Sum of a squared dot product I have a constant $d \times n$ matrix $\textbf{A}$, and a variable $d \times 1$ vector $\textbf{v}$ 
$$\sum_{i=0}^n (\mathbf{A}_i^\top \mathbf{v})^2$$
Is there a way to simplify this? Can I pull out any A's?
I know that I could use this property:
$$\sum_{i=0}^n\left( \mathbf{A}_i^\top\mathbf{v} \right) = \left( \sum_{i=0}^n\mathbf{A}_i^\top \right) \mathbf{v}$$
But I don't think that works for squares of dot products

It depends upon what you mean by the 'square'. If it is scalar product then it
equals:
 $$ \sum_i v^T A_i A_i^T v =  v^T \left( \sum_i  A_i A_i^T\right)  v = v^T B v $$
where $B$ is a semi-positive definite matrix. It will be definite positive under reasonable conditions on the $A_i$'s.
