Trying to follow a derivation in Lai & Xing's "Statistical Models and Methods for Financial Markets", regarding multivariate time series:
Let $$\Delta \vec{y}_t = \vec{\mu} + \Pi \vec{y}_{t-1}$$ (omitted some extra terms for simplicity, won't matter), be a VAR(1) process (vector-autoregressive), with $\vec{y}_t$ of dimension $n$.
Now there's a few special cases. If $\Pi$ is full rank and $\vec{\mu} = 0$, then the book says the time series is stationary (given a few other conditions).
If $\Pi$ is rank 0, on the other hand, then $\Pi = 0$ and really we have a random walk with drift.
The interesting case comes if $0 < \text{rank}(\Pi) < n.$ In this case, there exists a full-rank factorization with $n \times r$ matrices $\alpha, \beta$ such that $$\Pi = \alpha \beta^T$$ and both $\alpha$ and $\beta$ have rank $r$.
The claim now is that the column vectors of $\beta$ have the property that $\vec{\beta_i}^T \vec{y}_t$ is weakly stationary, i.e. the column-vectors of $\beta$ are cointegrating vectors.
I can't directly see that, so I thought I'd try to prove it. If we define $\vec{z}_t = \beta^T \vec{y}$, we have
$$\Delta \vec{z}_t = \beta^T \vec{\mu} + \beta^T \alpha \vec{z}_{t-1}$$
Now, I see that $\vec{z}_t$ will be a VAR(1) of dimension $r$, and it's matrix $\tilde \Pi := \beta^T \alpha$ is an $r \times r$ matrix, but I don't see how that shows it to be stationary. There's no guarantee that it'll be of full rank, right? What am I missing here?
I'd say that you're correct. There's no guarantee that $\beta^Ty_t$ is stationary. The following example illustrates this: $$ \Delta\begin{pmatrix} y_{1t}\\y_{2t} \end{pmatrix}= \begin{pmatrix} 0&0\\1&0 \end{pmatrix}\begin{pmatrix} y_{1t}\\y_{2t} \end{pmatrix}+\begin{pmatrix} \epsilon_{1t}\\\epsilon_{2t} \end{pmatrix}. $$ The full-rank decomposition would be $$ \begin{pmatrix} 0&0\\1&0 \end{pmatrix}=\begin{pmatrix} 0\\1 \end{pmatrix}\begin{pmatrix} 1&0 \end{pmatrix}\equiv\alpha\beta^T, $$ with $$ \beta^T\alpha=\begin{pmatrix} 1&0 \end{pmatrix}\begin{pmatrix} 0\\1 \end{pmatrix}=0. $$ Then $z_t=\beta^Ty_t=y_{1t}$, but $y_{1t}$ isn't stationary, it's a simple random walk $\Delta y_{1t}=\epsilon_{1t}$. So, if $\beta^T\alpha$ is degenerate then there could be less than $r$ cointegrating vectors (in the example above there's none). One can impose certain conditions in order to make $\mathop{\mathrm{rank}}\beta^T\alpha=r$. I can come up with the following: matrix $\Pi$ has to have the same algebraic and geometric multiplicities of zero eigenvalue. Or $\mathop{\mathrm{Im}}\Pi\cap\mathop{\mathrm{Ker}}\Pi=\{0\}$. Or something equivalent.