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Logarithm of a matrix

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Mathematical operation on invertible matrices

Inmathematics, alogarithm of a matrix is anothermatrix such that thematrix exponential of the latter matrix equals the original matrix. It is thus a generalization of the scalarlogarithm and in some sense aninverse function of thematrix exponential. Not all matrices have a logarithm and those matrices that do have a logarithm may have more than one logarithm. The study of logarithms of matrices leads toLie theory since when a matrix has a logarithm then it is in an element of aLie group and the logarithm is the corresponding element of the vector space of theLie algebra.

Definition

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Theexponential of a matrixA is defined by

eAn=0Ann!{\displaystyle e^{A}\equiv \sum _{n=0}^{\infty }{\frac {A^{n}}{n!}}}.

Given a matrixB, another matrixA is said to be amatrix logarithm ofB ifeA =B.

Because the exponential function is notbijective forcomplex numbers (e.g.eπi=e3πi=1{\displaystyle e^{\pi i}=e^{3\pi i}=-1}), numbers can have multiple complex logarithms, and as a consequence of this, some matrices may have more than one logarithm, as explained below. If the matrix logarithm ofB{\displaystyle B} exists and is unique, then it is written aslogB,{\displaystyle \log B,} in which caseelogB=B.{\displaystyle e^{\log B}=B.}

Power series expression

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IfB is sufficiently close to the identity matrix, then a logarithm ofB may be computed by means of thepower series

log(B)=log(I+(BI))=k=1(1)k+1k(BI)k=(BI)(BI)22+(BI)33{\displaystyle \log(B)=\log(I+(B-I))=\sum _{k=1}^{\infty }{\frac {(-1)^{k+1}}{k}}(B-I)^{k}=(B-I)-{\frac {(B-I)^{2}}{2}}+{\frac {(B-I)^{3}}{3}}-\cdots },

which can be rewritten as

log(B)=k=1(IB)kk=(IB)(IB)22(IB)33{\displaystyle \log(B)=-\sum _{k=1}^{\infty }{\frac {(I-B)^{k}}{k}}=-(I-B)-{\frac {(I-B)^{2}}{2}}-{\frac {(I-B)^{3}}{3}}-\cdots }.

Specifically, ifIB<1{\displaystyle \left\|I-B\right\|<1}, then the preceding series converges andelog(B)=B{\displaystyle e^{\log(B)}=B}.[1]

Example: Logarithm of rotations in the plane

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The rotations in the plane give a simple example. A rotation of angleα around the origin is represented by the 2×2-matrix

A=(cos(α)sin(α)sin(α)cos(α)).{\displaystyle A={\begin{pmatrix}\cos(\alpha )&-\sin(\alpha )\\\sin(\alpha )&\cos(\alpha )\\\end{pmatrix}}.}

For any integern, the matrix

Bn=(α+2πn)(0110),{\displaystyle B_{n}=(\alpha +2\pi n){\begin{pmatrix}0&-1\\1&0\\\end{pmatrix}},}

is a logarithm ofA.

Proof

log(A)=Bn {\displaystyle \log(A)=B_{n}~}  eBn=A{\displaystyle ~~e^{B_{n}}=A}

eBn=k=01k!Bnk {\displaystyle e^{B_{n}}=\sum _{k=0}^{\infty }{1 \over k!}B_{n}^{k}~} where

(Bn)0=1 I2,{\displaystyle (B_{n})^{0}=1~I_{2},}

(Bn)1=(α+2πn)(01+10),{\displaystyle (B_{n})^{1}=(\alpha +2\pi n){\begin{pmatrix}0&-1\\+1&0\\\end{pmatrix}},}

(Bn)2=(α+2πn)2(1001),{\displaystyle (B_{n})^{2}=(\alpha +2\pi n)^{2}{\begin{pmatrix}-1&0\\0&-1\\\end{pmatrix}},}

(Bn)3=(α+2πn)3(0+110),{\displaystyle (B_{n})^{3}=(\alpha +2\pi n)^{3}{\begin{pmatrix}0&+1\\-1&0\\\end{pmatrix}},}

(Bn)4=(α+2πn)4 I2{\displaystyle (B_{n})^{4}=(\alpha +2\pi n)^{4}~I_{2}}

...

k=01k!Bnk=(k=0(1)k2k!(α+2πn)2kk=0(1)k(2k+1)!(α+2πn)2k+1k=0(1)k(2k+1)!(α+2πn)2k+1k=0(1)k2k!(α+2πn)2k)=(cos(α)sin(α)sin(α)cos(α))=A .{\displaystyle \sum _{k=0}^{\infty }{1 \over k!}B_{n}^{k}={\begin{pmatrix}\sum _{k=0}^{\infty }{(-1)^{k} \over 2k!}(\alpha +2\pi n)^{2k}&-\sum _{k=0}^{\infty }{(-1)^{k} \over (2k+1)!}(\alpha +2\pi n)^{2k+1}\\\sum _{k=0}^{\infty }{(-1)^{k} \over (2k+1)!}(\alpha +2\pi n)^{2k+1}&\sum _{k=0}^{\infty }{(-1)^{k} \over 2k!}(\alpha +2\pi n)^{2k}\\\end{pmatrix}}={\begin{pmatrix}\cos(\alpha )&-\sin(\alpha )\\\sin(\alpha )&\cos(\alpha )\\\end{pmatrix}}=A~.}

qed.

Thus, the matrixA has infinitely many logarithms. This corresponds to the fact that the rotation angle is only determined up to multiples of 2π.

In the language of Lie theory, the rotation matricesA are elements of the Lie groupSO(2). The corresponding logarithmsB are elements of the Lie algebra so(2), which consists of allskew-symmetric matrices. The matrix

(0110){\displaystyle {\begin{pmatrix}0&1\\-1&0\\\end{pmatrix}}}

is a generator of theLie algebra so(2).

Existence

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The question of whether a matrix has a logarithm has the easiest answer when considered in the complex setting. A complex matrix has a logarithmif and only if it isinvertible.[2] The logarithm is not unique, but if a matrix has no negative realeigenvalues, then there is a unique logarithm that has eigenvalues all lying in the strip{zC | π<Im z<π}{\displaystyle \{z\in \mathbb {C} \ \vert \ -\pi <{\textit {Im}}\ z<\pi \}}. This logarithm is known as theprincipal logarithm.[3]

The answer is more involved in the real setting. A real matrix has a real logarithm if and only if it is invertible and eachJordan block belonging to a negative eigenvalue occurs an even number of times.[4] If an invertible real matrix does not satisfy the condition with the Jordan blocks, then it has only non-real logarithms. This can already be seen in the scalar case: no branch of the logarithm can be real at -1. The existence of real matrix logarithms of real 2×2 matrices is considered in a later section.

Properties

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IfA andB are bothpositive-definite matrices, then

trlog(AB)=trlog(A)+trlog(B).{\displaystyle \operatorname {tr} {\log {(AB)}}=\operatorname {tr} {\log {(A)}}+\operatorname {tr} {\log {(B)}}.}

Suppose thatA andB commute, meaning thatAB =BA. Then

log(AB)=log(A)+log(B){\displaystyle \log {(AB)}=\log {(A)}+\log {(B)}}

if and only ifarg(μj)+arg(νj)(π,π]{\displaystyle \operatorname {arg} (\mu _{j})+\operatorname {arg} (\nu _{j})\in (-\pi ,\pi ]}, whereμj{\displaystyle \mu _{j}} is aneigenvalue ofA{\displaystyle A} andνj{\displaystyle \nu _{j}} is the correspondingeigenvalue ofB{\displaystyle B}.[5] In particular,log(AB)=log(A)+log(B){\displaystyle \log(AB)=\log(A)+\log(B)} whenA andB commute and are bothpositive-definite. SettingB =A−1 in this equation yields

log(A1)=log(A).{\displaystyle \log {(A^{-1})}=-\log {(A)}.}

Similarly, for non-commutingA{\displaystyle A} andB{\displaystyle B}, one can show that[6]

log(A+tB)=log(A)+t0dz IA+zIBIA+zI+O(t2).{\displaystyle \log {(A+tB)}=\log {(A)}+t\int _{0}^{\infty }dz~{\frac {I}{A+zI}}B{\frac {I}{A+zI}}+O(t^{2}).}

More generally, a series expansion oflog(A+tB){\displaystyle \log {(A+tB)}} in powers oft{\displaystyle t} can be obtained using the integral definition of the logarithm

log(X+λI)log(X)=0λdzIX+zI,{\displaystyle \log {(X+\lambda I)}-\log {(X)}=\int _{0}^{\lambda }dz{\frac {I}{X+zI}},}

applied to bothX=A{\displaystyle X=A} andX=A+tB{\displaystyle X=A+tB} in the limitλ{\displaystyle \lambda \rightarrow \infty }.

Further example: Logarithm of rotations in 3D space

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A rotationR ∈ SO(3) inR{\displaystyle \mathbb {R} }3 is given by a 3×3orthogonal matrix.

The logarithm of such a rotation matrixR can be readily computed from the antisymmetric part ofRodrigues' rotation formula, explicitly inAxis angle. It yields the logarithm of minimalFrobenius norm, but fails whenR has eigenvalues equal to −1 where this is not unique.

Further note that, given rotation matricesA andB,

dg(A,B):=log(ATB)F{\displaystyle d_{g}(A,B):=\|\log(A^{\text{T}}B)\|_{F}}

is the geodesic distance on the 3D manifold of rotation matrices.

Calculating the logarithm of a diagonalizable matrix

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A method for finding logA for adiagonalizable matrixA is the following:

Find the matrixV ofeigenvectors ofA (each column ofV is an eigenvector ofA).
Find theinverseV−1 ofV.
Let
A=V1AV.{\displaystyle A'=V^{-1}AV.}
ThenA will be adiagonal matrix whose diagonal elements are eigenvalues ofA.
Replace each diagonal element ofA by its (natural) logarithm in order to obtainlogA{\displaystyle \log A'}.
Then
logA=V(logA)V1.{\displaystyle \log A=V(\log A')V^{-1}.}

That the logarithm ofA might be a complex matrix even ifA is real then follows from the fact that a matrix with real and positive entries might nevertheless have negative or even complex eigenvalues (this is true for example forrotation matrices). The non-uniqueness of the logarithm of a matrix follows from the non-uniqueness of the logarithm of a complex number.

Logarithm of a non-diagonalizable matrix

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

[1101].{\displaystyle {\begin{bmatrix}1&1\\0&1\end{bmatrix}}.}

For such matrices one needs to find itsJordan decomposition and, rather than computing the logarithm of diagonal entries as above, one would calculate the logarithm of theJordan blocks.

The latter is accomplished by noticing that one can write a Jordan block as

B=(λ10000λ10000λ100000λ100000λ)=λ(1λ100001λ100001λ1000001λ1000001)=λ(I+K){\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)}

whereK is a matrix with zeros on and under themain diagonal. (The number λ is nonzero by the assumption that the matrix whose logarithm one attempts to take is invertible.)

Then, by theMercator series

log(1+x)=xx22+x33x44+{\displaystyle \log(1+x)=x-{\frac {x^{2}}{2}}+{\frac {x^{3}}{3}}-{\frac {x^{4}}{4}}+\cdots }

one gets

logB=log(λ(I+K))=log(λI)+log(I+K)=(logλ)I+KK22+K33K44+{\displaystyle \log B=\log {\big (}\lambda (I+K){\big )}=\log(\lambda I)+\log(I+K)=(\log \lambda )I+K-{\frac {K^{2}}{2}}+{\frac {K^{3}}{3}}-{\frac {K^{4}}{4}}+\cdots }

Thisseries has a finite number of terms (Km is zero ifm is equal to or greater than the dimension ofK), and so its sum is well-defined.

Example. Using this approach, one finds

log[1101]=[0100],{\displaystyle \log {\begin{bmatrix}1&1\\0&1\end{bmatrix}}={\begin{bmatrix}0&1\\0&0\end{bmatrix}},}

which can be verified by plugging the right-hand side into the matrix exponential:exp[0100]=I+[0100]+12[0100]2=0+=[1101].{\displaystyle \exp {\begin{bmatrix}0&1\\0&0\end{bmatrix}}=I+{\begin{bmatrix}0&1\\0&0\end{bmatrix}}+{\frac {1}{2}}\underbrace {{\begin{bmatrix}0&1\\0&0\end{bmatrix}}^{2}} _{=0}+\cdots ={\begin{bmatrix}1&1\\0&1\end{bmatrix}}.}

Another, more standard, method by theHermite interpolation with the use ofconfluent Vandermonde matrix is presented by Higham.[7]

A functional analysis perspective

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A square matrix represents alinear operator on theEuclidean spaceRn wheren is the dimension of the matrix. Since such a space is finite-dimensional, this operator is actuallybounded.

Using the tools ofholomorphic functional calculus, given aholomorphic functionf defined on anopen set in thecomplex plane and a bounded linear operatorT, one can calculatef(T) as long asf is defined on thespectrum ofT.

The functionf(z) = logz can be defined on anysimply connected open set in the complex plane not containing the origin, and it is holomorphic on such a domain. This implies that one can define lnT as long as the spectrum ofT does not contain the origin and there is a path going from the origin to infinity not crossing the spectrum ofT (e.g., if the spectrum ofT is a circle with the origin inside of it, it is impossible to define lnT).

The spectrum of a linear operator onRn is the set of eigenvalues of its matrix, and so is a finite set. As long as the origin is not in the spectrum (the matrix is invertible), the path condition from the previous paragraph is satisfied, and lnT is well-defined. The non-uniqueness of the matrix logarithm follows from the fact that one can choose more than one branch of the logarithm which is defined on the set of eigenvalues of a matrix.

A Lie group theory perspective

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In the theory ofLie groups, there is anexponential map from aLie algebrag{\displaystyle {\mathfrak {g}}} to the corresponding Lie groupG

exp:gG.{\displaystyle \exp :{\mathfrak {g}}\rightarrow G.}

For matrix Lie groups, the elements ofg{\displaystyle {\mathfrak {g}}} andG are square matrices and the exponential map is given by thematrix exponential. The inverse maplog=exp1{\displaystyle \log =\exp ^{-1}} is multivalued and coincides with the matrix logarithm discussed here. The logarithm maps from the Lie groupG into the Lie algebrag{\displaystyle {\mathfrak {g}}}. Note that the exponential map is alocal diffeomorphism between a neighborhoodU of the zero matrix0_g{\displaystyle {\underline {0}}\in {\mathfrak {g}}} and a neighborhoodV of the identity matrix1_G{\displaystyle {\underline {1}}\in G}.[8]Thus the (matrix) logarithm is well-defined as a map,

log:GVUg.{\displaystyle \log :G\supset V\rightarrow U\subset {\mathfrak {g}}.}

An important corollary ofJacobi's formula then is

log(det(A))=tr(logA) .{\displaystyle \log(\det(A))=\mathrm {tr} (\log A)~.}

Constraints in the 2 × 2 case

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If a 2 × 2 real matrix has a negativedeterminant, it has no real logarithm. Note first that any 2 × 2 real matrix can be considered one of the three types of the complex numberz =x +yε, where ε2 ∈ { −1, 0, +1 }. Thisz is a point on a complex subplane of thering of matrices.[9]

The case where the determinant is negative only arises in a plane with ε2 =+1, that is asplit-complex number plane. Only one quarter of this plane is the image of the exponential map, so the logarithm is only defined on that quarter (quadrant). The other three quadrants are images of this one under theKlein four-group generated by ε and −1.

For example, leta = log 2 ; then cosha = 5/4 and sinha = 3/4.For matrices, this means that

A=exp(0aa0)=(coshasinhasinhacosha)=(1.250.750.751.25){\displaystyle A=\exp {\begin{pmatrix}0&a\\a&0\end{pmatrix}}={\begin{pmatrix}\cosh a&\sinh a\\\sinh a&\cosh a\end{pmatrix}}={\begin{pmatrix}1.25&0.75\\0.75&1.25\end{pmatrix}}}.

So this last matrix has logarithm

logA=(0log2log20){\displaystyle \log A={\begin{pmatrix}0&\log 2\\\log 2&0\end{pmatrix}}}.

These matrices, however, do not have a logarithm:

(3/45/45/43/4), (3/45/45/43/4), (5/43/43/45/4){\displaystyle {\begin{pmatrix}3/4&5/4\\5/4&3/4\end{pmatrix}},\ {\begin{pmatrix}-3/4&-5/4\\-5/4&-3/4\end{pmatrix}},\ {\begin{pmatrix}-5/4&-3/4\\-3/4&-5/4\end{pmatrix}}}.

They represent the three other conjugates by the four-group of the matrix above that does have a logarithm.

A non-singular 2 × 2 matrix does not necessarily have a logarithm, but it is conjugate by the four-group to a matrix that does have a logarithm.

It also follows, that, e.g., asquare root of this matrixA is obtainable directly from exponentiating (logA)/2,

A=(cosh((log2)/2)sinh((log2)/2)sinh((log2)/2)cosh((log2)/2))=(1.060.350.351.06) .{\displaystyle {\sqrt {A}}={\begin{pmatrix}\cosh((\log 2)/2)&\sinh((\log 2)/2)\\\sinh((\log 2)/2)&\cosh((\log 2)/2)\end{pmatrix}}={\begin{pmatrix}1.06&0.35\\0.35&1.06\end{pmatrix}}~.}

For a richer example, start with aPythagorean triple (p,q,r)and leta = log(p +r) − logq. Then

ea=p+rq=cosha+sinha{\displaystyle e^{a}={\frac {p+r}{q}}=\cosh a+\sinh a}.

Now

exp(0aa0)=(r/qp/qp/qr/q){\displaystyle \exp {\begin{pmatrix}0&a\\a&0\end{pmatrix}}={\begin{pmatrix}r/q&p/q\\p/q&r/q\end{pmatrix}}}.

Thus

1q(rppr){\displaystyle {\tfrac {1}{q}}{\begin{pmatrix}r&p\\p&r\end{pmatrix}}}

has the logarithm matrix

(0aa0){\displaystyle {\begin{pmatrix}0&a\\a&0\end{pmatrix}}} ,

wherea = log(p +r) − logq.

See also

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Notes

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  1. ^Hall 2015 Theorem 2.8
  2. ^Higham (2008), Theorem 1.27
  3. ^Higham (2008), Theorem 1.31
  4. ^Culver (1966)
  5. ^APRAHAMIAN, MARY; HIGHAM, NICHOLAS J. (2014)."The Matrix Unwinding Function, with an Application to Computing the Matrix Exponential".SIAM Journal on Matrix Analysis and Applications.35 (1): 97.doi:10.1137/130920137.
  6. ^Unpublished memo by S Adler (IAS)
  7. ^Higham 2008, pp. 4–7, Section 1.2.2.
  8. ^Hall 2015 Theorem 3.42
  9. ^Abstract Algebra/2x2 real matrices at Wikibooks

References

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