I have a vector $\mathbf{x}=\begin{pmatrix}\mathbf{x}_1\\\mathbf{x}_2\end{pmatrix}$ where $\mathbf{x}_1 \in \mathbb{R}^{n}$ and $\mathbf{x}_1 \in \mathbb{R}^{m}$.
My question is this: how do I write $\mathbf{x}$ in terms of some linear combination of $\mathbf{x}_1$ and $\mathbf{x}_2$?
For example, if $n=m$, then one can write $\mathbf{x}=\begin{pmatrix} 1 \\ 0\end{pmatrix} \otimes \mathbf{x}_1+ \begin{pmatrix} 0 \\ 1\end{pmatrix} \otimes \mathbf{x}_2$. This doesn't work when $n \neq m$ because the dimensions of $\begin{pmatrix} 1 \\ 0\end{pmatrix} \otimes \mathbf{x}_1$ and $ \begin{pmatrix} 0 \\ 1\end{pmatrix} \otimes \mathbf{x}_2$ are different.
Cheers in advance.
It is possible to manipulate and represent this using linear operators. One way is to zero pad each vector and manipulate the aggregate of the Kronecker product sum using a linear operator. The required sum is represented as: $$ \mathbf{x} = f\left( \sum_k^N \mathbf{b}_k \otimes \hat{\mathbf{x}}_k\right) = {\mathbf{P}^T\left(\sum_k^N \mathbf{b}_k \otimes \hat{\mathbf{x}}_k\right)} = \begin{pmatrix} \mathbf{x}_1 \\ \vdots \\ \mathbf{x}_k \end{pmatrix} $$ where $\mathbf{b}_k \in \mathbb{R}^{k}$ are indicator vectors with an $i$th non-zero scalar $a_k$, $\hat{\mathbf{x}}_k = \begin{bmatrix} \mathbf{x}_{k_{m_k \times 1}} \\\mathbf{0}_{(g - m_k) \times 1}\end{bmatrix}$, with $m_k$ as dimension of the $k$th vector, and $g = max(m_1,\ldots,m_k)$. The mapping $f : \mathbb{R}^{(k\times g)} \mapsto \mathbb{R}^h$ can be represented using operator $\mathbf{P}$.
Take an example with three vectors, $k=3$ $$ \mathbf{x}_1 \in \mathbb{R}^l, ~\mathbf{x}_2 \in \mathbb{R}^m,~\mathbf{x}_3 \in \mathbb{R}^n ,\text{with} ~g = max(l,m,n), \text{and} ~h = l+m+n $$ Let $\mathbf{x} = \begin{pmatrix} \mathbf{x}_1 \\ \mathbf{x}_2 \\ \mathbf{x}_3 \end{pmatrix}$. Since $l \neq m \neq n$, zero pad each vector, $$ \hat{\mathbf{x}}_1 = \begin{bmatrix} \mathbf{x}_1 \\ \mathbf{0}_{(g-l) \times 1} \end{bmatrix} \quad \hat{\mathbf{x}}_2 = \begin{bmatrix} \mathbf{x}_2 \\ \mathbf{0}_{(g-m) \times 1} \end{bmatrix} \quad \hat{\mathbf{x}}_3 = \begin{bmatrix} \mathbf{x}_3 \\ \mathbf{0}_{(g-n) \times 1} \end{bmatrix} $$ Now perform the Kronecker sum; $$ \begin{pmatrix} a_1 \\ 0 \\ 0\end{pmatrix} \otimes \hat{\mathbf{x}_1} + \begin{pmatrix} 0 \\ a_2 \\ 0\end{pmatrix} \otimes \hat{\mathbf{x}_2} + \begin{pmatrix} 0 \\ 0 \\ a_3\end{pmatrix} \otimes \hat{\mathbf{x}_3} = \begin{pmatrix} a_1\hat{\mathbf{x}_1} \\ a_2\hat{\mathbf{x}_2} \\ a_3\hat{\mathbf{x}_3}\end{pmatrix} = \hat{\mathbf{x}} $$ If you apply operator $\mathbf{P} = {\begin{pmatrix} \mathbf{I}_{l\times l} & \mathbf{0}_{l \times m} & \mathbf{0}_{l \times n} \\ \mathbf{0}_{(g-l)\times l} &\mathbf{0}_{(g-l)\times m} &\mathbf{0}_{(g-l)\times n} \\ \mathbf{0}_{m\times l} &\mathbf{I}_{m \times m} &\mathbf{0}_{m\times n}\\ \mathbf{0}_{(g-m)\times l} &\mathbf{0}_{(g-m)\times m} &\mathbf{0}_{(g-m)\times n}\\ \mathbf{0}_{n\times l} &\mathbf{0}_{n\times m} &\mathbf{I}_{n\times n} \\ \mathbf{0}_{(g-n)\times l} &\mathbf{0}_{(g-n)\times m} &\mathbf{0}_{(g-n)\times n}\end{pmatrix}}_{3g \times h} $, where $\mathbf{I}$ is identity matrix, and $\mathbf{0}$ is zero matrix. Then it can be seen that $$ \mathbf{x} = \mathbf{P}^T_{3g \times h} \hat{\mathbf{x}}_{3g \times 1} = {\begin{pmatrix} a_1\mathbf{x}_1 \\ a_2\mathbf{x}_2 \\ a_3\mathbf{x}_3\end{pmatrix}}_{h \times 1} $$ If you plug in $a_k =1$ you get back your unscaled vector. The biggest effort is probably in constructing the $\mathbf{P}$ matrix, which may not have nice properties.