# Orthogonalization

Sep 29, 2020
5 min read
Mar 11, 2022 15:59 UTC
Introducing the Gram-Schmidt process, a method for constructing an orthogonal basis given a non-orthogonal basis.

## Gram-Schmidt orthogonalization

The Gram-Schmidt process is a method for orthogonalizing a set of vectors. We start with a non-orthogonal basis $\{\boldsymbol{u}_1, \cdots, \boldsymbol{u}_k\}$ for $V$, and our goal is to find $\{\boldsymbol{v}_1, \cdots, \boldsymbol{v}_k\}$, an orthogonal basis for $V$.

1. The first member in the orthogonal basis is the same as the first member1 in the given basis: $\boldsymbol{v}_1 \triangleq \boldsymbol{u}_1$.

2. We need $\boldsymbol{v}_2$ such that $\boldsymbol{v}_2 \perp \boldsymbol{v}_1$, and $\mathcal{L}(\boldsymbol{u}_1, \boldsymbol{u}_2) = \mathcal{L}(\boldsymbol{v}_1, \boldsymbol{v}_2)$. We can show that $\boldsymbol{v}_2$ can be found by taking the difference between $\boldsymbol{u}_2$ and its projection on $\boldsymbol{v}_1$:

\begin{aligned} \boldsymbol{v}_2 &\triangleq \boldsymbol{u}_2 - P(\boldsymbol{u}_2 \mid \boldsymbol{v}_1) \\ &= \boldsymbol{u}_2 - \left( \frac{\boldsymbol{u}_2 \cdot \boldsymbol{v}_1}{\|\boldsymbol{v}_1\|^2} \right)\boldsymbol{v}_1 \end{aligned}

3. We find $\boldsymbol{v}_3$ such that $\boldsymbol{v}_3 \perp \mathcal{L}(\boldsymbol{v}_1, \boldsymbol{v}_2)$ and $\mathcal{L}(\boldsymbol{u}_1, \boldsymbol{u}_2, \boldsymbol{u}_3) = \mathcal{L}(\boldsymbol{v}_1, \boldsymbol{v}_2, \boldsymbol{v}_3)$. Similar to the last step,

\begin{aligned} \boldsymbol{v}_3 &\triangleq \boldsymbol{u}_3 - P(\boldsymbol{u}_3 \mid \mathcal{L}(\boldsymbol{v}_1, \boldsymbol{v}_2)) \\ &= \boldsymbol{u}_3 - \left( \frac{\boldsymbol{u}_3 \cdot \boldsymbol{v}_1}{\|\boldsymbol{v}_1\|^2} \right)\boldsymbol{v}_1 - \left( \frac{\boldsymbol{u}_3 \cdot \boldsymbol{v}_2}{\|\boldsymbol{v}_2\|^2} \right)\boldsymbol{v}_2 \end{aligned}

4. Continue this until we find

$$\boldsymbol{v}_k = \boldsymbol{u}_k - P(\boldsymbol{u}_k \mid \mathcal{L}(\boldsymbol{v}_1, \cdots, \boldsymbol{v}_{k-1}))$$

For example, say we have a linearly independent set of vectors

$$\begin{gathered} \boldsymbol{u}_1 = (1, 1, 1, 1)^\prime \\ \boldsymbol{u}_2 = (1, 1, 5, 5)^\prime \\ \boldsymbol{u}_3 = (4, 0, 12, 8)^\prime \end{gathered}$$

and we want to find an orthogonal basis $V = \mathcal{L}(\boldsymbol{u}_1, \boldsymbol{u}_2, \boldsymbol{u}_3) \subset \mathbb{R}^4$.

\begin{aligned} \boldsymbol{v}_1 &= (1, 1, 1, 1)^\prime \\ \boldsymbol{v}_2 &= (1, 1, 5, 5)^\prime - \frac{12}{4}(1, 1, 1, 1)^\prime = (-2, -2, 2, 2)^\prime \\ \boldsymbol{v}_3 &= (4, 0, 12, 8)^\prime - \frac{4 + 12 + 8}{4}(1, 1, 1, 1)^\prime - \frac{-8 + 24 + 16}{16}(-2, -2, 2, 2)^\prime = \cdots \end{aligned}

## Orthogonal complement of subspace

Suppose $V$ is a subspace of $\mathbb{R}^n$. The orthogonal complement of $V$ is defined as a set of all vectors that are orthogonal to $V$, often read “$V$ perp”:

$$V^\perp \triangleq \{\boldsymbol{x}: \boldsymbol{x} \in \mathbb{R}^n, \boldsymbol{x} \perp V \}$$

For example, if $V = \mathcal{L}\left(((1, 0)^\prime \right)$, $V^\perp = \mathcal{L}\left((0, 1)^\prime \right)$.

$V \cup V^\perp$ is not a subspace, but is a set in $\mathbb{R}^2$. $(3, 4)^\prime \in \mathbb{R}^2$ does not belong to $V \cup V^\perp$, but we can express it as

$$(3, 4)^\prime = \underbrace{(3, 0)^\prime}_{\in V} + \underbrace{(0, 4)^\prime}_{\in V^\perp}$$

Theorem: $V^\perp$ is a subspace of $\mathbb{R}^n$. To prove this we just need to show that it’s closed under linear combination. For any $\boldsymbol{x}_1, \boldsymbol{x}_2 \in V^\perp$,

$$\alpha_1 \boldsymbol{x}_1 + \alpha_2 \boldsymbol{x}_2 \in V^\perp \quad \forall \alpha_1, \alpha_2 \in \mathbb{R}$$

because $\alpha_1 \boldsymbol{x}_1 + \alpha_2 \boldsymbol{x}_2 \perp V$. For any $\boldsymbol{y} \in V$,

$$(\alpha_1 \boldsymbol{x}_1 + \alpha_2 \boldsymbol{x}_2) \cdot \boldsymbol{y} = \alpha_1 \boldsymbol{x}_1 \boldsymbol{y} + \alpha_2 \boldsymbol{x}_2 \boldsymbol{y} = 0$$

Theorem: For any vector $\boldsymbol{x} \in \mathbb{R}^n$, there exists $\boldsymbol{u}$ and $\boldsymbol{v}$ such that $\boldsymbol{x} = \boldsymbol{u} + \boldsymbol{v}$ where $\boldsymbol{u} \in V$ and $\boldsymbol{v} \in V^\perp$.

Let $\{\boldsymbol{u}_1, \cdots, \boldsymbol{u}_k \}$ be an orthogonal basis of $V$. By Gram-Schmidt, we can find

$$\{ \boldsymbol{u}_1, \cdots, \boldsymbol{u}_k, \boldsymbol{u}_{k+1}, \cdots, \boldsymbol{u}_n \}$$

to be an orthogonal basis of $\mathbb{R}^n$. Then, for any $\boldsymbol{x} \in \mathbb{R}^n$, we can write it as a linear combination of the basis vectors:

$$\boldsymbol{x} = \underbrace{\alpha_1 \boldsymbol{u}_1 + \cdots + \alpha_k \boldsymbol{u}_k}_{\text{lin. comb. of }\boldsymbol{u}_1, \cdots, \boldsymbol{u}_k \in V} + \underbrace{\alpha_{k+1} \boldsymbol{u}_{k+1} + \cdots + \alpha_n \boldsymbol{u}_n}_{\text{lin. comb. of } \boldsymbol{u}_{k+1}, \cdots, \boldsymbol{u}_n \in V^\perp}$$

This implies that $dim(V^\perp) = n-k = n - dim(V)$. We can also see that $(V^\perp)^\perp = V$.

For example, $V = \mathcal{L} \left( \boldsymbol{v}_1 = (1, 1, 1, 1)^\prime, \boldsymbol{v}_2 = (1, 1, -1, -1)^\prime \right)$. We first take two linearly independent vectors that are not in $V$, e.g. $(1, 0, 0, 0)^\prime$ and $(0, 0, 1, 0)^\prime$2. Then,

$$\left\{ (1, 1, 1, 1)^\prime, (1, 1, -1, -1)^\prime, (1, 0, 0, 0)^\prime, (0, 0, 1, 0)^\prime \right\}$$

is a basis of $\mathbb{R}^4$. Using G-S to orthogonalize them, we have

$$\begin{gathered} \boldsymbol{v}_3 = \left( 1, -1, 0, 0 \right)^\prime \\ \boldsymbol{v}_4 = \left( 0, 0, 1, -1 \right)^\prime \end{gathered}$$

and $V^\perp = \mathcal{L}(\boldsymbol{v}_3, \boldsymbol{v}_4)$. Take

\begin{aligned} \boldsymbol{x} &= (1, -2, 3, -4)^\prime \\ &= \boldsymbol{u} + \boldsymbol{v} \quad \boldsymbol{u} \in V, \boldsymbol{v} \in V^\perp \\ &= P(\boldsymbol{x} \mid V) + P(\boldsymbol{x} \mid V^\perp) \\ &= P(\boldsymbol{x} \mid \boldsymbol{v}_1) + P(\boldsymbol{x} \mid \boldsymbol{v}_2) + P(\boldsymbol{x} \mid \boldsymbol{v}_3) + P(\boldsymbol{x} \mid \boldsymbol{v}_4) \\ &= \left( \frac{1-2+3-4}{4}\boldsymbol{v}_1 + \frac{1-2-3+4}{4}\boldsymbol{v}_2 \right) + \left( \frac{1+2}{2}\boldsymbol{v}_3 + \frac{3 + 4}{2} \boldsymbol{v}_4 \right) \\ &= \left( -\frac{1}{2}(1, 1, 1, 1)^\prime + 0(1, 1, -1, -1)^\prime \right) + \left( \frac{3}{2}(1, -1, 0, 0)^\prime + \frac{7}{2} (0, 0, 1, -1)^\prime \right) \\ &= \left( -\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2} \right)^\prime + \left( \frac{3}{2}, -\frac{3}{2}, \frac{7}{2}, -\frac{7}{2} \right)^\prime \end{aligned}

Note that if we chose different LIN vectors from $(1, 0, 0, 0)^\prime$ and $(0, 0, 1, 0)^\prime$, our $\boldsymbol{v}_3$ and $\boldsymbol{v}_4$ may change, but the two projections are not going to change.

1. It doesn’t matter which one you start with, but the result may look different when different vectors are chosen. ↩︎

2. Tips for constructing this: use a lot of zeros. If you got this step wrong, the last step in G-S would give a vector that’s already seen. ↩︎

Related Posts
 Sep 21 Projection 7 min read Sep 14 Definitions in Arbitrary Linear Space 6 min read Aug 31 Vector Space 8 min read Aug 26 Matrices 8 min read Nov 18 Eigenvalues and Eigenvectors 20 min read