$\mathbf{M}$ is $n$-x-$n$ symmetric p.d., $\mathbf{Q}$ is $n$-x-$d$ (with $d<n$), and $\mathbf{u}$ is a $d$-x-$1$ vector. The columns of $\mathbf{Q}$ are linearly independent. Another $d$-x-$1$ vector $\mathbf{v}$ can be expressed as follows:
$$\mathbf{MQu=v}$$
The full rank matrix $\mathbf{M}$ can be considered an $n$-dimensional basis for $\mathbf{v}$ with coefficients $\mathbf{Qu}$.
$$\sum^n_{i=1}\mathbf{m}_i x_i=\mathbf{v}$$
(where $\mathbf{m}_i$ is the ith column of $\mathbf{M}$ and $x_i\in \mathbf{x=Qu}=\mathbf{q}_1 u_1+\mathbf{q}_2 u_2+...+\mathbf{q}_d u_d$)
But since the columns $\mathbf{q}_i\in\mathbf{Q}$ are known to be linearly independent, $\mathbf{MQ}$ can also be considered a $d$-dimensional basis for $\mathbf{v}$ with coefficients $\mathbf{u}$.
$$\sum^d_{i=1}\mathbf{Mq}_i u_i=\mathbf{v}$$
The first basis is full rank, while the second basis is rank deficient.
1) I was under the impression that a given vector $\mathbf{v}$ can be expressed in any basis, but that the dimension of the linear subspace of $\mathbb{R}^n$ defined by any such given basis remains the same. Here it seems that $\mathbf{v}$ can be expressed either in a basis spanning $d$ dimensions of $\mathbb{R}^n$, or a full rank basis spanning all $n$ of it. How to conceptually reconcile a vector that can be expressed in bases of different dimension?
2) Which basis should I use? Is one more useful or "correct" than the other? Or does it basically not matter?