# Linear Algebra - Orthogonal complement Vector Space

### Table of Contents

## Definition

Let U be a Linear Algebra - Vector Space (set of vector) of W. For each vector b in W, we can write b as the following projections: <MATH>b = b^{||U} + b^{\perp U}</MATH> where:

- <math>b^{||U}</math> is in U, and
- <math>b^{\perp U}</math> is orthogonal to every vector in U.

Let V be the set <math>V = \{b^{\perp U} : b \in W\}</math> . V is the orthogonal complement of U in W. Every vector in V is orthogonal to every vector in U

## Articles Related

## Direct sum

Every vector b in W can be written as the sum of a vector in U and a vector in V: <MATH>U \oplus V = W</MATH>

Proof: To show direct sum of U and V is defined, we need to show that the only in vector that is in both U and V is the zero vector. Any vector w in both U and V is orthogonal to itself. Thus <math>0 = \langle w,w \rangle = \|w\|^2</math> . By Property N2 of norms, that means w = 0.

## Example

### Gf2

Let U = Span {[1, 1, 0, 0], [0, 0, 1, 1]}. Let V denote the orthogonal complement of U in R4. What vectors form a basis for V?

Every vector in U has the form [a, a, b, b]. Therefore any vector of the form [c,−c, d,−d] is orthogonal to every vector in U.

Every vector in Span {[1,−1, 0, 0], [0, 0, 1,−1]} is orthogonal to every vector in U…. … so Span {[1,−1, 0, 0], [0, 0, 1,−1]} is a subspace of V, the orthogonal complement of U in R4.

And:

- <math>U \oplus V = \mathbb{R}^4 \text{so } dimU + dimV = 4</math>
- {[1, 1, 0, 0], [0, 0, 1, 1]} is linearly independent so dim U = 2… so dimV = 2
- {[1,−1, 0, 0], [0, 0, 1,−1]} is linearly independent so dimSpan {[1,−1, 0, 0], [0, 0, 1,−1]} is also 2….
- so Span {[1,−1, 0, 0], [0, 0, 1,−1]} = V.

### Basis for a null space

## Computation

We have:

- a basis u1, . . . , uk for U
- and a basis w1, . . . ,wn for W.

### One way

- Use orthogonalize(vlist) with vlist = [u1, . . . , uk ,w1, . . . ,wn]
- Write list returned as [u⇤1, . . . , u⇤k,w⇤1, . . . ,w⇤n]

These span the same space as input vectors u1, . . . , uk ,w1, . . . ,w⇤n, namely W, which has dimension n.

Therefore exactly n of the output vectors u⇤1, . . . , u⇤k ,w⇤1, . . . ,w⇤n are nonzero.

The vectors u⇤1 , . . . , u⇤k have same span as u1, . . . , uk and are all nonzero since u1, . . . , uk are linearly independent.

Therefore exactly n - k of the remaining vectors w⇤1, . . . ,w⇤n are nonzero.

- Every one of them is orthogonal to u1, . . . , un…
- so they are orthogonal to every vector in U…
- so they lie in the orthogonal complement of U.

By Direct-Sum Dimension Lemma, orthogonal complement has dimension n-k, so the remaining nonzero vectors are a basis for the orthogonal complement.