0

enter image description here

Above is an excerpt from page 437 of Chapter 11 about Kalman filters in the book "Control System Design" written by Bernard Friedland. The symbol $I$ denotes the identity matrix, $K$ denotes the gain matrix of the Kalman filter, and $C$ denotes the measurement matrix linking the $l$-vector $Y$ and $k$-vector $X$. Given that $C\hat K=I$, equation (11.56) demonstrates that $I-\hat K C$ is an idempotent matrix. However, I could not understand the inference about its rank and the similarity transformation equation (11.57).

Firstly, why from the full rank $l$ of C can we conclude that $I-\hat K C$ is also full rank $l$?

Secondly, why does its similarity transformation, equation (11.57), claim that the transformed matrix has $l$ 0 eigenvalues and $k-l$ 1 eigenvalues?

Thirdly, why the matrix $U_l$ is needed in equation (11.58) to build the matrix T?

8cold8hot
  • 217
  • Since $(\hat{K}C)^2=\hat{K}C$ we have $\hat{K}C(I-\hat{K}C)=0$ so I would expect the nullspace of $I-\hat{K}C$ to have rank at least $l$. – CyclotomicField May 26 '23 at 23:51
  • Thanks for the comment. So does it mean that I should work on proving that $I-\hat K C$ has the rank k-l instead of the full rank l? It seems that it would automatically solve the second question as well by combining the verdict in this thread: https://math.stackexchange.com/questions/101512/proving-the-trace-of-an-idempotent-matrix-equals-the-rank-of-the-matrix?noredirect=1&lq=1 – 8cold8hot May 28 '23 at 20:27
  • Yeah I believe if can show that it has rank $k-l$ that would be sufficient. – CyclotomicField May 28 '23 at 21:02

0 Answers0