I am implementing a fast Manhattan Frame estimation method based on this paper.
A crucial step to this algorithm is computing the real, symmetric $\mathbf{M}$ matrix parameterized by 15 variables as shown in the following screen shot
According to the paper, $\mathbf{M}$ must be a real, positive semi-definite matrix as Equation 29 of this paper states
$\mathbf{M} = \mathbf{Q} * \mathbf{D} * \mathbf{Q^T}$ where eigenvalues $\lambda_k \geq 0$ and $\mathbf{Q}$ is an orthogonal matrix
However, trying with a number of surface normal datasets, I end up getting matrix $\mathbf{D}$ of the following form
$ \mathbf{D} = \begin{pmatrix} -d_1 & 0 & 0 & 0 & 0 & 0 \\ 0 & -d_2 & 0 & 0 & 0 & 0 \\ 0 & 0 & d_3 & 0 & 0 & 0 \\ 0 & 0 & 0 & d_4 & 0 & 0 \\ 0 & 0 & 0 & 0 & d_5 & 0 \\ 0 & 0 & 0 & 0 & 0 & d_6 \\ \end{pmatrix} $
As you can see, this implies $\mathbf{M}$ is positive, symmetric indefinite matrix.
To mitigate this problem, one idea I have is to form $\mathbf{M_1} = |\mathbf{M}|$. Here $|\mathbf{M}|$ implies taking element-wise absolute value so that $ m_{ij} \in \mathbf{M_1} \geq 0$
My question is $\mathbf{M_1} = \mathbf{M_1^T}$ true for all cases?
A screenshot of how $\mathbf{M}$ and $\mathbf{D}$ came out from one of the surface normal dataset is shown below
As per the suggestion by @BenGrossmann, the problem with $\mathbf{M}$ matrix lied with the (5,5) and (6,6) elements. Solving this issue did not require me to use $\mathbf{M_1} = |\mathbf{M}|$.
Regarding the question, I am accepting the comment by @BenGrossmann on the original problem is $\mathbf{M_1} = \mathbf{M_1^T}$ true for all cases where $\mathbf{M_1} = |\mathbf{M}|$ given $\mathbf{M}$ is real and symmetric as No as at least one counter example exsists where $\mathbf{M_1}$ was shown to fail positive (semi)definite test.