Let $A$ be a $k \times 1$ random vector, and $\mathbf{A}$ be a $n \times k$ matrix of observations.
Letting $t \in \mathbb{R}^{k}$ be a vector of weights s.t. $||t||_2 = 1$, suppose we are interested in maximizing the variance of a linear combination of the columns of $\mathbf{A}$, i.e. $\mathbf{A}t$. I have read that this can be written as solving $$ \max_{||t||_2 = 1} \frac{t'(\mathbf{A' A})t}{t't}$$
However I'm a bit confused about where this came from. Would the problem not be solving the following instead? $$ \max_{||t||_2=1} \quad \text{Var}(\mathbf{A}t) = t'\text{Var}(\mathbf{A})t = t'(\mathbb{E}[AA'] - \mathbb{E}[A]\mathbb{E}[A'])t $$ Replacing the expectations with the sample analogue yields, \begin{align*} \max_{||t||_2=1} \quad &t'\left(\frac{1}{n}\sum_{i=1}^n A_i A_i' - (\frac{1}{n}\sum_{i=1}^n A_i) (\frac{1}{n}\sum_{i=1}^{n} A_i') \right)t \\ &\propto t'\left( \mathbf{A'A} - \frac{1}{n} \sum_{i=1}^{n}A_i \sum_{i=1}^{n}A_i' \right)t \\ &= t'\left( \mathbf{A'A} - \frac{1}{n} \sum_{i=1}^{n}\sum_{j=1}^{n}A_i A_j' \right)t \end{align*}
where $A_i$ is the $i$th row of observed $\mathbf{A}$. In the formulation above, a) what happened to the term involving the double summation and b) where did the $t't$ in the denominator come from?
If we further assumed $\mathbf{A}$ has mean 0, then I can see how a) drops out though that doesn't address b), and the formulation above also does not seem to mention the requirement of de-meaning.
