Masteries · Jason GlowneyArithmoí Book 1

Masteries

1.5.3 The Matrix Logarithm

We know that the matrix ex­po­nen­tial of a matrix MM is well-de­fined and ab­solute­ly con­ver­gent

exp⁡(M)=∑n=0∞Mnn!,\begin{array}{r} \exp{(M) = \sum_{n = 0}^{\infty}\frac{M^{n}}{n!}}, \end{array}
(1.5.18)

for which, the matrix log­a­rithm rep­re­sents the in­verse func­tion for the ex­po­nen­tial map­ping exp−1⁡(M)\exp^{- 1}(M). The matrix log­a­rithm, like the clas­si­cal log, is de­fined by a power series that only con­verges in a cer­tain neigh­bor­hood of the iden­ti­ty. To start, it is im­por­tant to first prove that (1+u)−1(1 + u)^{- 1} equals the fol­low­ing in­fi­nite series

11+u=1−u+u2−u3+u4−….\begin{array}{r} \frac{1}{1 + u} = 1 - u + u^{2} - u^{3} + u^{4} - \ldots. \end{array}
(1.5.19)

Set­ting SS equal to the above in­fi­nite series

S=1−u+u2−u3+u4−…,\begin{array}{r} S = 1 - u + u^{2} - u^{3} + u^{4} - \ldots, \end{array}
(1.5.20)

we sub­tract one from each side then divide both sides by uu

S−1u=−1+u−u2+u3−u4+…,\begin{array}{r} \frac{S - 1}{u} = - 1 + u - u^{2} + u^{3} - u^{4} + \ldots, \end{array}
(1.5.21)

which re­sults in a re­la­tion­ship in­volv­ing SS and uu, equal to the ini­tial in­fi­nite series

−(S−1u)=S.\begin{array}{r} - \left( \frac{S - 1}{u} \right) = S. \end{array}
(1.5.22)

We divide both sides of the above by SS, then mul­ti­ply­ing through by uu

−(S−1Su)=1→−(S−1S )=u,\begin{array}{r} - \left( \frac{S - 1}{Su} \right) = 1 \rightarrow - \left( \frac{S - 1}{S}\ \right) = u, \end{array}
(1.5.23)

which can be sim­pli­fied and solved for SS, to com­plete the proof

1S−1=u→1S=u+1→S=11+u.\begin{array}{r} \frac{1}{S} - 1 = u \rightarrow \frac{1}{S} = u + 1 \rightarrow S = \frac{1}{1 + u}. \end{array}
(1.5.24)
□\square

It turns out that we can de­scribe the log­a­rithm of xx as a re­la­tion­ship of this in­fi­nite series. For |x−1|≤1|x - 1| \leq 1, and x≠0x \neq 0, the log­a­rithm can be ex­pressed as the in­te­gral

ln⁡(x)=∫1x1t dt,\begin{array}{r} \ln(x) = \int_{1}^{x}{\frac{1}{t}\ dt}, \end{array}
(1.5.25)

where chang­ing the bounds of in­te­gra­tion and making a uu sub­sti­tu­tion, the in­te­gral can be ex­pressed, as

∫0x−111+u du,\begin{array}{r} \int_{0}^{x - 1}{\frac{1}{1 + u}\ du}, \end{array}
(1.5.26)

whose in­te­grand equals the in­fi­nite series of the proof

∫0x−111+u du=∫0x−1(1−u+u2−u3+u4−…) du.\begin{array}{r} \int_{0}^{x - 1}{\frac{1}{1 + u}\ du = \int_{0}^{x - 1}{\left( 1 - u + u^{2} - u^{3} + u^{4} - \ldots \right)\ du}}. \end{array}
(1.5.27)

Solv­ing this in­te­gral is straight­for­ward since it is in the form of a poly­no­mi­al, which can be ex­pressed as the ex­pand­ed in­fi­nite sum

=(x−1)−(x−1)22+(x−1)33−(x−1)44+…=∑k=1∞(−1)k−1(x−1)kk,\begin{array}{r} = (x - 1) - \frac{(x - 1)^{2}}{2} + \frac{(x - 1)^{3}}{3} - \frac{(x - 1)^{4}}{4} + \ldots = \sum_{k = 1}^{\infty}\frac{( - 1)^{k - 1}(x - 1)^{k}}{k}, \end{array}
(1.5.28)

which is equiv­a­lent to the Taylor series for ln⁡(x)\ln(x), around the iden­ti­ty. A change of vari­ables re­sults in the Mer­ca­tor series

ln⁡(1+x)=∑k=1∞(−1)k−1(x)kk=x−x22+x33−x44+…,\begin{array}{r} \ln(1 + x) = \sum_{k = 1}^{\infty}{\frac{( - 1)^{k - 1}(x)^{k}}{k} =}x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \frac{x^{4}}{4} + \ldots, \end{array}
(1.5.29)

valid for |x|≤1|x| \leq 1, and x≠−1x \neq - 1. This clas­si­cal in­fi­nite series can be ex­tend­ed to ma­tri­ces, with which we can define the log­a­rithm of a square matrix, as

ln⁡(ϵ+A)=A−A22+A33−A44+…,\begin{array}{r} \ln{(\epsilon + A) = A - \frac{A^{2}}{2} + \frac{A^{3}}{3} - \frac{A^{4}}{4} + \ldots}, \end{array}
(1.5.30)

for ϵ\epsilon, the iden­ti­ty matrix. The func­tion ln⁡(ϵ+A)\ln(\epsilon + A), is well-de­fined and con­tin­u­ous in the neigh­bor­hood of the iden­ti­ty ϵ\epsilon, and ab­solute­ly con­ver­gent for |A|<1|A| < 1.

The­o­rem (1.5.31) – The in­verse prop­er­ty of the matrix log­a­rithm will hold for any matrix eXe^{X}, within the unit dis­tance of the iden­ti­ty, as de­scribed by Still­well 1

ln⁡(eX)=X.\begin{array}{r} \ln\left( e^{X} \right) = X. \end{array}

Proof - As eX=ϵ+X1!+X22!+X33!+X44!+…,e^{X} = \epsilon + \frac{X}{1!} + \frac{X^{2}}{2!} + \frac{X^{3}}{3!} + \frac{X^{4}}{4!} + \ldots, and |eX−1|<1\left| e^{X} - 1 \right| < 1, we have

ln⁡(eX)=ln⁡(ϵ+(X1!+X22!+X33!+X44!+…))\begin{array}{r} \ln\left( e^{X} \right) = \ln\left( \epsilon + \left( \frac{X}{1!} + \frac{X^{2}}{2!} + \frac{X^{3}}{3!} + \frac{X^{4}}{4!} + \ldots \right) \right) \end{array}
=(X1!+X22!+…)−12(X1!+X22!+…)2+13(X1!+X22!+…)3−…,\begin{array}{r} = \left( \frac{X}{1!} + \frac{X^{2}}{2!} + \ldots \right) - \frac{1}{2}\left( \frac{X}{1!} + \frac{X^{2}}{2!} + \ldots \right)^{2} + \frac{1}{3}\left( \frac{X}{1!} + \frac{X^{2}}{2!} + \ldots \right)^{3} - \ldots, \end{array}

by way of the matrix log­a­rithm de­f­i­n­i­tion and the Mer­ca­tor series. As the series is ab­solute­ly con­ver­gent, the terms can be arranged by powers of XmX^{m}

ln⁡(eX)=X+(12!−12)X2+(13!−12+13)X3+….\begin{array}{r} \ln\left( e^{X} \right) = X + \left( \frac{1}{2!} - \frac{1}{2} \right)X^{2} + \left( \frac{1}{3!} - \frac{1}{2} + \frac{1}{3} \right)X^{3} + \ldots. \end{array}

The higher-pow­ered co­ef­fi­cients of XmX^{m} are zero

(12!−12)=0, \begin{array}{r} \left( \frac{1}{2!} - \frac{1}{2} \right) = 0,\ \end{array}

and we can say with cer­tain­ty that all the en­su­ing co­ef­fi­cients in the series will sum to zero since the same terms occur in the ex­pres­sion ln⁡(ex)\ln\left( e^{x} \right), for |ex−1|<1,\left| e^{x} - 1 \right| < 1, with sums equal to zero since ln⁡(ex)=x\ln\left( e^{x} \right) = x, under these con­di­tions. ∴ln⁡(eX)=X.\therefore\ln\left( e^{X} \right) = X.

□\square
  1. Still­well, John, "Naive Lie Theory", Springer 2008