Ensure that a specific matrix decomposition is unique

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Consider a matrix $\Pi$ which is an $m \times m$ real-valued matrix. We define $r := Rank(\Pi) \in (0,m] \cap \mathbb{N}$. We can always use two matrices of dimension $m \times r$ to decompose $\Pi$:

$\Pi = \alpha \beta^T.$

Obviously, if $\Pi$ is of full rank, we have $r=m$ and imposing that either matrix is the identity matrix ensures a unique decomposition. I also know how to do it for another peculiar case. If $m=2$ and $r=1$, we can choose $\beta_1 = 1$ and this will ensure a unique decomposition. In this case:

$\begin{bmatrix} \pi_{1,1} & \pi_{1,2} \\ \pi_{2,1} & \pi_{2,2} \end{bmatrix} = \begin{bmatrix} \pi_{1,1} \\ \pi_{2,1} \end{bmatrix} \begin{bmatrix} 1 & \frac{\pi_{2,2}}{\pi_{2,1}}\end{bmatrix}$.

Note that this implies $\beta_2 = \frac{\pi_{2,2}}{\pi_{2,1}} = \frac{\pi_{1,2}}{\pi_{1,1}}$, hence one column is a scalar multiple of the other and the rank of $\Pi$ is indeed $1$.

Now, how do you do it for arbitrary $m$ and $r$? My problem is that I am coding an algorithm for a vector error correction model which requires these two matrices $\alpha$ and $\beta$. Hence, I need a general way to impose normalizations so that I can define $\alpha$ and $\beta$ as unique real-valued matrices no matter the size or rank of $\Pi$.

Thanks in advance. Any help is appreciated.

EDIT (additional details):

Kilian and Lutkepohl (2016) propose in their chapter 3 to use the following normalization:

$\beta := \begin{bmatrix} I_r \\ \beta_{(m-r)} \end{bmatrix}$.

where $\beta_{(m-r)}$ is an $(m-r) \times r$ matrix. They say it is always possible to write $\Pi$ using this $\beta$, though it might require rearranging the variables in the VECM. The model is:

$\Delta Y_t = \Pi Y_{t-1} + \sum_{j=1}^p A_p \Delta Y_{t-j} + \epsilon_t$

where $\epsilon_t$ is second-order stationary, $Y_t$ is an $m \times 1 $ matrix and $\Delta Y_t := Y_t - Y_{t-1}$. The idea of this model is that although at least some or even all of the $m$ variables in $Y_t$ might not be second order stationary, some linear combinations thereof might be.