description/proof of that for linear map from Euclidean-normed Euclidean vectors space into itself, map is orthogonal iff canonical representative matrix is orthogonal
Topics
About: vectors space
The table of contents of this article
Starting Context
- The reader knows a definition of Euclidean-normed Euclidean vectors space.
- The reader knows a definition of orthogonal linear map.
- The reader knows a definition of canonical representative matrix of linear map between finite-product-of-copies-of-field vectors spaces.
- The reader knows a definition of orthogonal matrix.
- The reader admits the proposition that for any polynomial over any field with any finite number of variables, if the polynomial is constantly \(0\), the coefficients are \(0\).
Target Context
- The reader will have a description and a proof of the proposition that for any linear map from any Euclidean-normed Euclidean vectors space into itself, the map is orthogonal if and only if the canonical representative matrix is orthogonal.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(d\): \(\in \mathbb{N} \setminus \{0\}\)
\(f\): \(: \mathbb{R}^d \to \mathbb{R}^d\), \(\in \{\text{ the linear maps }\}\)
\(B\): \(= \text{ the canonical basis for } \mathbb{R}^d\), \(= \{b_1, ..., b_d\}\)
\(M\): \(= \text{ the representative matrix of } f \text{ with respect to } B\)
//
Statements:
\(f \in \{\text{ the orthogonal maps }\}\)
\(\iff\)
\(M \in \{\text{ the orthogonal matrices }\}\)
//
2: Note
Do not wantonly generalize this proposition as though "for any linear map from any Euclidean vectors space with any norm into itself, the map is orthogonal if and only if the representative matrix with respect to any basis is orthogonal", which is not true.
This proposition holds because the Euclidean norm and the canonical basis are used.
For example, let \(d = 2\), \(\mathbb{R}^d\) have any basis, \(B = \{b^1, b^2\}\), and \(f\) have the representative matrix with respect to \(B\), \(M = \begin{pmatrix} 1 & 0 \\ 0 & - 1 \end{pmatrix}\); also \(B' = \{1 / 2 (b^1 + b^2), b^1 - b^2\}\) is a basis for \(\mathbb{R}^d\) and let \(\mathbb{R}^d\) have the norm with respect to \(B'\) defined in the proposition that for any real or complex vectors space, taking any basis and the sum of the absolute components for each vector is a norm.
Then, \(M\) is orthogonal, but \(f\) is not orthogonal, because \(f (b^1 + b^2) = b^1 - b^2\), but \(\Vert b^1 + b^2 \Vert = \Vert 2 1 / 2 (b^1 + b^2) + 0 (b^1 - b^2) \Vert = \vert 2 \vert + \vert 0 \vert = 2\) and \(\Vert b^1 - b^2 \Vert = \Vert 0 (1 / 2 (b^1 + b^2)) + 1 (b^1 - b^2) \Vert = \vert 0 \vert + \vert 1 \vert = 1\), so, \(\Vert f (b^1 + b^2) \Vert \neq \Vert b^1 + b^2 \Vert\).
On the other hand, if the representative matrix with respect to a basis of an orthogonal \(f\) is orthogonal, the representative matrix with respect to another basis of \(f\) may not be orthogonal, by the proposition that for any module with any \(2\) bases of any same finite cardinality and any module endomorphism, the transition of the endomorphism matrices with respect to the change of the bases is this and the proposition that the conjugate of a unitary matrix by an invertible matrix is not necessarily unitary.
3: Proof
Whole Strategy: apply the proposition that for any polynomial over any field with any finite number of variables, if the polynomial is constantly \(0\), the coefficients are \(0\); Step 1: suppose that \(f\) is an orthogonal map; Step 2: see that \(M\) is an orthogonal matrix; Step 3: suppose that \(M\) is an orthogonal matrix; Step 4: see that \(f\) is an orthogonal map.
Step 1:
Let us suppose that \(f\) is an orthogonal map.
Step 2:
Let us see that \(M\) is an orthogonal matrix.
Let \(v \in \mathbb{R}^d\) be any.
\(\Vert f (v) \Vert^2 = (M v^t)^t M v^t\), by the definition of Euclidean norm, \(= v M^t M v^t = \Vert v \Vert^2\), because \(f\) is orthogonal, but \(= v v^t = \sum_{j \in \{1, ..., d\}} v^j v^j = \sum_{l \in \{1, ..., d\}, j \in \{1, ..., d\}} v^l \delta^l_j v^j\).
Let \(N := M^t M\).
\(v N v^t = \sum_{l \in \{1, ..., d\}, j \in \{1, ..., d\}} v^l N^l_j v^j\).
So, \(\sum_{l \in \{1, ..., d\}, j \in \{1, ..., d\}} v^l N^l_j v^j = \sum_{l \in \{1, ..., d\}, j \in \{1, ..., d\}} v^l \delta^l_j v^j\), and \(\sum_{l \in \{1, ..., d\}, j \in \{1, ..., d\}} (v^l N^l_j v^j - v^l \delta^l_j v^j) = 0\), so, \(\sum_{l \in \{1, ..., d\}, j \in \{1, ..., d\}} (N^l_j - \delta^l_j) v^l v^j = 0\).
That holds constantly with respect to \((v^1, ..., v^d)\), so, by the proposition that for any polynomial over any field with any finite number of variables, if the polynomial is constantly \(0\), the coefficients are \(0\), \(N^l_j - \delta^l_j = 0\), so, \(N^l_j = \delta^l_j\).
So, \(M^t M = I\).
So, \(M\) is an orthogonal matrix.
Step 3:
Let us suppose that \(M\) is an orthogonal matrix.
Step 4:
Let \(v \in \mathbb{R}^d\) be any.
\(\Vert f (v) \Vert^2 = \Vert M v^t \Vert^2 = (M v^t)^t M v^t = v M^t M v^t = v I v^t = v v^t = \Vert v \Vert^2\).
So, \(f\) is an orthogonal map.
