How to transpose $\left( \frac{\Sigma^{-1} 1 }{1^\top \Sigma^{-1} 1} \right)$? Matrix algebra simplification based on the transpose operator 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".
 A: 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 $w\in \Bbb R^m$.
A: Assuming your column of ones is a column vector, then the top is a column vector so the transpose is a row vector. So it's not a scalar. Write it as $$\left( \boldsymbol{\Sigma}^{-1} \boldsymbol{1} (\boldsymbol{1}^\top \boldsymbol{\Sigma}^{-1} \boldsymbol{1})^{-1}\right)^\top$$and it's much easier to see what's going on.
Now you can take out the denominator because it's a scalar. Use $(AB)^T=B^TA^T$ also.
Note that the denominator is a scalar so the transpose doesn't affect it.
Also, a covariance matrix is always symmetric, so you can drop the T on it after you finish.
