Matrix logarithm

In mathematics , the logarithm of a matrix is a generalization of the scalar logarithm on matrices . In a sense, it is an inverse function of the matrix exponential .

definition

A matrix is a logarithm of a given matrix if the matrix exponential of is: ${\displaystyle B}$ ${\displaystyle A}$ ${\displaystyle A}$ ${\displaystyle B}$

${\displaystyle e^{B}=A.\,}$

properties

A matrix has a logarithm if and only if it can be inverted . This logarithm can be a non-real matrix even if all entries in the matrix are real numbers. In this case the logarithm is not unique.

Calculation of the logarithm of a diagonalizable matrix

In the following a method is described to calculate for a diagonalizable matrix : ${\displaystyle \ln(A)}$ ${\displaystyle A}$

Find the matrix of eigenvectors of (each column of is an eigenvector of ). ${\displaystyle V}$ ${\displaystyle A}$ ${\displaystyle V}$ ${\displaystyle A}$

Calculate the inverse of . ${\displaystyle V^{-1}}$ ${\displaystyle V}$

Be

${\displaystyle A'=V^{-1}AV}$.

Then is a diagonal matrix whose diagonal elements are eigenvalues ​​of . ${\displaystyle A'}$ ${\displaystyle A}$

Replace each diagonal element of with its natural logarithm to get. Then applies . ${\displaystyle A'}$ ${\displaystyle \ln(A')}$ ${\displaystyle \ln(A)=V(\ln A')V^{-1}}$

The fact that the logarithm of can be complex, although it is real, results from the fact that a real matrix can have complex eigenvalues ​​(this applies, for example, to rotation matrices ). The ambiguity of the logarithm follows from the ambiguity of the logarithm of a complex number. ${\displaystyle A}$ ${\displaystyle A}$

Example: . How one now calculates is not clearly defined, since the natural logarithm has the branch intersection at −1. If one approaches the number with a positive imaginary part, then is ; if one approaches the number with the negative imaginary part, one obtains . Here you can see the ambiguity of the logarithm and also the not necessarily real-valued entries, although the matrix was real-valued. ${\displaystyle A={\begin{pmatrix}0&1\\1&0\end{pmatrix}},V={\frac {1}{\sqrt {2}}}{\begin{pmatrix}-1&1\\1&1\end{pmatrix}}=V^{-1},A'={\begin{pmatrix}-1&0\\0&1\end{pmatrix}}}$ ${\displaystyle \ln A'}$ ${\displaystyle -1=\mathrm {e} ^{\mathrm {i} \pi }}$ ${\displaystyle \ln(-1)=\lim \limits _{\epsilon \to 0}\ln(\mathrm {e} ^{\mathrm {i} (\pi +\epsilon )})=\mathrm {i} \pi }$ ${\displaystyle -1=\mathrm {e} ^{-\mathrm {i} \pi }}$ ${\displaystyle \ln(-1)=\lim \limits _{\epsilon \to 0}\ln(\mathrm {e} ^{-\mathrm {i} (\pi -\epsilon )})=-\mathrm {i} \pi }$

The logarithm of a non-diagonalizable matrix

The above algorithm does not work for non-diagonalizable matrices such as

${\displaystyle {\begin{pmatrix}1&1\\0&1\end{pmatrix}}.}$

For such matrices one must first determine the Jordan normal form . Instead of the logarithm of the diagonal entries, you have to calculate the logarithm of the Jordan blocks.

The latter is achieved by writing the Jordan matrix as

${\displaystyle B={\begin{pmatrix}\lambda &1&0&0&\cdots &0\\0&\lambda &1&0&\cdots &0\\0&0&\lambda &1&\cdots &0\\\vdots &\vdots &\vdots &\ddots &\ddots &\vdots \\0&0&0&0&\lambda &1\\0&0&0&0&0&\lambda \\\end{pmatrix}}=\lambda {\begin{pmatrix}1&\lambda ^{-1}&0&0&\cdots &0\\0&1&\lambda ^{-1}&0&\cdots &0\\0&0&1&\lambda ^{-1}&\cdots &0\\\vdots &\vdots &\vdots &\ddots &\ddots &\vdots \\0&0&0&0&1&\lambda ^{-1}\\0&0&0&0&0&1\\\end{pmatrix}}=\lambda (I+K)}$

where K is a matrix with zeros below and on the main diagonal . (The number λ is not equal to zero if one assumes that the matrix whose logarithm one wants to calculate is invertible.)

Through the formula

${\displaystyle \ln(1+x)=x-{\frac {x^{2}}{2}}+{\frac {x^{3}}{3}}-{\frac {x^{4}}{4}}+\cdots }$

you get

${\displaystyle \ln B=\ln {\big (}\lambda (I+K){\big )}=\ln(\lambda I)+\ln(I+K)=(\ln \lambda )I+K-{\frac {K^{2}}{2}}+{\frac {K^{3}}{3}}-{\frac {K^{4}}{4}}+\cdots }$

This series does not converge for a general matrix as it would for real numbers with magnitude smaller . This particular matrix, however, is a nilpotent matrix , so the series has a finite number of terms ( zero when denoting the rank of ). ${\displaystyle K}$ ${\displaystyle 1}$ ${\displaystyle K}$ ${\displaystyle K^{m}}$ ${\displaystyle m-1}$ ${\displaystyle K}$

This approach gives you

${\displaystyle \ln {\begin{pmatrix}1&1\\0&1\end{pmatrix}}={\begin{pmatrix}0&1\\0&0\end{pmatrix}}.}$

From the perspective of functional analysis

A square matrix represents a linear operator in Euclidean space . Since this space is finite-dimensional, every operator is bounded . ${\displaystyle \mathbb {R} ^{n}}$

Let be a holomorphic function on an open set in the complex plane and be a bounded operator. One can calculate if is defined on the spectrum of . ${\displaystyle f}$ ${\displaystyle T}$ ${\displaystyle f(T)}$ ${\displaystyle f(z)}$ ${\displaystyle T}$

The function can be defined on any simply connected open set in the complex plane that does not contain zero and is holomorphic on this definition set . It follows that if the spectrum does not contain zero and there is a path from zero to infinity that does not intersect the spectrum of (for example, if the spectrum of forms a circular line whose center is zero, then it is not possible to define). ${\displaystyle f(z)=\ln(z)}$ ${\displaystyle \ln(T)}$ ${\displaystyle T}$ ${\displaystyle T}$ ${\displaystyle T}$ ${\displaystyle \ln(T)}$

For the special case of Euclidean space, the spectrum of a linear operator is the set of eigenvalues ​​of the matrix, i.e. finite. As long as zero is not contained in the spectrum (i.e. the matrix is ​​invertible) and the above path condition is obviously fulfilled, it follows that it is well-defined. The ambiguity follows from the fact that one can choose more than one branch of the logarithm which is defined on the set of eigenvalues ​​of the matrix. ${\displaystyle \ln(T)}$

See also

• Cepstrum

Individual evidence

1. ^ Branch Cut Wolfram Research, accessed September 19, 2018.