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Suppose we have two matrix valued functions A, B where:

$A:\mathbb{R}^{2}\rightarrow \mathbb{R}^{nxn}$

$B:\mathbb{R}\rightarrow \mathbb{R}^{nxn}$

with $\forall \; x,y \in \mathbb{R}$:

$\frac{\delta A}{\delta x}=B(x)A(x,y)$

$\frac{\delta A}{\delta y}=-A(x,y)B(y)$

$A(x,x)=I$

Show that:

$A(x,y)A(y,z)=A(x,z)$

I'm really at a total loss here, the result seems intuitively wrong.

guest
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1 Answers1

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Derivate $A(x,y)A(y,z)$ in $y$ direction. This gives zero, using the product rule and the equations you supplied for partial derivations of $A$.

$\frac{∂}{∂y}\left(A(x, y)A(y,z)\right) = \left(\frac{∂}{∂y}A(x, y)\right)A(y,z) + A(x, y)\left(\frac{∂}{∂y}A(y,z)\right) = -A(x,y)B(y)A(y,z) + A(x, y)B(y)A(y,z) = 0$

So you can always replace $A(x,y)A(y,z) = A(x, x)A(x,z) = A(x, z)$ using $A(x, x) = I$.

Rolf Kreibaum
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  • Why doesn't $A(x, y)\frac{δ}{δy}A(y,z) = -A(x,y)B(z)A(y,z)$? – guest Feb 08 '15 at 17:12
  • Here the variable names may be misleading. You don't derivate into a particular global „y“ direction. You derivate into the direction where „y“ appears in $A(y,z)$, which is the first entry. And for the first entry you have the rule $\frac{δA}{δx}(x,y) = +B(x)A(x,y)$, but that rule uses other coordinate names. It is equivalent to $\frac{δA}{δy}(y,z) = +B(y)A(y,z)$ which is seen by replacing variable names. – Rolf Kreibaum Feb 08 '15 at 17:19
  • I thought $\frac{δ}{δy}A(x,y)$ means the partial derivative of the 2nd argument so if we replaced x,y,z with a,b,c $\frac{δ}{δy}A(a,b)=-A(a,b)B(b)$? – guest Feb 08 '15 at 17:29
  • Indeed $\frac{δ}{δy}A(x,y)$ refers to partial derivation of the second argument. This is used in the term $(\frac{δ}{δy}A(x,y))A(y,z)$. But replacing $x,y,z$ by $a,b,c$ would give you $\frac{δ}{δb}A(a,b)$, with a $δb$ in the denominator. – Rolf Kreibaum Feb 08 '15 at 17:36
  • Also, the notation for a partial derivation is actually using a $∂$ not a $δ$. I'll update my answer accordingly. – Rolf Kreibaum Feb 08 '15 at 17:37