Matrix Taylor series and differentiation: how do we show that if $X=A-tB$, $\frac{d^j}{dt^j}X^{-1}=(-1)^jj!X^{-1}(X'X^{-1})^j$?

Question

I cite this paper on matrix Taylor series.

The statement I am confused by is the following:

$$\frac{d^j}{dt^j}X^{-1}=(-1)^jj!\cdot X^{-1}(X'X^{-1})^j\tag{1}$$

Where $X=A-tB$ for two square matrices of the same order $A,B$.

Firstly, $X'=\frac{d}{dt}X(t)=-B$ - I think! Secondly: $$0=\frac{d}{dt}(I)=\frac{d}{dt}(XX^{-1})=X'X^{-1}+X(X^{-1})'\implies -X^{-1}XX^{-1}=(X^{-1})'\tag{2}$$

By the product rule, for which I can see remains true for matrix-by-scalar differentiation (but does it hold for scalar-by-matrix differentiation?)

I tried to show their derivative law - $(1)$ - by induction. The base case of $j=1$ holds by $(2)$. It then remains to show that if $(1)$ holds for some $j$, it holds too for $j+1$. The natural way to approach this is to simply differentiate $(1)$. I will be assuming the chain rule (and the power law for indices) holds for matrix-by-scalar differentiation - please correct me if it doesn't!

$$\begin{align}\frac{d}{dt}\left(\frac{d^j}{dt^j}X^{-1}\right)&=\frac{d}{dt}\Big[(-1)^j\cdot j!\cdot X^{-1}(X'X^{-1})^j\Big]\tag{3}\\&=(-1)^j\cdot j!\cdot\left(\left(\frac{d}{dt}X^{-1}\right)(X'X^{-1})^j+X^{-1}\frac{d}{dt}(X'X^{-1})^j\right)\tag{4}\\&=(-1)^j\cdot j!\cdot\left(-X^{-1}X'X^{-1}(X'X^{-1})^j+j\cdot X^{-1}(X'X^{-1})^{j-1}\cdot\frac{d}{dt}(X'X^{-1})\right)\tag{5}\\&=(-1)^{j+1}\cdot j!\cdot X^{-1}(X'X^{-1})^{j+1}+(-1)^{j}\cdot j!\cdot j\cdot X^{-1}(X'X^{-1})^{j-1}\\&\cdot(X''X^{-1}-X'X^{-1}X'X^{-1})\tag{6}\\&=(-1)^{j+1}\cdot j!\cdot X^{-1}(X'X^{-1})^{j+1}+(-1)^{j+1}\cdot(j+1)!\cdot X^{-1}(X'X^{-1})^{j+1}\tag{7}\end{align}$$

$(3)$ holds by inductive hypothesis, $(4)$ assumes the product rule. $(5)$ assumes both the chain rule and the power law, and $(7)$ uses the fact that $X''=\frac{d}{dt}(-B)=\mathbf{0}$. The strange thing is, my end result, $(7)$, is almost the expression for the case $j+1$, however, there is a malignant expression on the left which destroys the induction.

Where did I go wrong? And can I assume the laws I assumed for matrix calculus?

Answer 1

(5) is where you start to go wrong. Recall matrix multiplication is not-necessarily-commutative, so the law for differentiating powers is $$ \frac{\mathrm{d}}{\mathrm{d}t}M^n=\sum_{j=0}^{n-1}M^j\frac{\mathrm{d}M}{\mathrm{d}t}M^{n-1-j} $$ without the ability to slide all $M$s in front of the derivative.

So your (5) should be $$ \dots=(-1)^j\cdot j!\cdot\bigg(-X^{-1}X'X^{-1}(X′X^{-1})^j +X^{-1}\sum_{k=0}^{j-1}(X'X^{-1})^k(X'X^{-1})'(X'X^{-1})^{j-1-k}\bigg) $$ and now use $X''=0$ to get $(X'X^{-1})'=X''X^{-1}-X'(X^{-1})'=-X'(X^{-1})'$ and the base-case formula for $(X^{-1})'$ to get all summands in the big parentheses are equal.

Answer 2

The chain rule doesn't work the way you think it does for matrix powers. For example, suppose the matrix $M$ is a function of $t$. If we differentiate $M^3$ with respect to $t$, then the result is $$ \frac{d}{dt} \left( M^3 \right) = \frac{d}{dt} \left( M M M \right) = M' M M + M M' M + M M M' $$ which is not equal to $3 M' M^2$ unless $[M, M']= 0$.

As an alternate approach, try using the general Liebniz rule on the $j$th derivative of $X X^{-1}$: $$ 0 = \frac{d^j}{dt^j}(I) = \frac{d^j}{dt^j} (X X^{-1}) = \sum_{k = 0}^{j} {j \choose k} \frac{d^k}{dt^k} X \frac{d^{j-k}}{dt^{j-k}} X^{-1}. $$ This looks awful, but in fact almost all of the terms in this sum vanish because $X'' = 0$. You should be able to figure it out from there.