# Covariance matrix and persistence of excitation of input

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 $$r_{xx}(k)=\mathbf{E}[x(k)x^\top (k)]>0$$?

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