The standard branch of the matrix logarithm

Because of Euler’s identity ${e^{\pi i} + 1 = 0}$ , the complex exponential is not injective: ${e^{z + 2\pi i k} = e^z}$ for any complex {z} and integer {k} . As such, the complex logarithm ${z \mapsto \log z}$ is not well-defined as a single-valued function from ${{\bf C} \backslash \{0\}}$ to ${{\bf C}}$ . However, after making a branch cut, one can create a branch of the logarithm which is single-valued. For instance, after removing the negative real axis ${(-\infty,0]}$ , one has the standard branch ${\hbox{Log}: {\bf C} \backslash (-\infty,0] \rightarrow \{ z \in {\bf C}: |\hbox{Im} z| < \pi \}}$ of the logarithm, with ${\hbox{Log}(z)}$ defined as the unique choice of the complex logarithm of {z} whose imaginary part has magnitude strictly less than ${\pi}$ . This particular branch has a number of useful additional properties:

The standard branch ${\hbox{Log}}$ is holomorphic on its domain ${{\bf C} \backslash (-\infty,0]}$ .
One has ${\hbox{Log}( \overline{z} ) = \overline{ \hbox{Log}(z) }}$ for all in the domain ${{\bf C} \backslash (-\infty,0]}$ . In particular, if ${z \in {\bf C} \backslash (-\infty,0]}$ is real, then ${\hbox{Log} z}$ is real.
One has ${\hbox{Log}( z^{-1} ) = - \hbox{Log}(z)}$ for all in the domain ${{\bf C} \backslash (-\infty,0]}$ .

One can then also use the standard branch of the logarithm to create standard branches of other multi-valued functions, for instance creating a standard branch ${z \mapsto \exp( \frac{1}{2} \hbox{Log} z )}$ of the square root function. We caution however that the identity ${\hbox{Log}(zw) = \hbox{Log}(z) + \hbox{Log}(w)}$ can fail for the standard branch (or indeed for any branch of the logarithm).

One can extend this standard branch of the logarithm to ${n \times n}$ complex matrices, or (equivalently) to linear transformations ${T: V \rightarrow V}$ on an {n} -dimensional complex vector space {V} , provided that the spectrum of that matrix or transformation avoids the branch cut ${(-\infty,0]}$ . Indeed, from the spectral theorem one can decompose any such ${T: V \rightarrow V}$ as the direct sum of operators ${T_\lambda: V_\lambda \rightarrow V_\lambda}$ on the non-trivial generalised eigenspaces ${V_\lambda}$ of {T} , where ${\lambda \in {\bf C} \backslash (-\infty,0]}$ ranges in the spectrum of {T} . For each component ${T_\lambda}$ of {T} , we define

$\displaystyle \hbox{Log}( T_\lambda ) = P_\lambda( T_\lambda )$

where ${P_\lambda}$ is the Taylor expansion of ${\hbox{Log}}$ at ${\lambda}$ ; as ${T_\lambda-\lambda}$ is nilpotent, only finitely many terms in this Taylor expansion are required. The logarithm ${\hbox{Log} T}$ is then defined as the direct sum of the ${\hbox{Log} T_\lambda}$ .

The matrix standard branch of the logarithm has many pleasant and easily verified properties (often inherited from their scalar counterparts), whenever ${T: V \rightarrow V}$ has no spectrum in ${(-\infty,0]}$ :

(i) We have ${\exp( \hbox{Log} T ) = T}$ .
(ii) If ${T_1: V_1 \rightarrow V_1}$ and ${T_2: V_2 \rightarrow V_2}$ have no spectrum in ${(-\infty,0]}$ , then ${\hbox{Log}( T_1 \oplus T_2 ) = \hbox{Log}(T_1) \oplus \hbox{Log}(T_2)}$ .
(iii) If has spectrum in a closed disk in ${{\bf C} \backslash (-\infty,0]}$ , then ${\hbox{Log}(T) = P_z(T)}$ , where ${P_z}$ is the Taylor series of ${\hbox{Log}}$ around (which is absolutely convergent in ).
(iv) ${\hbox{Log}(T)}$ depends holomorphically on . (Easily established from (ii), (iii), after covering the spectrum of by disjoint disks; alternatively, one can use the Cauchy integral representation ${\hbox{Log}(T) = \frac{1}{2\pi i} \int_\gamma \hbox{Log}(z)(z-T)^{-1}\ dz}$ for a contour ${\gamma}$ in the domain enclosing the spectrum of .) In particular, the standard branch of the matrix logarithm is smooth.
(v) If ${S: V \rightarrow W}$ is any invertible linear or antilinear map, then ${\hbox{Log}( STS^{-1} ) = S \hbox{Log}(T) S^{-1}}$ . In particular, the standard branch of the logarithm commutes with matrix conjugations; and if is real with respect to a complex conjugation operation on (that is to say, an antilinear involution), then ${\hbox{Log}(T)}$ is real also.
(vi) If ${T^*: V^* \rightarrow V^*}$ denotes the transpose of (with ${V^*}$ the complex dual of ), then ${\hbox{Log}(T^*) = \hbox{Log}(T)^*}$ . Similarly, if ${T^\dagger: V^\dagger \rightarrow V^\dagger}$ denotes the adjoint of (with ${V^\dagger}$ the complex conjugate of ${V^*}$ , i.e. ${V^*}$ with the conjugated multiplication map ${(c,z) \mapsto \overline{c} z}$ ), then ${\hbox{Log}(T^\dagger) = \hbox{Log}(T)^\dagger}$ .
(vii) One has ${\hbox{Log}(T^{-1}) = - \hbox{Log}( T )}$ .
(viii) If ${\sigma(T)}$ denotes the spectrum of , then ${\sigma(\hbox{Log} T) = \hbox{Log} \sigma(T)}$ .

As a quick application of the standard branch of the matrix logarithm, we have

Proposition 1 Let be one of the following matrix groups: ${GL_n({\bf C})}$ , ${GL_n({\bf R})}$ , ${U_n({\bf C})}$ , , ${Sp_{2n}({\bf C})}$ , or ${Sp_{2n}({\bf R})}$ , where ${Q: {\bf R}^n \rightarrow {\bf R}}$ is a non-degenerate real quadratic form (so is isomorphic to a (possibly indefinite) orthogonal group for some ${0 \leq k \leq n}$ . Then any element of whose spectrum avoids ${(-\infty,0]}$ is exponential, that is to say ${T = \exp(X)}$ for some in the Lie algebra ${{\mathfrak g}}$ of .

Proof: We just prove this for {G=O(Q)} , as the other cases are similar (or a bit simpler). If ${T \in O(Q)}$ , then (viewing {T} as a complex-linear map on ${{\bf C}^n}$ , and using the complex bilinear form associated to {Q} to identify ${{\bf C}^n}$ with its complex dual ${({\bf C}^n)^*}$ , then {T} is real and ${T^{*-1} = T}$ . By the properties (v), (vi), (vii) of the standard branch of the matrix logarithm, we conclude that ${\hbox{Log} T}$ is real and ${- \hbox{Log}(T)^* = \hbox{Log}(T)}$ , and so ${\hbox{Log}(T)}$ lies in the Lie algebra ${{\mathfrak g} = {\mathfrak o}(Q)}$ , and the claim now follows from (i). $\Box$

Exercise 2 Show that ${\hbox{diag}(-\lambda, -1/\lambda)}$ is not exponential in ${GL_2({\bf R})}$ if ${-\lambda \in (-\infty,0) \backslash \{-1\}}$ . Thus we see that the branch cut in the above proposition is largely necessary. See this paper of Djokovic for a more complete description of the image of the exponential map in classical groups, as well as this previous blog post for some more discussion of the surjectivity (or lack thereof) of the exponential map in Lie groups.

For a slightly less quick application of the standard branch, we have the following result (recently worked out in the answers to this MathOverflow question):

Proposition 3 Let be an element of the split orthogonal group which lies in the connected component of the identity. Then ${\hbox{det}(1+T) \geq 0}$ .

The requirement that {T} lie in the identity component is necessary, as the counterexample ${T = \hbox{diag}(-\lambda, -1/\lambda )}$ for ${\lambda \in (-\infty,-1) \cup (-1,0)}$ shows.

Proof: We think of {T} as a (real) linear transformation on ${{\bf C}^{2n}}$ , and write {Q} for the quadratic form associated to {O(n,n)} , so that ${O(n,n) \equiv O(Q)}$ . We can split ${{\bf C}^{2n} = V_1 \oplus V_2}$ , where ${V_1}$ is the sum of all the generalised eigenspaces corresponding to eigenvalues in ${(-\infty,0]}$ , and ${V_2}$ is the sum of all the remaining eigenspaces. Since {T} and ${(-\infty,0]}$ are real, ${V_1,V_2}$ are real (i.e. complex-conjugation invariant) also. For {i=1,2} , the restriction ${T_i: V_i \rightarrow V_i}$ of {T} to ${V_i}$ then lies in ${O(Q_i)}$ , where ${Q_i}$ is the restriction of {Q} to ${V_i}$ , and

$\displaystyle \hbox{det}(1+T) = \hbox{det}(1+T_1) \hbox{det}(1+T_2).$

The spectrum of ${T_2}$ consists of positive reals, as well as complex pairs ${\lambda, \overline{\lambda}}$ (with equal multiplicity), so ${\hbox{det}(1+T_2) > 0}$ . From the preceding proposition we have ${T_2 = \exp( X_2 )}$ for some ${X_2 \in {\mathfrak o}(Q_2)}$ ; this will be important later.

It remains to show that ${\hbox{det}(1+T_1) \geq 0}$ . If ${T_1}$ has spectrum at {-1} then we are done, so we may assume that ${T_1}$ has spectrum only at ${(-\infty,-1) \cup (-1,0)}$ (being invertible, {T} has no spectrum at {0} ). We split ${V_1 = V_3 \oplus V_4}$ , where ${V_3,V_4}$ correspond to the portions of the spectrum in ${(-\infty,-1)}$ , {(-1,0)} ; these are real, {T} -invariant spaces. We observe that if ${V_\lambda, V_\mu}$ are generalised eigenspaces of {T} with ${\lambda \mu \neq 1}$ , then ${V_\lambda, V_\mu}$ are orthogonal with respect to the (complex-bilinear) inner product ${\cdot}$ associated with {Q} ; this is easiest to see first for the actual eigenspaces (since ${ \lambda \mu u \cdot v = Tu \cdot Tv = u \cdot v}$ for all ${u \in V_\lambda, v \in V_\mu}$ ), and the extension to generalised eigenvectors then follows from a routine induction. From this we see that ${V_1}$ is orthogonal to ${V_2}$ , and ${V_3}$ and ${V_4}$ are null spaces, which by the non-degeneracy of {Q} (and hence of the restriction ${Q_1}$ of {Q} to ${V_1}$ ) forces ${V_3}$ to have the same dimension as ${V_4}$ , indeed {Q} now gives an identification of ${V_3^*}$ with ${V_4}$ . If we let ${T_3, T_4}$ be the restrictions of {T} to ${V_3,V_4}$ , we thus identify ${T_4}$ with ${T_3^{*-1}}$ , since {T} lies in {O(Q)} ; in particular ${T_3}$ is invertible. Thus

$\displaystyle \hbox{det}(1+T_1) = \hbox{det}(1 + T_3) \hbox{det}( 1 + T_3^{*-1} ) = \hbox{det}(T_3)^{-1} \hbox{det}(1+T_3)^2$

and so it suffices to show that ${\hbox{det}(T_3) > 0}$ .

At this point we need to use the hypothesis that {T} lies in the identity component of {O(n,n)} . This implies (by a continuity argument) that the restriction of {T} to any maximal-dimensional positive subspace has positive determinant (since such a restriction cannot be singular, as this would mean that {T} positive norm vector would map to a non-positive norm vector). Now, as ${V_3,V_4}$ have equal dimension, ${Q_1}$ has a balanced signature, so ${Q_2}$ does also. Since ${T_2 = \exp(X_2)}$ , ${T_2}$ already lies in the identity component of ${O(Q_2)}$ , and so has positive determinant on any maximal-dimensional positive subspace of ${V_2}$ . We conclude that ${T_1}$ has positive determinant on any maximal-dimensional positive subspace of ${V_1}$ .

We choose a complex basis of ${V_3}$ , to identify ${V_3}$ with ${V_3^*}$ , which has already been identified with ${V_4}$ . (In coordinates, ${V_3,V_4}$ are now both of the form ${{\bf C}^m}$ , and ${Q( v \oplus w ) = v \cdot w}$ for ${v,w \in {\bf C}^m}$ .) Then ${\{ v \oplus v: v \in V_3 \}}$ becomes a maximal positive subspace of ${V_1}$ , and the restriction of ${T_1}$ to this subspace is conjugate to ${T_3 + T_3^{*-1}}$ , so that

$\displaystyle \hbox{det}( T_3 + T_3^{*-1} ) > 0.$

But since ${\hbox{det}( T_3 + T_3^{*-1} ) = \hbox{det}(T_3) \hbox{det}( 1 + T_3^{-1} T_3^{*-1} )}$ and ${ 1 + T_3^{-1} T_3^{*-1}}$ is positive definite, so ${\hbox{det}(T_3)>0}$ as required. $\Box$

The standard branch of the matrix logarithm

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