Vector subspace

Orthogonal complement

Given an inner product space (𝑉,𝕂,⟨⋅|⋅⟩), the orthogonal complement 𝐴⟂ of a subset 𝐴 ⊆𝑉 is the vector subspace of vectors orthogonal to those 𝐴 linalg

𝐴⟂={𝑣∈𝑉:(∀𝑎∈𝐴)[⟨𝑎|𝑣⟩=0]}

Proof of subspace

Clearly\Span⃗𝟎 ∈𝐴⟂. If 𝑣1,𝑣2 ∈𝐴⟂ and 𝛼,𝛽 ∈𝕂 then ⟨𝑎|𝛼𝑣1+𝛽𝑣2⟩ =𝛼⟨𝑎|𝑣1⟩ +𝛽⟨𝑎|𝑣2⟩ =0 for all 𝑎 ∈𝐴, and thus 𝛼𝑣1 +𝛽𝑣2 ∈𝐴⟂. Therefore 𝐴⟂ is a subspace.

Properties

Let 𝐴 ⊆𝑉 be an arbitrary subset. Then

1. 𝐴⟂ is topologically closed
2. 𝐴 ∩𝐴⟂ ={0}
3. 𝐵 ⊆𝐴 ⟹ 𝐴⟂ ⊆𝐵⟂
4. 𝐴 ⊆(𝐴⟂)⟂
5. If B𝜖(𝑣) ⊆𝐴 for some 𝑣 ∈𝑉, then 𝐴⟂ ={0}
6. 𝐴⟂ =(span⁡𝐴)⟂

Proof of 1–6

Note that the orthogonal complement of a singleton {𝑣} can be expressed as a preïmage

{𝑣}⟂=(⟨𝑣|)−1{0}

Since {0} is closed in 𝕂, and the inner product is continuous, it follows {𝑣}⟂ is closed. Now for an arbitrary set 𝐴,

𝐴⟂=⋂𝑎∈𝐴{𝑎}⟂

which is an intersection of closed sets and is therefore closed, proving ^S1.

Note if 𝑣 ∈𝐴 ∩𝐴⟂ then ⟨𝑣|𝑣⟩ =0, implying 𝑣 =0 by ^IP3, proving ^S2.

Let 𝐵 ⊆𝐴 ⊆𝑉 and 𝑣 ∈𝐴⟂. Then ⟨𝑣|𝑏⟩ =0 for any 𝑏 ∈𝐵 ⊆𝐴, so 𝑣 ∈𝐵⟂, proving ^S3.

Let 𝑎 ∈𝐴. Then by definition ⟨𝑎|𝑣⟩ =0 for any 𝑣 ∈𝐴⟂, so 𝑎 ∈(𝐴⟂)⟂, proving ^S4.

Without loss of generality 𝑣 =0, for ⟨𝐵𝜖(𝑣)|𝑎⟩ =0 iff ⟨B𝜖(𝑣)−𝑣|𝑎⟩ =⟨B𝜖(0)|𝑎⟩ =0. Now

B𝜖(0)={𝑣∈𝑉:⟨𝑣|𝑣⟩<𝜖}

and

⟨B𝜖(0)|𝑦⟩=0⟺⟨B𝜖(0)|𝜖𝑦2‖𝑦‖⟩=0

but since 𝜖𝑦2‖𝑦‖ ∈B𝜖(0), it follows from ^IP3 that 𝑦 =0, proving ^S5.

Let 𝑥 ∈𝐴⟂ and 𝑦 ∈span⁡𝐴, so 𝑦 =∑𝑛𝑖=1𝜆𝑖𝑎𝑖 for some {𝑎𝑖}𝑛𝑖=1 ⊆𝐴. Then

⟨𝑥|𝑦⟩=⟨𝑥|𝑛∑𝑖=1𝜆𝑖|𝑎𝑖⟩=0

proving ^S6.

Let 𝑊 ≤𝑉 be a vector subspace. Then

1. 𝑉 =𝑊 ⊕𝑊⟂ (Internal direct sum).
2. 𝑊 =(𝑊⟂)⟂.

Proof of 1–2

^V1 follows directly from ^S2, and ^V2 follows directly from ^S4.

Other properties include

• The orthogonal complement of an invariant subspace under a unitary operator is invariant

See also

• Orthogonal complement polarity

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