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If we let $\psi:\text{GL}(n,\Bbb R)\times\text{GL}(n,\Bbb R)\rightarrow \text{GL}(n,\Bbb R)$, by $\psi:(A,B)\mapsto AB$, then $\psi$ can be seen as a differentiable map, letting $(A)_{ij}={a}_{ij}$ and $(B)_{ij}={b}_{ij}$, the components of $\psi(A,B)$ are:

$\psi(A,B)_{ij}=\sum\limits_{k=1}^n a_{ik}\;b_{kj}$.

which is a degree two polynomial in $2n$ variables and so is smooth.

I am just wondering if the following is correct for $d_{id,id} \psi\;$, the derivative at the identity in each entry.

We look at $\psi(id+A,id+B)_{ij}=\sum\limits_{k=1}^n (id+ A)_{ik}\;(id+B)_{kj}=\sum\limits_{k=1}^n (\delta_{ik}+ a_{ik})\;(\delta_{kj}+b_{kj})$

$=\sum\limits_{k=1}^n (\delta_{ik}\delta_{kj}+ a_{ik}\delta_{kj}+\delta_{ik}b_{kj}+a_{ik}b_{kj})=1+a_{ij}+b_{ij}+\sum\limits_{k=1}^n a_{ik}b_{kj}$.

To find the derivative we take the linear terms in $a$'s and $b$'s.

So $d_{id,id}\psi(A,B)=A+B$ ?

snulty
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1 Answers1

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You're working too hard. You can make this computation much easier for yourself if you just systematically refuse to pick bases. At $(1, 1)$, matrix multiplication looks like

$$(1 + A)(1 + B) = 1 + A + B + AB$$

and after taking linear terms you get $A + B$. This argument works essentially without modification in any Banach algebra.

Qiaochu Yuan
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  • I suppose it'll be similar if I didn't work at the identity? Just with matrix coefficients on A and B? Thanks! I want to try differentiate the inverse map next if possible, just checking I have the right idea in a simpler case! – snulty Dec 05 '14 at 17:58
  • @snulty: yes, that's right. Differentiating the inverse works the same way: if the operator norm of $A$ is less than $1$ then $\frac{1}{1 + A} = 1 - A + A^2 \mp \dots $ and taking linear terms gets you $-A$. – Qiaochu Yuan Dec 05 '14 at 18:01
  • I've been meaning to look at the operator norm properly, so I will! But is there a way to differentiate $A^{-1}$ from the adjugate formula: $A^{-1}=\frac{1}{\det A}\text{adj}(A)$. Or is that too messy to bother with? – snulty Dec 05 '14 at 18:06
  • @snulty: there might be, but that sounds like strictly more work. Again, the argument I just gave works essentially without modification in any Banach algebra. – Qiaochu Yuan Dec 05 '14 at 18:07
  • Yeah you're probably right, I may give it a try out of interest though. Thanks for the follow up method too about inverses. I'll look at that! – snulty Dec 05 '14 at 18:13