description/proof of that for real vectors space with inner product and \(2\) linearly independent unit vectors with angle \(\theta\), unit vector on plane with angle \(\theta'\) from 1st vector is this
Topics
About: vectors space
The table of contents of this article
- Starting Context
- Target Context
- Orientation
- Main Body
- 1: Structured Description
- 2: Note 1
- 3: Proof
- 4: Note 2
Starting Context
- The reader knows a definition of inner product on real or complex vectors space.
Target Context
- The reader will have a description and a proof of the proposition that for any real vectors space with any inner product and any \(2\) linearly independent unit vectors with any angle \(\theta\), the unit vector on the plane with any angle \(\theta'\) from the 1st vector is this.
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:
\(\mathbb{R}\): with the canonical field structure
\(V\): \(\in \{\text{ the } \mathbb{R} \text{ vectors spaces }\}\) with any inner product, \(\langle \bullet, \bullet \rangle\)
\(u_1\): \(\in V\), with \(\langle u_1, u_1 \rangle = 1\)
\(u_2\): \(\in V\), with \(\langle u_2, u_2 \rangle = 1\)
\(u\): \(\in V\), \(= c u_1 + d u_2\), with \(\langle u, u \rangle = 1\)
\(\theta\): \(\in \mathbb{R}\) such that \(0 \lt \theta \lt \pi\)
//
Statements:
(
\(\langle u_1, u_2 \rangle = cos \theta\)
\(\land\)
\(\langle u_1, u \rangle = cos \theta'\) where \(0 \le \theta' \lt 2 \pi\)
\(\land\)
\(\langle u_2, u \rangle = cos (\theta' - \theta)\)
)
\(\implies\)
\(c = cos \theta' - (sin \theta' / sin \theta) cos \theta \land d = sin \theta' / sin \theta\)
//
\(\langle u_2, u \rangle = cos (\theta' - \theta)\) instead of \(cos (\theta' + \theta)\) means nothing but that \(\theta'\) is taken in the same direction with \(\theta\).
2: Note 1
The dimension of \(V\) can be any including infinity.
In other words, \(u\) is \(u_1\) rotated \(\theta'\) toward \(u_2\).
On any \(d\)-dimensional real vectors space, a rotation around a vector as "axis" is a '\(d - 1\)'-dimensional rotation; on the '\(d - 1\)'-dimensional subspace, a rotation around a vector as "axis" is a '\(d - 2\)'-dimensional rotation; ...; on the \(3\)-dimensional subspace, a rotation around a vector as "axis" is a \(2\)-dimensional rotation.
So, any \(2\)-dimensional rotation on any \(d\)-dimensional real vectors space requires some \(d - 2\) axes.
When one thinks of a \(2\)-dimensional rotation on a \(d\)-dimensional real vectors space, often, it is simpler to just think of the \(2\)-dimensional subspace than to think of the \(d - 2\) axes.
This proposition is thinking of the \(2\)-dimensional subspace spanned by \(u_1\) and \(u_2\).
3: Proof
Whole Strategy: Step 1: put \(u = c u_1 + d u_2\) into \(\langle u_1, u \rangle = cos \theta'\) and \(\langle u, u \rangle = 1\), and get the candidates for \(c\) and \(d\); Step 2: choose the unique \(c\) and \(d\).
Step 1:
Let us put \(u = c u_1 + d u_2\) into \(\langle u_1, u \rangle = cos \theta'\) and \(\langle u, u \rangle = 1\).
\(cos \theta' = \langle u_1, c u_1 + d u_2 \rangle = c \langle u_1, u_1 \rangle + d \langle u_1, u_2 \rangle = c + d cos \theta\).
\(1 = \langle u, u \rangle = \langle c u_1 + d u_2, c u_1 + d u_2 \rangle = c^2 \langle u_1, u_1 \rangle + c d \langle u_1, u_2 \rangle + d c \langle u_2, u_1 \rangle + d^2 \langle u_2, u_2 \rangle = c^2 + 2 c d cos \theta + d^2\).
As \(c = cos \theta' - d cos \theta\), \(1 = (cos \theta' - d cos \theta)^2 + 2 (cos \theta' - d cos \theta) d cos \theta + d^2 = d^2 (1 + cos^2 \theta - 2 cos^2 \theta) + d (- 2 cos \theta' cos \theta + 2 cos \theta' cos \theta) + cos^2 \theta' = d^2 (1 - cos^2 \theta) + cos^2 \theta' = d^2 sin^2 \theta + cos^2 \theta'\), so, \(d^2 = (1 - cos^2 \theta') / sin^2 \theta = sin^2 \theta' / sin^2 \theta\) (\(sin \theta \neq 0\) because \(0 \lt \theta \lt \pi\)), so, \(d = +- sin \theta' / sin \theta\).
\(c = cos \theta' -+ (sin \theta' / sin \theta) cos \theta\).
Step 2:
While Step 1 has gotten only candidates, \(c\) and \(d\) should be uniquely determined, because as \(\theta'\) is dictated to be unambiguously taken in the same direction with \(\theta\), \(u\) should be uniquely determined.
In other words, the condition, \(\langle u_2, u \rangle = cos (\theta' - \theta)\), should determine \(c\) and \(d\) uniquely.
\(\langle u_2, u \rangle = \langle u_2, c u_1 + d u_2 \rangle = c \langle u_2, u_1 \rangle + d \langle u_2, u_2 \rangle = c cos \theta + d = (cos \theta' -+ (sin \theta' / sin \theta) cos \theta) cos \theta +- sin \theta' / sin \theta = cos \theta' cos \theta -+ (sin \theta' / sin \theta) cos^2 \theta +- sin \theta' / sin \theta = cos \theta' cos \theta +- (sin \theta' / sin \theta - (sin \theta' / sin \theta) cos^2 \theta) = cos \theta' cos \theta +- sin \theta' / sin \theta (1 - cos^2 \theta) = cos \theta' cos \theta +- sin \theta' / sin \theta (sin^2 \theta) = cos \theta' cos \theta +- sin \theta' sin \theta\).
On the other hand, \(cos (\theta' - \theta) = cos (- \theta) cos \theta' - sin (- \theta) sin \theta' = cos \theta cos \theta' + sin \theta sin \theta'\).
So, \(c = cos \theta' - (sin \theta' / sin \theta) cos \theta\) and \(d = sin \theta' / sin \theta\) is the answer.
4: Note 2
Especially, when \(\theta = \pi / 2\), which means that \((u_1, u_2)\) is orthonormal, \(c = cos \theta'\) and \(d = sin \theta'\), and \(u = cos \theta' u_1 + sin \theta' u_2\).
For any vector on the \(2\)-dimensional subspace, it can be expressed as \(u = r u_1 = r ({u_1}^1 e_1 + {u_1}^2 e_2)\) for any orthonormal basis for the subspace, \((e_1, e_2)\). \(u_2\) can be taken to be \(u_2 = - {u_1}^2 e_1 + {u_1}^1 e_2\) to be orthonormal to \(u_1\). By this proposition, \(u_1\) is rotated \(\theta'\) to be \(cos \theta' u_1 + sin \theta' u_2 = cos \theta' ({u_1}^1 e_1 + {u_1}^2 e_2) + sin \theta' (- {u_1}^2 e_1 + {u_1}^1 e_2) = (cos \theta' {u_1}^1 - sin \theta' {u_1}^2) e_1 + (cos \theta' {u_1}^2 + sin \theta' {u_1}^1) e_2\), and \(u\) is rotated \(\theta'\) to be \(r ((cos \theta' {u_1}^1 - sin \theta' {u_1}^2) e_1 + (cos \theta' {u_1}^2 + sin \theta' {u_1}^1) e_2) = ((cos \theta' r {u_1}^1 - sin \theta' r {u_1}^2) e_1 + (cos \theta' r {u_1}^2 + sin \theta' r {u_1}^1) e_2)\), which means that the transformation matrix for the components with respect to the basis is \(\begin{pmatrix} cos \theta' & - sin \theta' \\ sin \theta' & cos \theta' \end{pmatrix}\).
For any vector, \(v \in V\), \(v = u + v'\), where \(v'\) is orthogonal to \(e_1\) and \(e_2\), and \(v\) is \(\theta'\) rotated to be \(((cos \theta' r {u_1}^1 - sin \theta' r {u_1}^2) e_1 + (cos \theta' r {u_1}^2 + sin \theta' r {u_1}^1) e_2) + v'\).
