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Assume that a discrete-time system can be described by the following state-space equations

$x(k+1)=Ax(k)+Bu(k)+w(k)$

where the input signal $u(k)$ is stationary and ergodic with $\mathbf{E}[u(k)]=0$.

Define then the covariance matrix

$ R(k) := \mathbf{E}\Bigg[ \begin{bmatrix} x(k)\\u(k) \end{bmatrix}\begin{bmatrix} x(k)\\u(k) \end{bmatrix}^\top\Bigg] = \begin{bmatrix} r_{xx}(k) & r_{xu}(k)\\ r_{xu}^\top (k) & r_{uu}(k) \end{bmatrix} $

In particular, if $u$ is persistently exciting of order $n$ then $R(k)>0$ and in particular for the Sylvester Theorem,

$R_{uu}(k)= \mathbf{E}\Bigg[ \begin{bmatrix} u(k)\\ \vdots \\u(k+n-1) \end{bmatrix}\begin{bmatrix} u(k)\\ \vdots \\u(k+n-1) \end{bmatrix}^\top\Bigg]>0$.

I have two statements/questions:

1) if $u$ is PE(n) then it is also PE(n-1) so is it true that

$r_{uu}(k)=\mathbf{E}[u(k)u^\top (k)] >0 \quad ?$

2) Knowing that $r_{uu}(k)>0$, is it possible to verify that also $r_{xx}(k)=\mathbf{E}[x(k)x^\top (k)]>0$?

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If your covariance matrix is PD, i.e. $R(k)>0$, for Sylvester criterion $r_{uu}(k)>0$ and so also $r_{xx}(k)>0$.

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