$x,m $ and $ \Lambda$ are all matrix,$\Lambda$ is covariance matrix
$e^{\frac{-(x-m)^T\Lambda ^{-1}(x-m)}{2}} \times e^{ju^Tx}$
its exponential part calculation is
$\frac{-(x^T \Lambda^{-1}x-x^T\Lambda^{-1}m-m^T\Lambda^{-1}x+m^T\Lambda^{-1}m)+2ju^Tx}{2}$
=$\frac{-(x^T \Lambda^{-1}x-2m^T\Lambda^{-1}x+m^T\Lambda^{-1}m)+2ju^Tx}{2}$
=$\frac{-(x-(m+j\Lambda u))^T\Lambda^{-1}(x-(m+j\Lambda u))+jm^Tu-\frac{1}{2}u^T\Lambda u}{2}$
1.For the first and second formula,why is $-x^T\Lambda^{-1}m-m^T\Lambda^{-1}x=-2m^T\Lambda^{-1}x$?
2.For the second and third formula,how does the second formula become to third formula?can someone teach me more detail?
If $\Lambda$ is symmetric, then $x^\top \Lambda^{-1} m = m^\top \Lambda^{-1} x$.
Take the first term in the third formula and expand it. (e.g. $(a+b)^\top\Lambda^{-1}(a+b) = a^\top \Lambda^{-1} a + 2 a^\top \Lambda^{-1} b + b^\top \Lambda^{-1} b$ etc.)