I have encountered interesting problem:
So we have a real square matrix given as $A = E + \epsilon B$ which is very similar to the indentity matrix and the $\epsilon ^2$ we can neglect then.
And my textbook says the polar decomposition looks like:
$(E + \epsilon \frac {1}{2}(B-B^T))(E + \epsilon \frac {1}{2}(B+B^T))$
and I wasn't able to get there by myself. I don't understand why that result is true.
I tried singular decomposition first, calculating $AA^T$ and its eingenvalues, but I got pure nonsense.
The polar decomposition is defined as: $A = (U \Sigma U^T)(UV^T)$ where $U,V$ are the unitary matrices corresponding to the eigenvectors of $AA^T$ and $\Sigma $ has their square roots of the eigenvalues on its diagonal.
Best Answer
Assuming $E$ is your funny notation for the identity matrix $I$.
We have $A\equiv\exp(A-I)\pmod{\varepsilon^2}$. Now decomposing $A-I$ into its symmetric and anti-symmetric components $$ A-I=\underbrace{\frac{(A-I)+(A-I)^T}2}_{S}+\underbrace{\frac{(A-I)-(A-I)^T}2}_{T}=\varepsilon\frac{B+B^T}{2}+\varepsilon\frac{B-B^T}{2} $$ and thus Zassenhaus formula gives $\exp(A-I)=\exp(S+T)=\exp(S)\exp(T)\exp(-\frac12[S,T]+\dots)$. Hence $$ A\equiv\exp(S)\exp(T)\pmod{\varepsilon^2} $$ which is your polar decomposition.