0
$\begingroup$

There are matrices; $x_k$, $u_k$, $A_k$, $B_k$, and $Q_k$ whose dimension are $[n\times1]$, $[m\times1]$, $[n\times n]$, $[n\times m]$, and $[n\times n]$, respectively, such that $$x_{k+1}=A_kx_k+B_ku_k+w_k.$$ Another condition is that the matrices $Q_k$ are positive semidefinite symmetric.


My textbook jumped from the equation (1) to the equation (2), regarding very natural and of course.

$$ \mathbf{E}\left(A_{N-1}x_{N-1}+B_{N-1}u_{N-1}+w_{N-1}\right)'Q_N\left(A_{N-1}x_{N-1}+B_{N-1}u_{N-1}+w_{N-1}\right) \tag1 $$

$$ u'_{N-1}B'_{N-1}Q_NB_{N-1}u_{N-1}+2x'_{N-1}A'_{N-1}Q_NB_{N-1}u_{N-1}+x'_{N-1}A'_{N-1}Q_NA_{N-1}x_{N-1}+\mathbf{E}\left(w'_{N-1}Q_Nw_{N-1}\right) \tag 2 $$


But I don't say before that the matrices $w_k$ whose each element has zero-mean distribution. I can follow almost steps except that how can it become $$x'_{N-1}A'_{N-1}Q_NB_{N-1}u_{N-1}+u'_{N-1}B'_{N-1}Q_NA_{N-1}x_{N-1} = 2x'_{N-1}A'_{N-1}Q_NB_{N-1}u_{N-1}.$$

I can guess it could be from the semi-definite symmetric property of $Q_k$. Can someone explain with a detail? Thank you.

$\endgroup$
0
$\begingroup$

The trick here is that the $Q$ matrix is symmetric, so the two terms on the left side of your equation are transposes of each other. Since both terms are scalars (1 by 1 matrices), they're equal, and you immediately get the expression on the right hand side.

$\endgroup$
  • $\begingroup$ If $Q$ is symmetric, $A'QB=B'QA$? $\endgroup$ – Danny_Kim Jul 21 '16 at 1:40
  • 1
    $\begingroup$ No. If $Q$ is symmetric, then $ (A^{T}QB)^{T}=B^{T}QA$. If it also happens that $A^{T}QB$ is a 1 by 1 matrix, then $A^{T}QB=(A^{T}QB)^{T}=B^{T}QA$. $\endgroup$ – Brian Borchers Jul 21 '16 at 3:06
  • $\begingroup$ Ahha, very clear for me! Thank you very much. $\endgroup$ – Danny_Kim Jul 21 '16 at 5:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.