# Matrix exponential Differentiation

We have the equation $e^X = \sum_{k=0}^\infty{1 \over k!}X^k.$, where X is a matrix of dimension $3 \times 3$ .

Now I have a function $f(x)=C_1x+C_2*\frac{x^2}{2}$ where $C_1,C_2,f(x)$ has dimension $3 \times 3$ . $C_1,C_2$ both are constant matrices, not commutative and x is a scalar variable

It has been known that

$\frac{d}{dx} \exp{(x A)}=\frac{d}{dx}\sum_{k=0}^\infty A^k \frac{x^k}{k!}=\sum_{k=1}^\infty A^k \frac{x^{k-1}}{(k-1)!} = A\cdot \exp(x A)$, where A is a constant $3 \times 3$ matrix

Doubts ** ---

1) If for example $Y = C_1x+C_2*\frac{x^2}{2}$ is a function x where Y is $3 \times 3$ matrix what is $\frac{\mathrm{d} \{Y^{4}\} }{\mathrm{d} x}$? Is it $4*Y^3*(C_1+C_2 *x)$ ?

2) What is $\frac{\mathrm{d} \{e^{f(x)}\} }{\mathrm{d} x}$? In my case f(x) is not a constant it a matrix that varies with x.

• This question, along with your more recent one, are both coming out of issues from your question here. It would probably be more useful if you incorporated these issues into that second link rather than separating them like this. – Semiclassical Jul 16 '14 at 13:33
• sure sir..will do – Nirvana Jul 16 '14 at 13:35

No. If the matrices $C_1$ and $C_2$ do not commute, then you cannot differentiate the composite matrix function in such a simple way. For example, if $Y=f(x)=C_1x+C_2x^2/2$, then, by Leibniz' formula (the prime stands for differentiation withe respect to $x$), $$(Y^4)'=Y'Y^3+YY'Y^2+Y^2Y'Y+Y^3Y'\ne 4Y^3Y', \tag1$$ because $Y$ does not commute with $Y'$. To obtain a concise formula, one can use Feynman ordering of operators: $$(Y^4)'=\frac{\overset 3{Y^4}-\overset 1{Y^4}}{\overset 3{Y}-\overset 1{Y}}\overset 2{Y'}.$$ Here the precise meaning of the rhs is as follows. You take the function $$g(y,z,w)=\frac{h(y)-h(w)}{y-w}z,\quad \text{where h(y)=y^4,}$$ and make the substitution $y\mapsto \overset 3{Y}$, $z\mapsto\overset 2{Y'}$, $w\mapsto \overset 3{Y}$. The numbers over the operators indicate the order in which they act: the larger the number, the more to the left the operator stands in products. Note that $$g(y,z,w)=y^3z+y^2zw+yzw^2+zw^3,$$ so that after the substitution we just obtain (1). Likewise, $$(e^{Y})'=\frac{e^{\overset 3{Y}}-e^{\overset 1{Y}}}{\overset 3{Y}-\overset 1{Y}}\overset 2{Y'}.$$
• Matrices are not commutative...What do you mean by $\overset 3{Y^4}$ .. Means what is 3 and stands for ? Both are powers? – Nirvana Jul 16 '14 at 13:09
• No, $3$ is not a power, it is the Feynman ordering number. Best shown by examples: $\overset1A\overset2B=BA$; $\overset1{X^2}\overset3{Y^3}\overset2Z=Y^3ZX^2$ etc. Also, $e^{A+B}\ne e^Ae^B$ if $A$ and $B$ do not commute, but $e^{\overset1A+\overset2B}=e^Be^A$. – Vladimir Jul 16 '14 at 13:11
• The fraction means this. Take the function $zf(x,y)=z(e^x-e^y)/(x-y)$ (on the diagonal $x=y$, take $f(x,x)=(e^x)'=e^x$ by continuity) and substitute $x\to \overset3Y$, $z\to\overset2{Y'}$, and $y\to\overset1Y$, thus obtaining $f(\overset3Y,\overset1Y)\overset2{Y'}$. You can get details and rigorous definitions in V.P. Maslov Operational Methods or in Nazaikinskii, Sternin, Shatalov Methods of Noncommutative Analysis – Vladimir Jul 21 '14 at 9:08
1) $f(t)=Y(t)^4=Y(t)Y(t)Y(t)Y(t)$, then the Leibniz product rule can be invoked to yeild \begin{align*} f'(t)&=Y'(t)Y(t)Y(t)Y(t)+Y(t)Y'(t)Y(t)Y(t)+Y(t)Y(t)Y'(t)Y(t)+Y(t)Y(t)Y(t)Y'(t)\\ &=\sum_{j=1}^4 Y(t)^{j-1}Y'(t)Y(t)^{4-j}. \end{align*} 2) We can argue similarly for the exponential function $f(X)=e^X$. In order to determine $f'(X)$ is suffices to consider rays of the type $\phi(t)=f(X+tY)$. For such rays, by the same reasoning, $$f'(X)Y=\phi'(t)=\sum_{n=1}^\infty \frac{1}{n!}\sum_{j=1}^nX^{j-1}YX^{n-j}.$$ Now, if $g$ is a map into the matrices, the chain rule applies, yeilding $$(f\circ g)'(t)=f'(g(t))g'(t).$$ Observe that $f'(X)Y$ need not coincide with $f(X)Y$ when $X$ and $Y$ do not commute (i.e. $f'(X)\neq f(X)$). I suppose the answer by Vladimir gives you a more general solution to problems of this type.