# Kalman Filter Derivation - Why does taking the derivative of the trace give minimum error and not maximum?

When deriving the Discrete Kalman Filter, there is an intermediate step where you take the derivative of the trace of $P_k$ and set it equal to 0:

$P_k = E[e_k e_k^T]$
$= P_k^- + K_kH_kP_k^-H_k^-K_k^T - P_k^-H_k^-K_k^T- K_kH_kP_k^-+K_kR_kK_k^T$

$d(trP_k)\over dK_k$ $= 2K_kH_kP_k^-H_k^T - 2(H_kP_k^-)^T +2K_kP_k = 0$

Why does using this method give us the minimum mean-squared error?

In terms of the system state $x_i$, we know the trace of $P_k$ is
$\sum E[(x_i-\hat x_i)^2] = \sum E[(x_i^2-2\hat x_ix_i^2+\hat x_i^2)]$

So why does setting the derivative of this equal to 0 and solving for $K_k$ give us a optimal Kalman Gain Matrix $K_k$ with minimal error, rather than maximal?

• I believe it should follow from convexity of squared error as a function of the gain matrix $K_k$. – Mark Jan 14 at 6:29