If $X \in \mathcal{X} = \{x_1,x_2\}$, for some $x_1,x_2 \in \mathbb{R}^{n \times 1}$ and $Y \in \mathcal{Y} = \{y_1,y_2\}$, for some $y_1,y_2 \in \mathbb{R}^{m \times 1}$, can I say that the stacked vector $Z = (X^T,Y^T)^T$ belongs to $\mathcal{X} \times \mathcal{Y}$ ?

I think not, because $Z = (X^T,Y^T)^T \in \mathcal{Z} = \left \{\ \begin{pmatrix} x_1 \\ y_1 \end{pmatrix}, \begin{pmatrix} x_1 \\ y_2 \end{pmatrix}, \begin{pmatrix} x_2 \\ y_1 \end{pmatrix}, \begin{pmatrix} x_2 \\ y_2 \end{pmatrix} \right\}$, and

$ \mathcal{X} \times \mathcal{Y} = \{ (x_1,y_1), (x_1,y_2), (x_2,y_1), (x_2,y_2) \}. $

Then, is there a way to write $\mathcal{Z}$ in terms if $\mathcal{X}$ and $\mathcal{Y}$ ?


Firstly, it seems that you've already defined the transpose $$(\cdot)^T : \bigcup_{n=1}^\infty (\mathbb{R}^{n}\cup\mathbb{R}^{n \times 1}) \rightarrow \bigcup_{n=1}^\infty (\mathbb{R}^{n}\cup \mathbb{R}^{n \times 1})$$

for all row and column vectors (I'm just using plain $\mathbb{R}^n$ for the "rows" rather than $\mathbb{R}^{1 \times n}$).

So the image of subsets $A \subseteq \bigcup_{n=1}^\infty (\mathbb{R}^{n}\cup \mathbb{R}^{n \times 1})$ under $(\cdot)^T$ are defined already as

$$(\cdot)^T(A) = \{x^T \mid x \in A\}$$

so for readability you might as well define

$$A^T = (\cdot)^T(A)$$

This means that you can define $\mathcal{Z}$ as

$$\mathcal{Z} = (\mathcal{X}^T \times \mathcal{Y}^T)^T$$


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