Positive Definitness of Noise Covariance Matrix in Kalman Filter
If a covariance matrix is positive definite it has no zero eigenvalue, so it is invertible. 

If you had an exact measurement, it would not have a covariance with itself(variance) or any other measurements, that will probably make it positive semi-definite, with a zero eigenvalue associated with that exact sensor.

For a covariance matrix

$$ A = \begin{pmatrix} A_1 &amp;0 &amp;A_2\\ 0 &amp; 0 &amp; 0\\ A_3 &amp; 0 &amp;A_4\end{pmatrix},$$

The eigenvalues will be determined by

$$ det(sI-A) = det\begin{pmatrix} sI-A_1 &amp;0 &amp;-A_2\\ 0 &amp; s &amp; 0\\ -A_3 &amp; 0 &amp;sI-A_4\end{pmatrix},$$

$$ det(sI-A) = (-1)^ks \,det\begin{pmatrix} sI-A_1 &amp;-A_2\\-A_3 &amp;sI-A_4\end{pmatrix},$$

So having an exact measurement would lead you to have a zero eigenvalue associated with that measurement. And the covariance matrix would not be invertible.