How does the following matrix algebra reduce dimensionally
$$\left( \frac{\boldsymbol{\Sigma}^{-1} \boldsymbol{1} }{\boldsymbol{1}^\top \boldsymbol{\Sigma}^{-1} \boldsymbol{1}}\right)^\top$$
given that $\boldsymbol{\Sigma}$ is the covariance matrix and $\boldsymbol{1}$ is a vector of ones?
I think the solution is a vector, like how its un-transposed self was a vector, but I'm more interested in how the transpose operator changes the dimensions of each individual vector and matrix in the parentheses, derivation-wise. which explains the question title being "how to", instead of "what is".
Best Answer
There are essentially two ways:
Through the properties of matrix multiplication and $(AB)^\top=B^\top A^\top$ and $(A^{-1})^\top=(A^\top)^{-1}$: $$\left(\frac{\Sigma^{-1}1}{1^\top\Sigma^{-1} 1}\right)^\top=(\Sigma ^{-1} 1( 1^\top\Sigma^{-1} 1)^{-1})^\top=(1^\top(\Sigma^\top)^{-1}1)^{-1} 1^\top(\Sigma^\top)^{-1}=\frac{1^\top(\Sigma^\top)^{-1}}{1^\top(\Sigma^\top)^{-1}1}$$
Using the fact that transponse is linear with respect to the action of the scalars over matrices plus the same identities $$\left(\frac{\Sigma^{-1}1}{1^\top\Sigma^{-1} 1}\right)^\top=\frac{\left(\Sigma^{-1}1\right)^\top}{1^\top\Sigma^{-1} 1}=\frac{1^\top(\Sigma^\top)^{-1}}{1^\top\Sigma^{-1} 1}$$
Now, here you should be in a case where $\Sigma^\top=\Sigma$ because it's a covariance matrix, but since the discussion is general we've just re-discovered the fact that $v^\top Aw=w^\top A^\top v$ for all $A\in\Bbb R^{n\times m}$, $v\in \Bbb R^n$ and $w\in \Bbb R^m$.