description/proof of that for 3-dimensional Euclidean vectors space, rotation matrix w.r.t. orthonormal basis around any axis is this
Topics
About: vectors space
The table of contents of this article
Starting Context
- The reader knows a definition of Euclidean inner product on Euclidean vectors space.
- The reader admits the proposition that for any finite-dimensional vectors space and any vectors space endomorphism, the transition of the endomorphism matrix with respect to any change of bases is this.
- The reader admits the Laplace expansion of determinant of matrix.
Target Context
- The reader will have a description and a proof of the proposition that for the 3-dimensional Euclidean vectors space, the rotation matrix with respect to any orthonormal basis around any axis 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:
\(V\): \(= \mathbb{R}^3\), \(= \text{ the Euclidean vectors space }\), with the Euclidean inner product
\(B\): \(\in \{\text{ the orthonormal bases for } V\}\), \(= \{b_1, b_2, b_3\}\)
\(n\): \(\in V\), \(= n^1 b_1 + n^2 b_2 + n^3 b_3\), such that \(\langle n, n \rangle = 1\)
\(f\): \(: V \to V\), \(= \text{ the } \theta \text{ -rotation around } n\)
\(N\): \(= \text{ the components matrix of } f \text{ with respect to } B\)
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Statements:
\(N = M N' M^{-1} = \begin{pmatrix} (1 - {n^1}^2) cos \theta + {n^1}^2 & - n^1 n^2 cos \theta - n^3 sin \theta + n^1 n^2 & n^2 sin \theta - n^3 n^1 cos \theta + n^3 n^1 \\ - n^1 n^2 cos \theta + n^3 sin \theta + n^1 n^2 & (1 - {n^2}^2) cos \theta + {n^2}^2 & - n^2 n^3 cos \theta - n^1 sin \theta + n^2 n^3 \\ - n^3 n^1 cos \theta - n^2 sin \theta + n^3 n^1 & - n^2 n^3 cos \theta + n^1 sin \theta + n^3 n^2 & ({n^1}^2 + {n^2}^2) cos \theta + {n^3}^2 \end{pmatrix}\), where \(M = \begin{pmatrix} 1 / \sqrt{{n^1}^2 + {n^2}^2} n^2 & 1 / \sqrt{{n^1}^2 + {n^2}^2} n^3 n^1 & n^1 \\ - 1 / \sqrt{{n^1}^2 + {n^2}^2} n^1 & 1 / \sqrt{{n^1}^2 + {n^2}^2} n^2 n^3 & n^2 \\ 0 & - 1 / \sqrt{{n^1}^2 + {n^2}^2} ({n^1}^2 + {n^2}^2) & n^3 \end{pmatrix}\) and \(N' = \begin{pmatrix} cos \theta & - sin \theta & 0 \\ sin \theta & cos \theta & 0 \\ 0 & 0 & 1 \end{pmatrix}\)
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2: Proof
Whole Strategy: apply the proposition that for any finite-dimensional vectors space and any vectors space endomorphism, the transition of the endomorphism matrix with respect to any change of bases is this; Step 1: take an orthonormal basis, \(B'\), with \(b'_3 = n\); Step 2: get the components matrix of \(f\) with respect to \(B'\); Step 3: apply the proposition that for any finite-dimensional vectors space and any vectors space endomorphism, the transition of the endomorphism matrix with respect to any change of bases is this.
Step 1:
\(B\)'s being orthonormal means that \(\langle b_j, b_l \rangle = \delta_{j, l}\), then, \(\langle \overline{v}^1 b_1 + \overline{v}^2 b_2 + \overline{v}^3 b_3, v^1 b_1 + v^2 b_2 + v^3 b_3 \rangle = \overline{v}^1 v^1 + \overline{v}^2 v^2 + \overline{v}^3 v^3\).
Let us take an orthonormal basis, \(B' = (b'_1, b'_2, b'_3)\), with \(b'_3 = n\): that condition has not determined it uniquely, yet.
Let \(b'_1\) be \(1 / \sqrt{{n^1}^2 + {n^2}^2} (n^2 b_1 - n^1 b_2 + 0 b_3)\), which certainly satisfies \(\langle b'_3, b'_1 \rangle = 0\) and \(\langle b'_1, b'_1 \rangle = 0\).
Let \(b'_2\) be \(b'_3 \times b'_1 = 1 / \sqrt{{n^1}^2 + {n^2}^2} (n^3 n^1 b_1 + n^2 n^3 b_2 - ({n^1}^2 + {n^2}^2) b_3)\).
\((b'_1, b'_2, b'_3) = (b_1, b_2, b_3) M\) where \(M = \begin{pmatrix} 1 / \sqrt{{n^1}^2 + {n^2}^2} n^2 & 1 / \sqrt{{n^1}^2 + {n^2}^2} n^3 n^1 & n^1 \\ - 1 / \sqrt{{n^1}^2 + {n^2}^2} n^1 & 1 / \sqrt{{n^1}^2 + {n^2}^2} n^2 n^3 & n^2 \\ 0 & - 1 / \sqrt{{n^1}^2 + {n^2}^2} ({n^1}^2 + {n^2}^2) & n^3 \end{pmatrix}\).
Step 2:
Let us get the components matrix of \(f\) with respect to \(B'\), \(N'\).
As it is the \(\theta\)-rotation around the \(b'_3\) axis, \(N' = \begin{pmatrix} cos \theta & - sin \theta & 0 \\ sin \theta & cos \theta & 0 \\ 0 & 0 & 1 \end{pmatrix}\): each \((r cos \theta_0, r sin \theta_0, z)^t\) is mapped to \((r cos (\theta_0 + \theta), r sin (\theta_0 + \theta), z)^t = (r (cos \theta_0 cos \theta - sin \theta_0 sin \theta), r (sin \theta_0 cos \theta + cos \theta_0 sin \theta), z)^t = (cos \theta (r cos \theta_0) - sin \theta (r sin \theta_0), sin \theta (r cos \theta_0) + cos \theta (r sin \theta_0), z)^t\).
Step 3:
Let us apply the proposition that for any finite-dimensional vectors space and any vectors space endomorphism, the transition of the endomorphism matrix with respect to any change of bases is this.
\(N = M N' M^{-1}\).
Let us get \(M^{-1}\) by the the Laplace expansion of determinant of matrix.
\(det M = 1\): while that can be gotten by the diligent computation, it is a well-known fact because it is the transformation matrix between orthonormal bases.
\(M^{-1} = \begin{pmatrix} 1 / \sqrt{{n^1}^2 + {n^2}^2} n^2 & - 1 / \sqrt{{n^1}^2 + {n^2}^2} n^1 & 0 \\ 1 / \sqrt{{n^1}^2 + {n^2}^2} n^3 n^1 & 1 / \sqrt{{n^1}^2 + {n^2}^2} n^2 n^3 & - 1 / \sqrt{{n^1}^2 + {n^2}^2} ({n^1}^2 + {n^2}^2) \\ n^1 & n^2 & n^3 \end{pmatrix}\).
\(N' M^{-1} = \begin{pmatrix} 1 / \sqrt{{n^1}^2 + {n^2}^2} (n^2 cos \theta - n^3 n^1 sin \theta) & - 1 / \sqrt{{n^1}^2 + {n^2}^2} (n^1 cos \theta + n^2 n^3 sin \theta) & 1 / \sqrt{{n^1}^2 + {n^2}^2} ({n^1}^2 + {n^2}^2) sin \theta \\ 1 / \sqrt{{n^1}^2 + {n^2}^2} (n^3 n^1 cos \theta + n^2 sin \theta) & 1 / \sqrt{{n^1}^2 + {n^2}^2} (n^2 n^3 cos \theta - n^1 sin \theta) & - 1 / \sqrt{{n^1}^2 + {n^2}^2} ({n^1}^2 + {n^2}^2) cos \theta \\ n^1 & n^2 & n^3 \end{pmatrix}\).
\(M N' M^{-1} = \begin{pmatrix} (1 - {n^1}^2) cos \theta + {n^1}^2 & - n^1 n^2 cos \theta - n^3 sin \theta + n^1 n^2 & n^2 sin \theta - n^3 n^1 cos \theta + n^3 n^1 \\ - n^1 n^2 cos \theta + n^3 sin \theta + n^1 n^2 & (1 - {n^2}^2) cos \theta + {n^2}^2 & - n^2 n^3 cos \theta - n^1 sin \theta + n^2 n^3 \\ - n^3 n^1 cos \theta - n^2 sin \theta + n^3 n^1 & - n^2 n^3 cos \theta + n^1 sin \theta + n^3 n^2 & ({n^1}^2 + {n^2}^2) cos \theta + {n^3}^2 \end{pmatrix}\).
3: Note
Let us see that the formula does not contradict some special cases.
When \((n^1, n^2, n^3) = (1, 0, 0)\), \(N = \begin{pmatrix} 1 & 0 & 0 \\ 0 & cos \theta & - sin \theta \\ 0 & sin \theta & cos \theta \end{pmatrix}\).
When \((n^1, n^2, n^3) = (0, 1, 0)\), \(N = \begin{pmatrix} cos \theta & 0 & sin \theta \\ 0 & 1 & 0 \\ - sin \theta & 0 & cos \theta \end{pmatrix}\).
When \(\theta = 0\), \(N = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\).
