Matrix exponential Differentiation

Question

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.

Answer 1

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, $$ (e^{Y})'=\frac{e^{\overset 3{Y}}-e^{\overset 1{Y}}}{\overset 3{Y}-\overset 1{Y}}\overset 2{Y'}. $$

Answer 2

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.