Common factor in sum of transpose matrices

Question

I have the following linear matrix inequality:

$(B K_1)^T + BK_1 < -2A$

where B is 2x1, $K_1$ is 1x2 and A is 2x2.

Is it possible to find $K_1$ as a common factor in the left-hand side of the inequality? Thanks!

Answer 1

I assume here that the $A$ matrix $A$ is symmetric. It is known that such a matrix inequality has solutions if and only if $BB^T>0$ or $B_{\bot}^TAB_\bot<0$ where $B_\bot$ is a basis of left null-space of $B$.

When this is the case, all solutions $K_1$ write

$$K_1= B_r^+M+Z-B_r^+B_rZ$$

where $B=:B_\ell B_r$ where $B_\ell$ and $B_r$ are full rank factors, $M:=-R^{-1}B_\ell^T+R^{-1/2}L\Phi^{-1/2}$, $L$ is an arbitrary matrix such that $||L||<1$, and $R$ is an arbitrary symmetric positive definite matrix such that $$\Phi:=(B_\ell R^{-1}B_\ell^T-2A)^{-1}>0.$$

In all the expressions above, the superscript $+$ denotes the Moore-Penrose pseudoinverse.

For more details, see Theorem 2.3.12 in Skelton, Iwasaki and Grigoriadis, "A Unified Algebraic Approach to Linear Control Design".