Masteries · Jason GlowneyArithmoí Book 1

Masteries

1.1.7 Group Exponentiation

The el­e­ments in the groups dis­cussed so far have been finite. We can ex­po­nen­ti­ate ma­tri­ces sim­i­lar­ly to how it is done for single vari­ables, making in­fi­nite de­grees of free­dom for ro­ta­tions in ℝn and ℂn\mathbb{R}^{n}\ and\ \mathbb{C}^{n}. We define ex­po­nen­ti­a­tion for a matrix M by

eM=∑n=0∞Mnn!=(M00!+M11!+M22!+M33!+…),\begin{array}{r} e^{M} = \sum_{n = 0}^{\infty}\frac{M^{n}}{n!} = \left( \frac{M^{0}}{0!} + \frac{M^{1}}{1!} + \frac{M^{2}}{2!} + \frac{M^{3}}{3!} + \ldots \right), \end{array}
(1.1.14)

where M0!\frac{M}{0!} equals the iden­ti­ty matrix with the same degree as M. We use the ex­am­ple

M=(0−xx0),\begin{array}{r} M = \begin{pmatrix} 0 & - x \\ x & 0 \end{pmatrix}, \end{array}
(1.1.15)

which would result in the fol­low­ing ex­po­nen­tial map­ping ex­pan­sion

eM=(1001)+(0−xx0)+(−x2200−x22)+(0x36−x360)+(x42400x424)+(0−x5120x51200)+….\begin{array}{r} e^{M} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} + \begin{pmatrix} 0 & - x \\ x & 0 \end{pmatrix} + \begin{pmatrix} \frac{- x^{2}}{2} & 0 \\ 0 & \frac{- x^{2}}{2} \end{pmatrix} + \begin{pmatrix} 0 & \frac{x^{3}}{6} \\ - \frac{x^{3}}{6} & 0 \end{pmatrix} + \begin{pmatrix} \frac{x^{4}}{24} & 0 \\ 0 & \frac{x^{4}}{24} \end{pmatrix} + \begin{pmatrix} 0 & \frac{- x^{5}}{120} \\ \frac{x^{5}}{120} & 0 \end{pmatrix} + \ldots. \end{array}
(1.1.16)

Group­ing the above terms by ad­di­tion, we find the in­fi­nite series for trigono­met­ric sine and cosine func­tions appear

eM=(1−x22+x424+…−x+x36+…x−x36+…1−x22+x424+…)=(cos⁡(x)−sin⁡(x)sin⁡(x)cos⁡(x)).\begin{array}{r} e^{M} = \begin{pmatrix} 1 - \frac{x^{2}}{2} + \frac{x^{4}}{24} + \ldots & - x + \frac{x^{3}}{6} + \ldots \\ x - \frac{x^{3}}{6} + \ldots & 1 - \frac{x^{2}}{2} + \frac{x^{4}}{24} + \ldots \end{pmatrix} = \begin{pmatrix} \cos(x) & - \sin(x) \\ \sin(x) & \cos(x) \end{pmatrix}. \end{array}
(1.1.17)

For small­er square ma­tri­ces n×nn \times n, like the above, ex­po­nen­ti­a­tion is fea­si­ble, but for larger ma­tri­ces this be­comes chal­leng­ing, and many al­go­rithms exist to cal­cu­late them. For a di­ag­o­nal­iz­able matrix MM, it is pos­si­ble to de­com­pose it into its eigen­val­ues arranged on the di­ag­o­nal matrix D, and by the re­spec­tive eigen­vec­tors of MM, form­ing the or­tho­nor­mal column vec­tors in a matrix V. The matrix M can then be rep­re­sent­ed as

M=VDV−1.\begin{array}{r} M = VDV^{- 1}. \end{array}
(1.1.18)

Using the matrix M from above, we can de­com­pose it into its eigen­vec­tors and eigen­val­ues by the de­fined ma­tri­ces V,D\ V,D, to ex­press the de­com­po­si­tion of MM, as

M=VDV−1=(−ii11)(−ix00ix)(i212−i212).\begin{array}{r} M = VDV^{- 1} = \begin{pmatrix} - i & i \\ 1 & 1 \end{pmatrix}\begin{pmatrix} - ix & 0 \\ 0 & ix \end{pmatrix}\begin{pmatrix} \frac{i}{2} & \frac{1}{2} \\ - \frac{i}{2} & \frac{1}{2} \end{pmatrix}. \end{array}
(1.1.19)

Using this de­com­po­si­tion, the matrix ex­po­nen­tial be­comes easier to cal­cu­late since ex­po­nen­ti­a­tion is then only re­quired for di­ag­o­nal matrix D. The matrix ex­po­nen­tial eMe^{M} can now be ex­pressed in terms of the de­com­po­si­tion

eM=VeDV−1=(−ii11)(e−ix00eix)(i212−i212)=(cos⁡(x)−sin⁡(x)sin⁡(x)cos⁡(x)).\begin{array}{r} e^{M} = Ve^{D}V^{- 1} = \begin{pmatrix} - i & i \\ 1 & 1 \end{pmatrix}\begin{pmatrix} e^{- ix} & 0 \\ 0 & e^{ix} \end{pmatrix}\begin{pmatrix} \frac{i}{2} & \frac{1}{2} \\ - \frac{i}{2} & \frac{1}{2} \end{pmatrix} = \begin{pmatrix} \cos(x) & - \sin(x) \\ \sin(x) & \cos(x) \end{pmatrix}. \end{array}
(1.1.20)

At this point, we can apply what we have learned to­wards a matrix group ex­am­ple. We start with a simple rep­re­sen­ta­tion of the abelian cyclic group Z3Z_{3}, where abelian de­fines that the matrix prod­ucts com­mute. The group M is de­fined as the set of three per­mu­ta­tions

m1=(100010001),m2=(010001100),m3=(001100010).\begin{array}{r} m_{1} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix},m_{2} = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix},m_{3} = \begin{pmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}. \end{array}
(1.1.21)

for which, mm serves as our iden­ti­ty per­mu­ta­tion, all el­e­ments are com­mu­ta­tive (abelian), in­vers­es exist for each el­e­ment, and the group is closed under mul­ti­pli­ca­tion. Ex­am­ples of the group prop­er­ties for MM are

Iden­ti­ty:

m1m3=m3m1=m3→(100010001)(001100010)=(001100010)(100010001)=(001100010).\begin{array}{r} m_{1}m_{3} = m_{3}m_{1} = m_{3} \rightarrow \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}. \end{array}
(1.1.22)

In­verse:

m1−1=(100010001),   m2−1=(001100010),   m3−1=(010001100):  ∀mj−1∈M.\begin{array}{r} m_{1}^{- 1}{= \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix},\ \ \ m}_{2}^{- 1} = \begin{pmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix},\ \ \ m_{3}^{- 1} = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix}:\ \ \forall m_{j}^{- 1} \in M. \end{array}
(1.1.23)

Clo­sure:

m2m3=(010001100)(001100010)=(100010001)=m1,m_{2}m_{3} = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix}\begin{pmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = m_{1},
m2m2=(010001100)(010001100)=(001100010)=m3,m_{2}m_{2} = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix}\begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix} = m_{3},
m3m3=(001100010)(001100010)=(010001100)=m2.\begin{array}{r} m_{3}m_{3} = \begin{pmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}\begin{pmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix} = m_{2.} \end{array}
(1.1.24)

Com­mu­ta­tive:

m2m3=m3m2=m1→(010001100)(001100010)=(001100010)(010001100)=(100010001).\begin{array}{r} m_{2}m_{3} = m_{3}m_{2} = m_{1} \rightarrow \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix}\begin{pmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}\begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}. \end{array}
(1.1.25)

We are con­sid­er­ing a cyclic group, where the prod­ucts of the per­mu­ta­tions m2,m3m_{2}{,m}_{3} cycle through all the el­e­ments of the group. The per­mu­ta­tions m2,m3m_{2}{,m}_{3} will also cycle by the linear trans­for­ma­tion of the el­e­ments of a vector when left mul­ti­plied. Taking a vector v=[y1,y2,y3]′v = \lbrack y_{1},y_{2},y_{3}\rbrack' , we then left mul­ti­ply by m2m_{2}, find­ing

m2v=(010001100)(y1y2y3)=(y2y3y1).\begin{array}{r} m_{2}v = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix}\begin{pmatrix} y_{1} \\ y_{2} \\ y_{3} \end{pmatrix} = \begin{pmatrix} y_{2} \\ y_{3} \\ y_{1} \end{pmatrix}. \end{array}
(1.1.26)

So, we see that for the prod­uct m2vm_{2}v, the el­e­ments in vv have cycled upward by the linear trans­for­ma­tion..Fur­ther, for each mj∈Mm_{j} \in M, we can define some yjmj=Yjy_{j}m_{j} = Y_{j}, for yj∈ℝy_{j}\in \mathbb{R}, to yield

Y1=(y1000y1000y1),   Y2=(0y2000y2y200),   Y3=(00y3y3000y30).\begin{array}{r} Y_{1} = \begin{pmatrix} y_{1} & 0 & 0 \\ 0 & y_{1} & 0 \\ 0 & 0 & y_{1} \end{pmatrix},\ \ \ Y_{2} = \begin{pmatrix} 0 & y_{2} & 0 \\ 0 & 0 & y_{2} \\ y_{2} & 0 & 0 \end{pmatrix},\ \ \ Y_{3} = \begin{pmatrix} 0 & 0 & y_{3} \\ y_{3} & 0 & 0 \\ 0 & y_{3} & 0 \end{pmatrix}. \end{array}
(1.1.27)

Each yjy_{j} value is an ar­gu­ment of the YjY_{j} ma­tri­ces above and each YjY_{j} can then be ex­po­nen­ti­at­ed to form a matrix group rep­re­sen­ta­tion. For the ex­am­ple eY2e^{Y_{2}}, Y2Y_{2} is di­ag­o­nal­iz­able, there­fore, it can be rep­re­sent­ed as

Y2=V2D2V2−1,\begin{array}{r} Y_{2} = V_{2}D_{2}V_{2}^{- 1}, \end{array}
(1.1.28)

for V2V_{2}, the eigen­vec­tor matrix of Y2,Y_{2},and D2D_{2}, the di­ag­o­nal matrix of eigen­val­ues of Y2Y_{2}. The sim­pli­fied ex­po­nen­tial map­ping eY2e^{Y_{2}}, then be­comes

eY2=V2eD2V2−1.\begin{array}{r} e^{Y_{2}} = V_{2}e^{D_{2}}V_{2}^{- 1}. \end{array}
(1.1.29)

Cal­cu­lat­ing V2V_{2} and D2D_{2}, gives us

D2=(y2000y2(−12−i32)000y2(−12+i32)),   V2=(1(−12−i32)(−12+i32)1(−12+i32)(−12−i32)111),\begin{array}{r} D_{2} = \begin{pmatrix} y_{2} & 0 & 0 \\ 0 & y_{2}\left( - \frac{1}{2} - \frac{i\sqrt{3}}{2} \right) & 0 \\ 0 & 0 & y_{2}\left( - \frac{1}{2} + \frac{i\sqrt{3}}{2} \right) \end{pmatrix},\ \ \ V_{2} = \begin{pmatrix} 1 & \left( - \frac{1}{2} - \frac{i\sqrt{3}}{2} \right) & \left( - \frac{1}{2} + \frac{i\sqrt{3}}{2} \right) \\ 1 & \left( - \frac{1}{2} + \frac{i\sqrt{3}}{2} \right) & \left( - \frac{1}{2} - \frac{i\sqrt{3}}{2} \right) \\ 1 & 1 & 1 \end{pmatrix}, \end{array}
(1.1.30)

and using these de­com­po­si­tion com­po­nents, the matrix ex­po­nen­tial eY2e^{Y_{2}} will have the matrix group rep­re­sen­ta­tion,

eY2=(ey23+e−y22(2cos⁡(3y22)3)ey23−e−y22(cos⁡(3y22)3−sin⁡(3y22)3)ey23−e−y22(cos⁡(3y22)3+sin⁡(3y22)3)ey23−e−y22(cos⁡(3y22)3+sin⁡(3y22)3)ey23+e−y22(2cos⁡(3y22)3)ey23−e−y22(cos⁡(3y22)3−sin⁡(3y22)3)ey23−e−y22(cos⁡(3y22)3−sin⁡(3y22)3)ey23−e−y22(cos⁡(3y22)3+sin⁡(3y22)3)ey23+e−y22(2cos⁡(3y22)3)).\begin{array}{r} e^{Y_{2}} = \begin{pmatrix} \frac{e^{y_{2}}}{3} + e^{- \frac{y_{2}}{2}}\left( \frac{2\cos\left( \sqrt{3}\frac{y_{2}}{2} \right)}{3} \right) & \frac{e^{y_{2}}}{3} - e^{- \frac{y_{2}}{2}}\left( \frac{\cos\left( \sqrt{3}\frac{y_{2}}{2} \right)}{3} - \frac{\sin\left( \sqrt{3}\frac{y_{2}}{2} \right)}{\sqrt{3}} \right) & \frac{e^{y_{2}}}{3} - e^{- \frac{y_{2}}{2}}\left( \frac{\cos\left( \sqrt{3}\frac{y_{2}}{2} \right)}{3} + \frac{\sin\left( \sqrt{3}\frac{y_{2}}{2} \right)}{\sqrt{3}} \right) \\ \frac{e^{y_{2}}}{3} - e^{- \frac{y_{2}}{2}}\left( \frac{\cos\left( \sqrt{3}\frac{y_{2}}{2} \right)}{3} + \frac{\sin\left( \sqrt{3}\frac{y_{2}}{2} \right)}{\sqrt{3}} \right) & \frac{e^{y_{2}}}{3} + e^{- \frac{y_{2}}{2}}\left( \frac{2\cos\left( \sqrt{3}\frac{y_{2}}{2} \right)}{3} \right) & \frac{e^{y_{2}}}{3} - e^{- \frac{y_{2}}{2}}\left( \frac{\cos\left( \sqrt{3}\frac{y_{2}}{2} \right)}{3} - \frac{\sin\left( \sqrt{3}\frac{y_{2}}{2} \right)}{\sqrt{3}} \right) \\ \frac{e^{y_{2}}}{3} - e^{- \frac{y_{2}}{2}}\left( \frac{\cos\left( \sqrt{3}\frac{y_{2}}{2} \right)}{3} - \frac{\sin\left( \sqrt{3}\frac{y_{2}}{2} \right)}{\sqrt{3}} \right) & \frac{e^{y_{2}}}{3} - e^{- \frac{y_{2}}{2}}\left( \frac{\cos\left( \sqrt{3}\frac{y_{2}}{2} \right)}{3} + \frac{\sin\left( \sqrt{3}\frac{y_{2}}{2} \right)}{\sqrt{3}} \right) & \frac{e^{y_{2}}}{3} + e^{- \frac{y_{2}}{2}}\left( \frac{2\cos\left( \sqrt{3}\frac{y_{2}}{2} \right)}{3} \right) \end{pmatrix}. \end{array}
(1.1.31)

In the matrix group above, we now have an in­fi­nite number of linear trans­for­ma­tions based on the abil­i­ty to set an in­fi­nite number of values for the ar­gu­ment y∈ℝy\in \mathbb{R}. The matrix group acts on v,v,as we showed pre­vi­ous­ly by left mul­ti­pli­ca­tion, and the trans­for­ma­tion yields a result that re­mains ∈ℝ3\in \mathbb{R}^{3}

T(v)=eY2v | T:ℝ3→ℝ3.\begin{array}{r} T(v) = e^{Y_{2}}v\ |\ T:\mathbb{R}^{3} \rightarrow \mathbb{R}^{3}. \end{array}
(1.1.32)