CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-16

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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

1.1L1L2

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

2.1step 1.1algebra

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

3.1step 2.1L3∎

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

Depends on

Used by

Dependency tree · two levels

36 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on 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