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$A(s)$: Matrix whose each component is a function of s.

$B(t)$: Matrix whose each component is a function of t.

$A(0)=B(0)=1$.

${\frac{\partial}{\partial s}}|_{s=0}{\frac{\partial}{\partial t}}|_{t=0}\{A(s)^{-1}B(t)^{-1}A(s)B(t)\}$

$=\left({\frac{\partial}{\partial s}}|_{s=0}A(s)^{-1}\right)\left({\frac{\partial}{\partial t}}|_{t=0}B(t)^{-1}\right)+\left({\frac{\partial}{\partial t}}|_{t=0}B(t)^{-1}\right)\left({\frac{\partial}{\partial s}}|_{s=0}A(s)\right)$

$=\left({\frac{\partial}{\partial s}}|_{s=0}A(s)\right)\left({\frac{\partial}{\partial t}}|_{t=0}B(t)\right)-\left({\frac{\partial}{\partial t}}|_{t=0}B(t)\right)\left({\frac{\partial}{\partial s}}|_{s=0}A(s)\right)$

Please tell me why this calculation is valid.

Igarashi
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1 Answers1

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The key property is the formula for the derivative of products of functions holds also for the case of matrices.

$$ \frac{\partial}{\partial t} (C(t)D(t)) = \left(\frac{\partial}{\partial t} C(t)\right) \cdot D(t) + C(t) \cdot \frac{\partial}{\partial t} D(t). \tag{$\sharp$} $$ (But here we cannot change the order of the product, because we consider matrices. The proof is straightforward.)

If $D(t) = C(t)^{-1}$, then, $$ O = \left(\frac{\partial}{\partial t} C(t)\right) \cdot C(t)^{-1} + C(t) \cdot \frac{\partial}{\partial t} C(t)^{-1}, $$ where $O$ denotes the zero matrix.

If additionally $C(0) = I$ (the identity matrix), then, $$ O = \frac{\partial}{\partial t}|_{t=0} C(t) + \frac{\partial}{\partial t}|_{t=0} C(t)^{-1}. $$

The last equality you wrote follows from this.

By ($\sharp$), $$ \frac{\partial}{\partial t}|_{t=0} (A(s)^{-1} B(t)^{-1} A(s) B(t)) = A(s)^{-1} \cdot \frac{\partial}{\partial t}|_{t=0} (B(t)^{-1} A(s) B(t)).$$ In the same manner, by using that $B(0) = B(0)^{-1} = I$, $$\frac{\partial}{\partial t}|_{t=0} (B(t)^{-1} A(s) B(t)) = \left(\frac{\partial}{\partial t}|_{t=0} B(t)^{-1}\right) A(s) B(0) + B(0)^{-1} \cdot \frac{\partial}{\partial t}|_{t=0} (A(s) B(t))$$

$$ = \left(\frac{\partial}{\partial t}|_{t=0} B(t)^{-1}\right) A(s) + A(s) \cdot \left(\frac{\partial}{\partial t}|_{t=0} B(t)\right).$$

Thus we see that $$ \frac{\partial}{\partial t}|_{t=0} (A(s)^{-1} B(t)^{-1} A(s) B(t)) = A(s)^{-1} \cdot \left(\frac{\partial}{\partial t}|_{t=0} B(t)^{-1}\right) A(s) + \left(\frac{\partial}{\partial t}|_{t=0} B(t)\right).$$

Hence, $$ \frac{\partial}{\partial s}|_{s=0} \frac{\partial}{\partial t}|_{t=0} (A(s)^{-1} B(t)^{-1} A(s) B(t)) = \frac{\partial}{\partial s}|_{s=0} \left( A(s)^{-1} \cdot \left(\frac{\partial}{\partial t}|_{t=0} B(t)^{-1}\right) A(s) \right).$$

By using the assumption that $A(0) = A(0)^{-1} = I$, we see that $$\frac{\partial}{\partial s}|_{s=0} \left( A(s)^{-1} \cdot \left(\frac{\partial}{\partial t}|_{t=0} B(t)^{-1}\right) A(s) \right) = \left(\frac{\partial}{\partial s}|_{s=0} A(s)^{-1}\right) \cdot \left(\frac{\partial}{\partial t}|_{t=0} B(t)^{-1}\right) + \left(\frac{\partial}{\partial t}|_{t=0} B(t)^{-1}\right) \cdot \left(\frac{\partial}{\partial s}|_{s=0} A(s)\right) $$ in the same manner. Thus we have the first equality.

Kazuki OKAMURA
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