General notation: Capital boldface letters are matrices, lowercase boldface letters are vectors, and non-boldface lowercase letters are scalars.
$\mathbf{B}$ is an $n$ x $d$ (where often $d<n$) matrix of basis vectors
$$\mathbf{B}=[\mathbf{b}_1,\mathbf{b}_2,...,\mathbf{b}_d]$$
in which to express the n x 1 vector $\mathbf{v}$ with coefficients $\mathbf{a}=[a^1,a^2,...,a^d]$:
$$\mathbf{v}=\mathbf{Ba}=\sum_{i=1}^d \mathbf{b}_i a^i \tag{eq. 1*}$$
And $\mathbf{B}$ is defined
$$\mathbf{B}=\mathbf{S^{-1}Q} \tag{eq. 2*}$$
Where $\mathbf{S}$ is a square, full rank, symmetric matrix, and $\mathbf{Q}$ is an $n$ x $d$ matrix; and we know
$$\mathbf{Q'v}=\mathbf{D(x)y} \tag{eq. 3*}$$
(Where the apostrophe indicates transpose, and the notation $\mathbf{D(u)}$ indicates a diagonal matrix whose diagonal elements are the elements of the vector $\mathbf{u}$.)
The vectors $\mathbf{x,y}$ are d x 1.
Let $\mathbf{\tilde{B}}$ be the ($n$ x $d$) matrix of reciprocal basis vectors such that $\mathbf{\tilde{B}}'\mathbf{B}=\mathbf{I}$.
($\mathbf{I}$ being a $d$x$d$ identity matrix)
In other words, $\mathbf{\tilde{B}}=[\mathbf{b}^1,\mathbf{b}^2,...,\mathbf{b}^d]$ and $\mathbf{b}_i \cdot \mathbf{b}^j=\delta^j_i$.
Then premultiplying eq. 1* by the transpose of $\mathbf{\tilde{B}}$ gives:
$$\mathbf{\tilde{B}}'\mathbf{v}=\mathbf{a} \tag{eq. 4*}$$
Now, my question follows. Considering eq. 2*, an expression for $\mathbf{\tilde{B}}$ that is consistent with eq. 4* is:
$$\mathbf{\tilde{B}}=\mathbf{QD(a) D(x)^{-1} D(y)^{-1}} \tag{eq. 5*}$$
This is where I have my question. Is this step valid? On the one hand eq. 5* is perfectly consistent with eqs. 2*, 3* and 4*, but on the other hand it feels pretty arbitrary to do this, and leads to some questionable results.
For eg., if eq. 5* is a valid expression for $\mathbf{\tilde{B}}$, then it follows that
$$\mathbf{B}=\mathbf{\tilde{Q} D(a)^{-1} D(x) D(y)}$$
(Where $\mathbf{\tilde{Q}}$ is the ($n$ x $d$) matrix of reciprocal vectors to $\mathbf{Q}$ such that $\mathbf{\tilde{Q}}'\mathbf{Q}=\mathbf{I}$)
Which I can then set equal to the definition in eq. 2*:
$$\mathbf{S^{-1}Q}=\mathbf{\tilde{Q} D(a)^{-1} D(x) D(y)}$$
Which I can then pre-multiply by $\mathbf{Q}'$ to get:
$$\mathbf{Q'\mathbf{S^{-1}Q}= D(a)^{-1} D(x) D(y)}$$
Which would mean that the elements of the diagonal matrix on the RHS are the eigenvalues of $\mathbf{S^{-1}}$, and that the matrix $\mathbf{Q}$ is not unique, but could consist of one of ${n\choose d}$ possible permutations of the eigenvectors of $\mathbf{S^{-1}}$ taken $d$ at a time (because $\mathbf{Q}$ is $n$ x $d$). Moreover, since $\mathbf{S}^{-1}$ is symmetric, then Eq. 5* would imply that the columns of $\mathbf{Q}$ are pairwise orthogonal, and that therefore $\mathbf{\tilde{Q}}=\mathbf{Q}'$.