Jtdylktuy

Let $tilde Sigma=text{diag}(tilde sigma_i)$ be a diagonal matrix, with $tilde sigma_i>0$. ($1 le i le n$).

Suppose that $A$ is a real invertible $n times n$ matrix with positive determinant, satisfying $A-O(A)=tilde Sigma$, where $O(A)=A(sqrt{A^TA})^{-1}$ is the orthogonal polar factor of $A$, i.e. $A=OP$ for some special orthogonal matrix $O$ and symmetric positive-definite matrix $P$.

Is it true that $A=text{diag}(tilde sigma_i+1)$? (In that case $O(A)=text{Id}$).

Writing $A=USigma V^T$ (SVD), the equation $A-O(A)=tilde Sigma$ becomes

$$ U(Sigma -text{Id}) V^T=tilde Sigma.$$

Taking the transpose of the equation, we also have

$$ V(Sigma -text{Id}) U^T=tilde Sigma.$$

Combining these two equations, we then have

$$ U(Sigma -text{Id})^2 U^T=tilde Sigma^2.$$

Considering the eigenvalues of both sides, we deduce that $(sigma_i-1)^2=tilde sigma_{tau(i)}^2$, where $tau in S_n$ is a permutation. Let $v_i$ be the $i$-th column of $U^T$. If all the $tilde sigma_i$ are distinct, then
$(Sigma -text{Id})^2 v_i=tilde sigma_{tau(i)}^2 v_i$, which implies $v_i in text{span}{e_{tau(i)}}$. Since the columns of $U^T$ are orthonormal, we have $v_i=pm e_{alpha(i)}$, i.e. $U^T$ must be a signed permutation matrix.

The same reasoning can be applied to $V$. Thus, $A=USigma V^T$ must be diagonal.
(Is this really true? I am not so sure now).

A counterexample is $A=-tfrac12 I$ and $tildeSigma=tfrac12 I$ for even $n.$ Here $O(A)=-I.$

$A$ is not necessarily diagonal. For an example it should work to take $n=4,$ $tildeSigma=tfrac12I,$ and $A=UDU^T$ where $D=operatorname{diag}(tfrac32,tfrac32,-tfrac12,-tfrac12)$ and $U$ is a rotation by forty-five degrees in the y-z plane.

It is true that $O^2=I$ and $O$ commutes with $tildeSigma.$ The situation could be described as: after an orthogonal change of basis, $tildeSigma$ is still diagonal and $A$ is diagonal with $A_{ii}=tildeSigma_{ii}pm 1$ (and the $-1$ case can only occur when $tildeSigma_{ii}<1.$)

Write $A=OP$ with $Oin SO_n$ and $P$ symmetric positive definite.

There's a $Vin SO_n$ such that the matrix $P':=V^TPV$ is diagonal. Write $S'$ for the result of replacing the diagonal entries of $P'-I$ by their sign $pm 1$ (note $P-I$ is non-singular, so this is well defined) and write $|P'-I|$ for the result of replacing the diagonal entries of $P'-I$ by their absolute value. Define $S=VS'V^T$ and $|P-I|=V|P'-I|V^T.$ Note that $P-I$ and $S$ and $|P-I|$ all commute - they are simultaneously diagonalized by conjugation by $V$ on the right. And $S|P-I|=P-I.$

$S$ is symmetric and satisfies $S^2=I$ so is actually orthogonal. Applying the uniqueness of the orthogonal polar decomposition to $OS|P-I|=tildeSigma$ gives $OS=I,$ which gives $O=S.$

It is worth noting that even if $A$ is diagonal, then the solution is not unique:

Indeed, take $A_1=text{diag}(3/2,5/4,2)$,$A_2=text{diag}(-1/2,-3/4,2)$ which both correspond to $tilde Sigma=text{diag}(1/2,1/4,1)$.

Indeed, note that for diagonal matrix $D=text{diag}(d_i)$ in $GL^+$, $O(D)=text{diag}(text{sgn} (d_i))$.

Furthermore, if $Sigma=lambda Id$, then for every solution $A$, and every orthogonal matrix $U$, $U^TAU$ is also a solution. This follows immediately from the fact that $O(AU)=O(A)U,O(UA)=UO(A)$. (These properties are due to the fact that $O(A)$ is simultaneously the left and right orthogonal polar factor of $A$).

Thus, if $A-O(A)=tilde Sigma=lambda Id$, then

$$U^TAU-O(U^TAU)=U^TAU-U^TO(A)U=U^T(A-O(A))U=U^Ttilde Sigma U=tilde Sigma,$$

since $Sigma$ commutes with all matrices.