1067: For 3-Dimensional Euclidean Vectors Space, Rotation Matrix w.r.t. Orthonormal Basis Around Any Axis Is This

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description/proof of that for 3-dimensional Euclidean vectors space, rotation matrix w.r.t. orthonormal basis around any axis is this

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Topics

About: vectors space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Proof
• 3: Note

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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.

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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.

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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.

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Main Body

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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 $⟨n,n⟩=1$
$f$: $:V\to V$, $=\text{ the }\theta \text{ -rotation around }n$
$N$: $=\text{ the components matrix of }f\text{ with respect to }B$
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Statements:
$N=M{N}^{\prime }{M}^{-1}=\left(\begin{array}{ccc}(1-{n^1}^2)cos\theta +{n^1}^2& -n^1{n}^{2}cos\theta -n^{3}sin\theta +n^1{n}^2& n^{2}sin\theta -n^3{n}^{1}cos\theta +n^3{n}^1\\ -n^1{n}^{2}cos\theta +n^{3}sin\theta +n^1{n}^2& (1-{n^2}^2)cos\theta +{n^2}^2& -n^2{n}^{3}cos\theta -n^{1}sin\theta +n^2{n}^3\\ -n^3{n}^{1}cos\theta -n^{2}sin\theta +n^3{n}^1& -n^2{n}^{3}cos\theta +n^{1}sin\theta +n^3{n}^2& ({n^1}^2+{n^2}^2)cos\theta +{n^3}^2\end{array}\right)$, where $M=\left(\begin{array}{ccc}1/\sqrt{{n^1}^2+{n^2}^2}{n}^2& 1/\sqrt{{n^1}^2+{n^2}^2}{n}^3{n}^1& n^1\\ -1/\sqrt{{n^1}^2+{n^2}^2}{n}^1& 1/\sqrt{{n^1}^2+{n^2}^2}{n}^2{n}^3& n^2\\ 0& -1/\sqrt{{n^1}^2+{n^2}^2}({n^1}^2+{n^2}^2)& n^3\end{array}\right)$ and $N^{\prime }=\left(\begin{array}{ccc}cos\theta & -sin\theta & 0\\ sin\theta & cos\theta & 0\\ 0& 0& 1\end{array}\right)$
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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, $B^{\prime }$, with $b_3^{\prime }=n$; Step 2: get the components matrix of $f$ with respect to $B^{\prime }$; 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 $⟨b_j,b_l⟩={\delta }_{j,l}$, then, $⟨{\bar{v}}^1{b}_1+{\bar{v}}^2{b}_2+{\bar{v}}^3{b}_3,v^1{b}_1+v^2{b}_2+v^3{b}_3⟩={\bar{v}}^1{v}^1+{\bar{v}}^2{v}^2+{\bar{v}}^3{v}^3$.

Let us take an orthonormal basis, $B^{\prime }=(b_1^{\prime },b_2^{\prime },b_3^{\prime })$, with $b_3^{\prime }=n$: that condition has not determined it uniquely, yet.

Let $b_1^{\prime }$ be $1/\sqrt{{n^1}^2+{n^2}^2}(n^2{b}_1-n^1{b}_2+0b_3)$, which certainly satisfies $⟨b_3^{\prime },b_1^{\prime }⟩=0$ and $⟨b_1^{\prime },b_1^{\prime }⟩=0$.

Let $b_2^{\prime }$ be $b_3^{\prime }\times b_1^{\prime }=1/\sqrt{{n^1}^2+{n^2}^2}(n^3{n}^1{b}_1+n^2{n}^3{b}_2-({n^1}^2+{n^2}^2)b_3)$.

$(b_1^{\prime },b_2^{\prime },b_3^{\prime })=(b_1,b_2,b_3)M$ where $M=\left(\begin{array}{ccc}1/\sqrt{{n^1}^2+{n^2}^2}{n}^2& 1/\sqrt{{n^1}^2+{n^2}^2}{n}^3{n}^1& n^1\\ -1/\sqrt{{n^1}^2+{n^2}^2}{n}^1& 1/\sqrt{{n^1}^2+{n^2}^2}{n}^2{n}^3& n^2\\ 0& -1/\sqrt{{n^1}^2+{n^2}^2}({n^1}^2+{n^2}^2)& n^3\end{array}\right)$.

Step 2:

Let us get the components matrix of $f$ with respect to $B^{\prime }$, $N^{\prime }$.

As it is the $\theta$-rotation around the $b_3^{\prime }$ axis, $N^{\prime }=\left(\begin{array}{ccc}cos\theta & -sin\theta & 0\\ sin\theta & cos\theta & 0\\ 0& 0& 1\end{array}\right)$: each $(rcos{\theta }_0,rsin{\theta }_0,z{)}^t$ is mapped to $(rcos({\theta }_0+\theta ),rsin({\theta }_0+\theta ),z{)}^t=(r(cos{\theta }_{0}cos\theta -sin{\theta }_{0}sin\theta ),r(sin{\theta }_{0}cos\theta +cos{\theta }_{0}sin\theta ),z{)}^t=(cos\theta (rcos{\theta }_0)-sin\theta (rsin{\theta }_0),sin\theta (rcos{\theta }_0)+cos\theta (rsin{\theta }_0),z{)}^t$.

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.

$N=M{N}^{\prime }{M}^{-1}$.

Let us get $M^{-1}$ by the the Laplace expansion of determinant of matrix.

$detM=1$: while that can be gotten by the diligent computation, it is a well-known fact because it is the transformation matrix between orthonormal bases.

$M^{-1}=\left(\begin{array}{ccc}1/\sqrt{{n^1}^2+{n^2}^2}{n}^2& -1/\sqrt{{n^1}^2+{n^2}^2}{n}^1& 0\\ 1/\sqrt{{n^1}^2+{n^2}^2}{n}^3{n}^1& 1/\sqrt{{n^1}^2+{n^2}^2}{n}^2{n}^3& -1/\sqrt{{n^1}^2+{n^2}^2}({n^1}^2+{n^2}^2)\\ n^1& n^2& n^3\end{array}\right)$.

$N^{\prime }{M}^{-1}=\left(\begin{array}{ccc}1/\sqrt{{n^1}^2+{n^2}^2}(n^{2}cos\theta -n^3{n}^{1}sin\theta )& -1/\sqrt{{n^1}^2+{n^2}^2}(n^{1}cos\theta +n^2{n}^{3}sin\theta )& 1/\sqrt{{n^1}^2+{n^2}^2}({n^1}^2+{n^2}^2)sin\theta \\ 1/\sqrt{{n^1}^2+{n^2}^2}(n^3{n}^{1}cos\theta +n^{2}sin\theta )& 1/\sqrt{{n^1}^2+{n^2}^2}(n^2{n}^{3}cos\theta -n^{1}sin\theta )& -1/\sqrt{{n^1}^2+{n^2}^2}({n^1}^2+{n^2}^2)cos\theta \\ n^1& n^2& n^3\end{array}\right)$.

$M{N}^{\prime }{M}^{-1}=\left(\begin{array}{ccc}(1-{n^1}^2)cos\theta +{n^1}^2& -n^1{n}^{2}cos\theta -n^{3}sin\theta +n^1{n}^2& n^{2}sin\theta -n^3{n}^{1}cos\theta +n^3{n}^1\\ -n^1{n}^{2}cos\theta +n^{3}sin\theta +n^1{n}^2& (1-{n^2}^2)cos\theta +{n^2}^2& -n^2{n}^{3}cos\theta -n^{1}sin\theta +n^2{n}^3\\ -n^3{n}^{1}cos\theta -n^{2}sin\theta +n^3{n}^1& -n^2{n}^{3}cos\theta +n^{1}sin\theta +n^3{n}^2& ({n^1}^2+{n^2}^2)cos\theta +{n^3}^2\end{array}\right)$.

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3: Note

Let us see that the formula does not contradict some special cases.

When $(n^1,n^2,n^3)=(1,0,0)$, $N=\left(\begin{array}{ccc}1& 0& 0\\ 0& cos\theta & -sin\theta \\ 0& sin\theta & cos\theta \end{array}\right)$.

When $(n^1,n^2,n^3)=(0,1,0)$, $N=\left(\begin{array}{ccc}cos\theta & 0& sin\theta \\ 0& 1& 0\\ -sin\theta & 0& cos\theta \end{array}\right)$.

When $\theta =0$, $N=\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)$.

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References

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