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.

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

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

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

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

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