1.5.5 The Matrix-Matrix Exponential

In this sec­tion, we in­tro­duce the matrix-matrix ex­po­nen­tial and sev­er­al re­lat­ed proofs as out­lined in the work of Bar­radas and Cohen¹. The matrix ex­po­nen­tial of an­oth­er matrix is de­fined, as

for $X$, a normal and non-sin­gu­lar matrix, $Y$, any com­plex matrix, and both are square $n \times n$ ma­tri­ces. As ma­tri­ces can be non-com­mu­ta­tive, there is a dis­tinc­tion be­tween left matrix ex­po­nen­ti­a­tion$_{}^{Y}X$, and right matrix ex­po­nen­ti­a­tion $X^{Y}.$

A spec­tral the­o­rem is a result in math­e­mat­ics and linear al­ge­bra that per­tains to when a matrix or other op­er­a­tor can be di­ag­o­nal­ized, which we have al­ready uti­lized for the matrix ex­po­nen­tial map­ping and matrix log­a­rithm in the pre­vi­ous sec­tions. Spec­tral the­o­rem was gen­er­al­ized by the math­e­mati­cian John von Neu­mann, and his work is piv­otal in the field. Ex­am­ples for which spec­tral theory ap­plies are to self-ad­joint op­er­a­tors and normal op­er­a­tors on Hilbert spaces, and in this sec­tion will focus on the more gen­er­al class of ma­tri­ces de­fined as normal, where normal de­fines that a matrix $X$ com­mutes with its con­ju­gate trans­pose

Fur­ther, it can be stated that a matrix is normal if and only if it is uni­tar­i­ly di­ag­o­nal­iz­able, as

for $U$, a uni­tary matrix of column vec­tors of the or­tho­nor­mal eigen­vec­tors of $X,$and $D$, a di­ag­o­nal matrix of the eigen­val­ues of $X$.

The­o­rem (1.5.74) – All the el­e­ments of the Lie al­ge­bra $\mathfrak{s}_{3\mathbb{C}}^{\Delta}$ are uni­tar­i­ly di­ag­o­nal­iz­able and there­fore normal.

Proof – Using an ex­am­ple from each of the sub­groups of el­e­ments $\mathcal{X},\mathcal{Y}$, we start with the $\mathcal{X}_{1}$ el­e­ment with three matrix en­tries $a,b,c$, re­spec­tive­ly, in rows $1,2,3$, which take any value from the set $\left\{ \pm 1, \pm i, \pm ( - 1)^{\left( \frac{1}{4} \right)}, \pm ( - 1)^{\left( \frac{3}{4} \right)} \right\}$

Based on the sym­me­try of the trans­pose and con­ju­gate op­er­a­tions, all en­tries for $\mathcal{X}_{1}\mathcal{X}_{1}^{*},\mathcal{X}_{1}^{*}\mathcal{X}_{1}$ are equal to the iden­ti­ty el­e­ment result $\epsilon$, shown by taking the di­ag­o­nal matrix form

there­fore, $\mathcal{X}_{1}\mathcal{X}_{1}^{*} = \mathcal{X}_{1}^{*}\mathcal{X}_{1} = \epsilon$. Based on the sim­i­lar sym­me­tries for the $\mathcal{X}_{2},\mathcal{X}_{3}$ el­e­ments, the same re­sults will hold, and we can state $\mathcal{X}\mathcal{X}^{*} = \mathcal{X}^{*}\mathcal{X}$. We use the ex­am­ple for $\mathcal{Y}_{2}$ from the $\mathcal{Y}$ sub­group, having three matrix en­tries $a,b,c$, re­spec­tive­ly, in rows $1,2,3$, which take any value from the set $\left\{ \pm 1, \pm i, \pm ( - 1)^{\left( \frac{1}{4} \right)}, \pm ( - 1)^{\left( \frac{3}{4} \right)} \right\}$

Based on the sym­me­try of the $\mathcal{Y}$ sub­group's trans­pose op­er­a­tion $\mathcal{Y}_{2}^{T} = \mathcal{Y}_{3},\mathcal{Y}_{3}^{T} = \mathcal{Y}_{2}$, and the con­ju­gate op­er­a­tion, all en­tries for $\mathcal{Y}_{2}\mathcal{Y}_{2}^{*},\mathcal{Y}_{2}^{*}\mathcal{Y}_{2}$ are equal to the iden­ti­ty el­e­ment result $\epsilon$, taking the di­ag­o­nal matrix form

there­fore, $\mathcal{Y}_{2}\mathcal{Y}_{2}^{*} = \mathcal{Y}_{2}^{*}\mathcal{Y}_{2} = \epsilon$. Based on the sim­i­lar sym­me­tries of $\mathcal{Y}_{3}$, its trans­pose and con­ju­ga­tion op­er­a­tions, and the fact that $\mathcal{Y}_{1}$ is di­ag­o­nal, we have that $\mathcal{Y}\mathcal{Y}^{*} = \mathcal{Y}^{*}\mathcal{Y}$, and there­fore, the set of all el­e­ments $\mathcal{X},\mathcal{Y} \in\mathfrak{s}_{3\mathbb{C}}^{\Delta}$ are normal and uni­tar­i­ly di­ag­o­nal­iz­able.

$\square$

We can now state the fol­low­ing the­o­rems re­gard­ing the matrix-matrix ex­po­nen­tial and prove them:

The­o­rem (1.5.75) – If a matrix $X$ is normal and $Y$ is any matrix, $X^{Y}$ and $_{}^{Y}X$will have the same set of eigen­val­ues.

Proof – For any two ma­tri­ces $X,Y$, the matrix prod­ucts $XY,YX$ will have the same set of eigen­val­ues, as such, $\ln\left( e^{X} \right)Y$ has the same set of eigen­val­ues as $Yln\left( e^{X} \right)$. By the spec­tral map­ping the­o­rem, the eigen­val­ues of $e^{X}$ are the ex­po­nen­tials of the eigen­val­ues of $X$, $\therefore$ $_{}^{Y}X$ has the same set of eigen­val­ues as $X^{Y}.$

$\square$

The­o­rem (1.5.76) – If both ma­tri­ces $X,Y$ are normal and $XY = YX$ (com­mute), then $X^{Y}$,$_{}^{Y}X$ are normal, $\ln(X)Y = Yln(X)$, and $X^{Y} =_{}^{Y}X.$

Proof – If $U$ is a uni­tary matrix of column vec­tors rep­re­sent­ing the common set of or­tho­nor­mal eigen­vec­tors of $X,Y$, so that $XU = UD,YU = UE$, for $D$, the di­ag­o­nal matrix of eigen­val­ues of $X$, and $E$, the di­ag­o­nal set of eigen­val­ues of $Y$, then

for $X^{Y} = e^{\ln(X)Y} = UD^{E}U^{*}$, and there­fore, $X^{Y}$ is normal. Given any $Z,$ for $e^{Z}$as non-sin­gu­lar, with $Z = \ln(X)Y$, it fol­lows that $X^{Y}$ is non-sin­gu­lar, there­fore, $X^{Y}$ is normal and non-sin­gu­lar. To prove that $\ln(X)Y = Yln(X)$, we have that the di­ag­o­nal ma­tri­ces $\ln(D),E$com­mute

and the com­mu­ta­tive re­la­tion­ship $X^{Y} =_{}^{Y}X$, is also a direct con­se­quence of the above.

$\square$

The­o­rem (1.5.77) – The ex­po­nen­tial prod­uct for­mu­la: If $X$ is a normal and non-sin­gu­lar matrix, and $Y,Z$ are com­plex $n \times n$ ma­tri­ces, then

It also holds that if the set $\{ X,Y,Z\}$ is a normal com­mut­ing set of ma­tri­ces, then ${X^{Y + Z} = X}^{Y}X^{Z}$and $X^{Y + Z}$, are non-sin­gu­lar and normal. The same will hold for matrix-matrix ex­po­nen­tials of the form $_{}^{Y + Z}X$.

Proof – Sophus Lie's ex­po­nen­tial prod­uct for­mu­la is stated that for any $B,C$

where taking, $B = \ln(X)Y$, then $C = \ln(X)Z,$ yields the first as­ser­tion. Let­ting $\left\{ X,Y,Z \right\}$, be a set of com­mut­ing, normal, and non­sin­gu­lar ma­tri­ces, and using the common set of or­tho­nor­mal eigen­vec­tors that is guar­an­teed to exist as this is a si­mul­ta­ne­ous­ly di­ag­o­nal­iz­able set, then it is straight­for­ward that $Y + Z$ is normal and that $X(Y + Z) = (Y + Z)X.$ As $X^{Y}$ was normal and non­sin­gu­lar for $Y$ normal, this im­plies that $X^{Y + Z}$ is normal and non-sin­gu­lar as well. By taking $B = \ln(X)Y$, and $C = \ln(X)Z$, it fol­lows that $BC = CB$, this im­plies $e^{B + C} = e^{B}e^{C}$, and re­duces fur­ther to $X^{Y + Z} = X^{Y}X^{Z}.$

$\square$

1. Bar­radas, Igna­cio et Cohen, Joel, "It­er­at­ed Ex­po­nen­ti­a­tion, Matrix-Matrix Ex­po­nen­ti­a­tion, and En­tropy", The Jour­nal of Math­e­mat­i­cal Analy­sis and Ap­pli­ca­tions, Vol. 183, No. 1, April 1994.