Given $k$ column vectors $$x_1,x_2,\dots,x_k$$ each in $\mathcal{R}^{d}$, suppose we want to obtain a single column vector $\bar{x} \in \mathcal{R^{kd}}$ which is the vertical concatenation of the $x_i$ such that

$$ \bar{x} = x_1 \oplus x_2 \oplus \dots \oplus x_k=\begin{pmatrix} x_1\\x_2\\\vdots\\x_k \end{pmatrix} $$

Is there a clever way to express this operation using linear algebra (i.e. using Matrix-Matrix operations)? For instance, placing the $x_i$ as the columns of some matrix, and applying a transformation.1

One immediate issue is that the underlying concatenation operation is not a typical linear operation...

1 If $M$ is this matrix, it can be vectorized columnwise in MATLAB using $M(:)$ operation

  • $\begingroup$ I think the best way to express it is as you have done: $\overline{x} = x_1 \oplus x_2 \oplus \ldots \oplus x_k$. $\endgroup$ – David Mehrle Jul 21 '14 at 3:00
  • $\begingroup$ I don't see why so many people want things expressed algebraically when such is bound to be tedious, complicated, artificial, and just using language would be much more natural. $\endgroup$ – blue Jul 21 '14 at 3:12
  • $\begingroup$ @blue, I am just interested. $\endgroup$ – jII Jul 21 '14 at 3:15

This may or may not be helpful.

For $m \in \{1, \dots, k\}$, let $A_m = [(a_m)_{i,j}]$ be the $kd\times d$ matrix given by

$$(a_m)_{i,j} = \begin{cases}1 & i - d(m-1) = j\\ 0 & \text{otherwise}. \end{cases}$$

Then, if you set $\overline{x} = F(x_1, \dots, x_k)$, we have $F(x_1, \dots, x_k) = A_1x_1 + \dots + A_kx_k$.

Despite the definition above, $A_m$ is quite a simple matrix. It consists of $(m-1)$ copies of the $d\times d$ zero matrix stacked vertically, followed by a single copy of the $d\times d$ identity matrix, then $k - m$ copies of the $d\times d$ zero matrix.

  • $\begingroup$ Quite creative! $\endgroup$ – jII Jul 21 '14 at 3:25

For the record, this is called matrix vectorization: a linear transformation which converts the matrix into a column vector, denoted by vec(A). For the properties, and hopefully a representation you're after, see the link and references therein.


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