Polar decomposition of $A = E + \epsilon B$

Question

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.

Answer 1

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.

Answer 2

Given a real square matrix $A$ with positive determinant, its polar decomposition is given by $A=RU$, where $U$ is a positive definite symmetric matrix and $R$ is a properly orthogonal matrix.

The proof of the theorem shows that $U=\sqrt{A^TA}$ and $R=AU^{-1}$, so $$ U = \sqrt{A^TA} = \sqrt{(E+\epsilon B)(E+\epsilon B^T)} \approx \sqrt{E+\epsilon(B+B^T)} \approx E+\frac{1}{2}\epsilon(B+B^T) $$ and $$ R = AU^{-1} \approx (E+\epsilon B)\left(E-\frac{1}{2}\epsilon(B+B^T)\right) \approx E+\frac{1}{2}\epsilon(B-B^T) $$