Let $$ A=\left[ \begin{matrix} A_1& A_3\\ 0& A_2\\ \end{matrix} \right]\quad \text{and} \quad B=\left[ \begin{array}{c} B_1\\ B_2\\ \end{array} \right], $$and for any eigenvalue $s$ of $A$, we have $$ \mathrm{rank}\,\left[ A-sI_n,B \right] =n. $$ Prove there are a real matrix $K\in \mathbb{R}^{r\times n}$ and invertible matrix $T\in \mathbb{R}^{n\times n}$, such that $$ T\left( A+BK \right) T^{-1}=\left[ \begin{matrix} A_1& 0\\ 0& \bar{A}_2\\ \end{matrix} \right] ,\qquad TB=\left[ \begin{array}{c} \bar{B}_1\\ B_2\\ \end{array} \right], $$ where $A\in \mathbb{R}^{n\times n},B\in \mathbb{R}^{n\times r}$. Meanwhile, $\bar{A}_2$ and $\bar{B}_1$ is real matrix of the proper dimension, $\bar{A}_2$ and $A_1$ Have eigenvalues that are not identical to each other.

This problem is about controllability of linear system and pole assignment, so I think we may try controllability canonical form, or let $K=(K_1,K_2)$, then determine the $K_1,K_2$ and $T$, but it seems so difficult.

  • $\begingroup$ What are the constraints on $A_1$, $A_2$, $A_3$, $B_1$ and $B_2$? Because if they are all filled with zeros then all eigenvalues will always be equal to zero as well. Or is it also allowed that $\bar{A}_2\,\in\,\mathbb{R}^{0\times0}$? $\endgroup$ – Kwin van der Veen Sep 30 '17 at 12:11
  • $\begingroup$ @KwinvanderVeen It is impossible? We get s=0, then the rank isn't $n$? $\endgroup$ – Eufisky Sep 30 '17 at 12:19
  • $\begingroup$ I skipped over your part about the Hautus test/lemma, so the pair $(A,B)$ is controllable. This already implies that you can place the eigenvalues/poles anywhere you like and it should be possible to separate them into $A_1$ and $A_2$ with a similarity transformation. But a formal proof might take a bit more work. $\endgroup$ – Kwin van der Veen Sep 30 '17 at 17:19

The statement about the rank of $\left[A - s\,I, B\right]$ implies that the pair $(A,B)$ is controllable. Therefore the poles of the resulting matrix should be able to be placed anywhere. The similarity transformation should allow us to separate the eigenvalues/modes and therefore achieve the stated goal

$$ T \left(A + B\,K\right) T^{-1} = \begin{bmatrix} A_1 & 0 \\ 0 & \bar{A}_2 \end{bmatrix}, \quad T\,B = \begin{bmatrix} \bar{B}_1 \\ B_2 \end{bmatrix}, $$

given that

$$ A = \begin{bmatrix} A_1 & A_3 \\ 0 & A_2 \end{bmatrix}, \quad B = \begin{bmatrix} B_1 \\ B_2 \end{bmatrix}. $$

For solving this it is easier to separate the problem in smaller problems. For this I will define $T$ and $K$ as

$$ T = \begin{bmatrix} T_1 & T_2 \\ T_3 & T_4 \end{bmatrix}, \quad K = \begin{bmatrix} K_1 & K_2 \end{bmatrix}. $$

The last goal requires

$$ T\,B = \begin{bmatrix} T_1\,B_1 + T_2\,B_2 \\ T_3\,B_1 + T_4\,B_2 \end{bmatrix} = \begin{bmatrix} \bar{B}_1 \\ B_2 \end{bmatrix}. $$

The bottom half of this goal might have infinitely many solution if there is any overlap in the span of $B_1$ and $B_2$. But this problem should be solvable in general, in which case $T_3=0$ and $T_4=I$ should always solve it. Using this then the inverse of $T$ can shown to be

$$ T^{-1} = \begin{bmatrix} T_1 & T_2 \\ 0 & I \end{bmatrix}^{-1} = \begin{bmatrix} T_1^{-1} & -T_1^{-1}\,T_2 \\ 0 & I \end{bmatrix}. $$

Since nothing is specified about $\bar{B}_1$ then $T_1$ and $T_2$ could be anything for now as long as $T_1$ is invertible. The left hand side of the first goal can now be written as

$$ T \left(A + B\,K\right) T^{-1} = \begin{bmatrix} T_1\,A_1\,T_1^{-1} + \bar{B}_1\,K_1\,T_1^{-1} & T_1\,A_3 + T_2\,A_2 - T_1\,A_1\,T_1^{-1}\,T_2 + \bar{B}_1\left(K_2 - K_1\,T_1^{-1}\,T_2\right) \\ B_2\,K_1\,T_1^{-1} & A_2 + B_2\left(K_2 - K_1\,T_1^{-1}\,T_2\right) \end{bmatrix}. $$

It can be shown that the top and bottom left half can be set equal to the goal by using $T_1=I$ and $K_1=0$. This allows the top and bottom right of the first goal equation to be simplified to

$$ \begin{bmatrix} A_3 + T_2\,A_2 - A_1\,T_2 + \bar{B}_1\,K_2 \\ A_2 + B_2\,K_2 \end{bmatrix} = \begin{bmatrix} 0 \\ \bar{A}_2 \end{bmatrix}. $$

For the bottom half of this equation you could just use a pole placement algorithm to find $K_2$, such that none of the poles match those of $A_1$. This should be possible since the pair $(A_2,B_2)$ should be controllable. The top half can then be rewritten as

$$ A_3 + T_2\,\bar{A}_2 - A_1\,T_2 + B_1\,K_2 = 0, $$

which can be transformed into a Sylvester equation

$$ A\,X + X\,B = C, $$

with $X = T_2$, $A = -A_1$, $B = \bar{A}_2$ and $C = -A_3 - B_1\,K_2$. This equation has an unique solution for $X$ when $A$ and $-B$ do not have a common eigenvalue. This is an identical constraint as mentioned by your problem statement.

So to solve this problem you can first do a pole placement with the pair $(A_2,B_2)$ to find $K_2$, avoiding the eigenvalues of $A_1$. And then solve a Sylvester equation after substituting in this obtained $K_2$ in order to find $T_2$. Using these values then the final solution can then be expressed using

$$ T = \begin{bmatrix} I & T_2 \\ 0 & I \end{bmatrix}, \quad K = \begin{bmatrix} 0 & K_2 \end{bmatrix}. $$


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