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This is part of a larger proof on the orthonormal group viewed as a manifold. I have never done anything with matrix calculus and am trying to find $df$ where $$ f(A)=AA^T $$ where $$ f:M_{n\times n}\to S(n) $$ wehre $S(n)$ are the symmetric $n\times n$ matrices. So we should have $$ df:M_{n\times n}\to T_pS(n) $$ where $p\in S(n)$.

operatorerror
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1 Answers1

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Hint: If you already know that $df$ exists, you can compute it as the directional derivative $$ df_A(V) = \lim_{t \to 0} \frac{f(A+tV) - f(A)}{t} = \lim_{t \to 0} \frac{(A+tV)(A+tV)^T - A A^T}{t} = \cdots $$

Viktor Vaughn
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    this is perfect, thank you! What requires knowing that $df$ exists? It seems that if this limit exists for an arbitrary $V$ it should be differentiable no? – operatorerror Apr 13 '17 at 00:51
  • Even in multivariable calculus on $\mathbb{R}^3$, there are functions all of whose directional derivatives exist, but still fail to be differentiable. (See here for an example.) Basically, directional derivatives are computed along lines, but the derivative along other curves may still not exist. – Viktor Vaughn Apr 13 '17 at 01:22
  • ah of course, my fault – operatorerror Apr 13 '17 at 01:24