39 - Orthogonal Maps

Orthogonal Linear Map

Let $\alpha :V\to V$ be a linear map of a inner product space $V$

$$\alpha \text{ is orthogonal }\text{ if }⟨\alpha (v),\alpha (w)⟩=⟨v,w⟩$$

for all $v,w\in V$

Orthogonal Matrices preserve dot product

Let $\alpha :{\mathbb{R}}_{\mathrm{col}}^n\to {\mathbb{R}}_{\mathrm{col}}^n$ be a linear map
Let $A$ denote the matrix of $\alpha$ with respect to the standard basis

$$A\text{ is orthogonal with respect to the dot product}⟺A\text{ is an orthogonal matrix}$$

Proof

1. $⟹$
   Suppose that $\alpha$ is orthogonal with respect to the dot product
   Denote the standard basis as ${\mathbf{e}}_1,\cdots ,{\mathbf{e}}_n$
   Then

$${\delta }_{ij}={\mathbf{e}}_i\cdot {\mathbf{e}}_j=\alpha ({\mathbf{e}}_i)\cdot \alpha ({\mathbf{e}}_j)=(A{\mathbf{e}}_i{)}^T(A{\mathbf{e}}_j)={\mathbf{e}}_i^T{A}^{T}A{\mathbf{e}}_j$$

As ${\mathbf{e}}_i^T{A}^{T}A{\mathbf{e}}_j$ is the $(i,j)$th entry of $A^{T}A$ and ${\delta }_{ij}$ is the $(i,j)$th entry of $I_n$ then

$$A^{T}A=I_n$$

2. $⟸$
   Suppose that $A$ is orthogonal then $A^{T}A=I_n$

$$\begin{array}{rl}\alpha \left(\sum _{i=1}^n{u}_i{\mathbf{e}}_i\right)\cdot \alpha \left(\sum _{j=1}^n{v}_j{\mathbf{e}}_j\right)& =\sum _{i=1}^n\sum _{j=1}^n{u}_i{v}_j\alpha ({\mathbf{e}}_i)\cdot \alpha ({\mathbf{e}}_j)\\ & =\sum _{i=1}^n\sum _{j=1}^n{u}_i{v}_j{\delta }_{ij}\\ & =\sum _{i=1}^n{u}_i{v}_i\\ & \\ & =\left(\sum _{i=1}^n{u}_i{\mathbf{e}}_i\right)\cdot \left(\sum _{j=1}^n{v}_j{\mathbf{e}}_j\right)\end{array}$$

Hence $\alpha$ is orthogonal

Relation between Orthogonal Maps and Inner Product Spaces

An orthogonal map is an isometry of an inner product space

Proof

Suppose $\alpha$ is orthogonal then

$$d(\alpha (u),\alpha (w){)}^2=||\alpha (v-w)|{|}^2=⟨\alpha (v-w),\alpha (v-w)⟩=⟨v-w,v-w⟩=||v-w|{|}^2=d(v,w{)}^2$$

Hence $\alpha$ is an isometry

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Orthonormal Set

Let $V$ be an inner product space
If $\{v_1,\cdots ,v_k\}\subseteq V$ is a orthonormal set if

$$⟨v_i,v_j⟩={\delta }_{ij}$$

for all $i,j$
i.e. the vectors are unit length and mutually perpendicular

Orthonormal Sets are linearly independent in a inner product space

In an inner product space $V$, a orthonromal sets is linearly independent

Proof

Suppose $\{v_1,\cdots ,v_k\}$ is orthonormal and ${\alpha }_1,\cdots ,{\alpha }_k\in \mathbb{R}$ such that

$${\alpha }_1{v}_1+\cdots +{\alpha }_k{v}_k=0_V$$

For $1\le i\le k$ we havea

$$\begin{array}{rl}0=⟨0_V,v_i⟩& =⟨{\alpha }_1{v}_1+\cdots +{\alpha }_k{v}_k,v_i⟩\\ & ={\alpha }_1⟨v_1,v_i⟩+\cdots +{\alpha }_k⟨v_k,v_i⟩\\ & ={\alpha }_i\end{array}$$

Hence ${\alpha }_1=\cdots ={\alpha }_k=0$

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Equivalent Properties of an Inner Product

For $X\in {\mathcal{M}}_{n\times n}(\mathbb{R})$
Let ${\mathbb{R}}^n$ be a inner product space using the dot product

1. $X{X}^T=I_n$
2. $X^{T}X=I_n$
3. Rows of $X$ form an orthonormal basis of ${\mathbb{R}}^n$
4. Columns of $X$ form an orthonormal basis of ${\mathbb{R}}_{\mathrm{col}}^n$
5. For all $\mathbf{x},\mathbf{y}\in {\mathbb{R}}_{\mathrm{col}}^n$ then $X\mathbf{x}\cdot X\mathbf{y}=\mathbf{x}\cdot \mathbf{y}$