensional inner product space, $V = W \oplus W^{⊥}$ ).

[L2]

Dimensions add across an internal direct sum of finite-dimensional subspaces (If $V = ⨁_{i < n} U_{i}$ with every $U_{i}$ finite-dimensional, then $V$ is finite-dimensional and $\dim_{F} V = \sum_{i < n} \dim_{F} U_{i}$ ; in particular $\dim_{F} (U \oplus W) = \dim_{F} U + \dim_{F} W$ ).

[L3]

If one finite-dimensional subspace is contained in another and their dimensions agree, the two subspaces are equal (If $\dim_{F} V = n$ and $U$ is a linear subspace of $V$ , then $U$ is finite-dimensional, $\dim_{F} U \leq n$ , and $\dim_{F} U = n$ if and only if $U = V$ ).

Proof

technique · direct

1.1

Apply [L2] to [L1] to obtain $\dim V = \dim W + \dim W^{⊥}$ .

L1L2

2.1

Conjugate symmetry shows $W \subseteq W^{⊥ ⊥}$ . Apply step 1.1 first to $W$ and then to $W^{⊥}$ to get $\dim W^{⊥ ⊥} = \dim V - \dim W^{⊥} = \dim W$ .

step 1.1algebra

3.1

The inclusion and equal dimensions in step 2.1 imply $W^{⊥ ⊥} = W$ by [L3]. The same reasoning covers both endpoint subspaces.

step 2.1L3∎

Depends on

Used by

ker T^*=(imT)^⊥ and imT^*=(ker T)^⊥ in finite dimension Theorem

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 73 results over 25 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Sheldon Axler, Linear Algebra Done Right, 4th ed., results 6.51 and 6.52 (standard reference, not scraped)
Sergei Treil, Linear Algebra Done Wrong, Proposition 5.3.6 (standard reference, not scraped)

In finite dimension, W⊥⊥=WW⊥⊥=W and dim⁡W+dim⁡W⊥=dim⁡VdimW+dimW⊥=dimV