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 $A=U\Sigma V^T$ (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, $A=U\Sigma V^T$ must be diagonal. (Is this really true? I am not so sure now).


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

$A$ is not necessarily diagonal. For an example it should work to take $n=4,$ $\tilde\Sigma=\tfrac12I,$ 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 $\tilde\Sigma.$ The situation could be described as: after an orthogonal change of basis, $\tilde\Sigma$ is still diagonal and $A$ is diagonal with $A_{ii}=\tilde\Sigma_{ii}\pm 1$ (and the $-1$ case can only occur when $\tilde\Sigma_{ii}<1.$)

Write $A=OP$ with $O\in SO_n$ and $P$ symmetric positive definite.

There's a $V\in SO_n$ 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 $OS|P-I|=\tilde\Sigma$ gives $OS=I,$ which gives $O=S.$

  • $\begingroup$ Thanks. Indeed, I forgot about the possibility of $A_{ii}=\tilde \Sigma_{ii}-1$. I still wonder though: Is it true that when all the $\tilde \sigma_i$ are distinct, the solution $A$ must be diagonal? Also, you could take any rotation $U$ in your example, right? (the exact angle does not matter I think, as you can see in my answer below). $\endgroup$ – Asaf Shachar Jan 1 at 9:49
  • $\begingroup$ @AsafShachar: (1) Yes, if the $\tilde \sigma_i$ are distinct then $O$ and hence $A$ must be diagonal, since they commute with $\tilde \Sigma$ according to the argument in the second half of my answer. (2) Yes, I was just being concrete - anything that isn't a multiple of 90° should work $\endgroup$ – Dap Jan 1 at 10:26
  • $\begingroup$ Thank you very much! I see why your argument implies $S=O$ so $O^2=id$, and $\tilde \Sigma=|P-I|$ but how do we deduce that $O$ and $A$ commute with $\tilde \Sigma$? (or equivalently why $O$ commute with $A$?) $\endgroup$ – Asaf Shachar Jan 1 at 19:48
  • 1
    $\begingroup$ @AsafShachar: $|P-I|$ and $O$ are simultaneously diagonalizable. I have edited to more explicitly define $S$ and $|P-I|,$ which should make it easier to see that these all commute. $\endgroup$ – Dap Jan 2 at 7:50

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

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

since $\Sigma$ commutes with all matrices.


Your Answer

