1

I have an orthogonal matrix $R$ and I am trying to differentiate the term $R^{\top}R$ wrt $R$. However, I am stuck at this point because the matrices are not symmetric. Anybody has an idea or trick. I also know that $\text{det}\left(R\right) = \text{det}\left(R^{\top}\right) = 1$. I have searched around and did not find anything.

\begin{array}{c} f = {R^ \top }R\\ \partial f = \left( {\partial {R^ \top }} \right)R + {R^ \top }\partial R\\ {\rm{vec}}\left( {\partial f} \right) = {\rm{vec}}\left( {\left( {\partial {R^ \top }} \right)R} \right) + {\rm{vec}}\left( {{R^ \top }\partial R} \right)\\ = {\rm{vec}}\left( {{R^ \top } \otimes {I_m}} \right){\rm{vec}}\left( {\partial {R^ \top }} \right) + {\rm{vec}}\left( {{I_m} \otimes {R^ \top }} \right){\rm{vec}}\left( {\partial R} \right) \end{array}

Kumar
  • 332

1 Answers1

2

$R^T R =I$ for any orthogonal $R$ so the derivative within the set of orthogonal matrices is zero.

If we want the derivative within all matrices, then consider $$ (R+E)^T (R+E) = R^T R + E^T R + R^T E + EE^T = I + E^T R + R^T E +{\cal O}(||E||^2) $$ So the derivative is the map $$ E \mapsto E^T R + R^T E. $$

Mark Joshi
  • 5,604
  • Why is the derivative within the set of orthogonal matrices zero ? – Kumar Oct 06 '17 at 02:00
  • because the function is constant on the set of such matrices – Mark Joshi Oct 06 '17 at 04:38
  • Thanks. For the general case, can E be any matrix ? Your forumula looks similar to $$\lim_{h \to 0} \frac{{f\left( {R + hE} \right) - f\left( E \right)}}{h}$$ – Kumar Oct 06 '17 at 11:53
  • $E$ is any matrix. The difference in the two formulations is that mine is the total derivative whereas yours is the directional derivative (sometimes called Gateaux). A derivative is the linear map, $A,$ such that $$f(y) = f(x) + A.(x-y) + o(||x-y||).$$ Here $x$ and $y$ are matrices. – Mark Joshi Oct 06 '17 at 22:13