Opposite determinant in Autonne-Takagi factorization

Question

Let us consider a complex symmetric matrix in $M_2(\mathbb C)$ \begin{equation} A = \begin{pmatrix} x_1+ix_2 & x_3 \\ x_3 & -x_1+ix_2 \end{pmatrix} \end{equation} where the $x_i\in \mathbb R,\;\;i=1,2,3$. The Autonne-Takagi factorization theorem tells that a unitary matrix $U$ exisits such that \begin{equation} U^T A U=D \end{equation} where \begin{equation} D =\begin{pmatrix} d_1& 0 \\ 0 & d_2 \end{pmatrix} \end{equation} is a diagonal non negative matrix, with real entries. Given that \begin{equation} D^2 = D^\dagger D = (U^T A U)^\dagger U^T A U=U^\dagger A^\dagger (UU^\dagger)^TA U=U^\dagger (A^\dagger A) U \end{equation} we can see that $d_1^2$ and $d_2^2$ are the eigenvalues of the Hermitian matrix $A^\dagger A$. In our case, this matrix is already diagonal, as \begin{equation} A^\dagger A =\begin{pmatrix} x_1^2+x_2^2+x_3^2& 0 \\ 0 & x_1^2+x_2^2+x_3^2 \end{pmatrix} \end{equation} Given that the theorem tells us that $D$ is non negative, I would take $d_{1,2}=+\sqrt{x_1^2+x_2^2+x_3^2}$. However, if we compute the determinant of the factorization, we would obtain \begin{equation} \det(U^T)\det A \det U = \det A = - x_1^2-x_2^2-x_3^2 =\det D = x_1^2+x_2^2+x_3^2 \end{equation} hence the only solution would be $x_i=0$, $i=1,2,3$. What am I missing?

Answer 1

The factorization tells you that $U^\intercal AU = D$ for a suitable unitary matrix $U$. But $U$ is in general complex (even if $A$ is real), as can be easily seen by some examples. Hence, $U^\intercal U \neq I$ in general (this is only the case if $U$ is real). This implies that the equality

$$\det(U^\intercal AU) = \det(U^\intercal U) \det(A) = \det(A)$$

does not hold in general for if $\det(U^\intercal U) \neq 1$. Consider for example the matrix $U = \operatorname{diag}( i, 1)$, then $U$ is unitary with $\det(U) = \mathrm i$, but $\det(U^\intercal U) = - 1$.

What you indeed always can say is that $\lvert \det (U) \rvert = 1$, therefore it follows that

$$\lvert \det(U^\intercal AU) \rvert = \lvert \det(A) \rvert= \det(D).$$

Your given example is an example for the last equality.