# Is the solution to $A-O(A)=\tilde \Sigma$ unique?

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 $$O=S.$$

• 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). – Asaf Shachar Jan 1 at 9:49
• @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 – Dap Jan 1 at 10:26
• 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 $A$?) – Asaf Shachar Jan 1 at 19:48
• @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. – 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.