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Jun 10, 2019 at 16:55 vote accept Continue2Learn
Jun 10, 2019 at 16:47 comment added Continue2Learn Thanks. Closing the answer with my understanding that $(x_{a}\Lambda_{aa}\mu_{a}^{T}) = {(x_{a}^{T}\Lambda_{aa}\mu_{a})}^{T}$ (eq. (5)). And, L.H.S. = R.H.S. "Only because" the operation equals scalar values. Note: above reason, in current context, does not mean $\(AB)\^{T} = A\^{T}B\^{T}$ as Transpose operation is conducted on only one term of equation 5, LHS.
Jun 10, 2019 at 16:26 comment added Siong Thye Goh $x_a = x_a^T$ is not true. What I used is suppose $x^TAy=y^TA^Tx$ since both sides are scalars.
Jun 10, 2019 at 16:17 comment added Continue2Learn and please, the reason for $x_{a}$ = $x_{a}^{T}$ and $\mu_{a}$ = $\mu_{a}^{T}$. Thanks
Jun 10, 2019 at 16:10 comment added Siong Thye Goh $\Sigma$ is a positive definite symmetric matrix right? Hence $\Lambda$ is symmetrical as well.
Jun 10, 2019 at 16:03 comment added Continue2Learn Almost there! but missing something elementary - in eqn(5) and eqn(7), we're considering $A^{T}$ = A, (and moving ahead with calculation). This is something I am unable to recall...need help there
Jun 10, 2019 at 15:50 history answered Siong Thye Goh CC BY-SA 4.0