Transformation of vector into a symmetric matrix !!! Let us consider a transformation $T$, which converts a vector into a symmetric matrix, e.g.
$X_h  = [x_1, x_2, x_3]^T$, then $T(X_h) = \begin{bmatrix}x_1&x_2\\x_2&x_3\end{bmatrix}$.
I want to have this type of decomposition:
$$T(QX_h) = \bar{Q} T(X_h) \bar{Q}^T.$$
Here, $Q\in \mathbb{R}^{m\times \frac{q(q+1)}{2}}$, $\bar{Q}\in \mathbb{R}^{p \times q}$, $T(X_h)\in \mathbb{R}^{q \times q}$ and $X_h \in \mathbb{R}^{q(q+1)/2}$.
For example:
Consider a known constant matrix $Q \in \mathrm{R}^{10 \times 3}$. A vector  $X_h = [x_{11}, x_{12}, x_{22}]^T \in \mathrm{R}^3$.
Decomposition must be like,
$$T(QX_h) = \bar{Q}T(X_h)\bar{Q}^T = \bar{Q}  \begin{bmatrix}x_{11}&x_{12}\\x_{12}&x_{22}\end{bmatrix} \bar{Q}^T$$
where, $T(QX_h)\in \mathrm{R}^{4 \times 4}$, $\bar{Q}\in \mathrm{R}^{4 \times 2}$ and $T(X_h)\in \mathrm{R}^{2\times 2}$
Given $Q$, I want to find $\bar{Q}$ such that the equation holds for all $X_h$.
Is such decomposition possible?
 A: I will focus on the case where $Q$ is $3\times 3$ and $\bar Q$ is $2 \times 2$. In this case, such a decomposition is not necessarily possible.
In terms of the vectorization operator, we can write
$$
T(X) = \operatorname{vec}^{-1}\left(\pmatrix{1&0&0\\0&1&0\\0&1&0\\0&0&1}X_h\right).
$$
So, on the one hand, we can write
$$
\operatorname{vec}(T(QX_h)) = \operatorname{vec} \circ \operatorname{vec}^{-1}\left(\pmatrix{1&0&0\\0&1&0\\0&1&0\\0&0&1}QX_h\right)
= \pmatrix{1&0&0\\0&1&0\\0&1&0\\0&0&1}QX_h.
$$
On the other, we have
$$
\operatorname{vec}(\bar Q T(X_h)\bar Q^T) = [\bar Q \otimes \bar Q]\operatorname{vec}(T(X_h)) = [\bar Q \otimes \bar Q]\pmatrix{1&0&0\\0&1&0\\0&1&0\\0&0&1} X_h,
$$
where $\otimes$ denotes the Kronecker product. Thus, given a matrix $Q$, $\bar Q$ is suitable if and only if
$$
[\bar Q \otimes \bar Q]\pmatrix{1&0&0\\0&1&0\\0&1&0\\0&0&1} = 
\pmatrix{1&0&0\\0&1&0\\0&1&0\\0&0&1}Q.
$$
Note that the first column of the left hand side will always have the form $(a^2,ab,ab,b^2)$, where $(a,b)$ is the first column of $\bar Q$. Thus, there can be no $\bar Q$ corresponding to
$$
Q = \pmatrix{1&0&1\\0&0&0\\1&0&1},
$$
since the corresponding column on the right hand side will be $(1,0,0,1)$.
