Question regarding two properties of controllable subspace in control theory In Mathematical Control Theory II: Behavioral Systems and Robust Control

Two claims: 


*

*$\langle A+BK | im B \rangle = \langle A | im B \rangle$

*$\langle A | im B \rangle$ is the smallest $A$-invariant subspace containing $B$
I am struggling to prove these two properties:



*

*$\langle A+BK | im B \rangle =  im B + (A+BK) im B + (A+BK)^2 im B + \ldots = c_1 B + (A+BK) c_2 B + \dots = c_1 B + c_2AB + c_2BKB + \ldots$


If  $\langle A+BK | im B \rangle = \langle A | im B \rangle$ is true, then the above term BKB vanishes or some how gets absorbed into another $B$. How can we show that this is the case?



*To show that $\langle A | im B \rangle$  is the smallest A-invariant subspace of $im B$


Suppose there exist an $A$-invariant subspace $V$ such that $im B \subseteq V$, then we must show $\langle A | im B \rangle \subseteq V$. Does anyone know how to proceed from here?
 A: Take $x \in \operatorname{Im} B \subseteq V$. Since $V$ is $A$-invariant, $Ax \in V$. Since $x$ is arbitrary $A \operatorname{Im} B \subseteq V$. Remember that $V$ is a subspace, so $\operatorname{Im} B + A\operatorname{Im} B \subseteq V$. Similarly, since $Ax \in V$, then $A^2x \in V$ and so on. So
$$\operatorname{Im} B + A\operatorname{Im} B + A^2 \operatorname{Im} B + \dots \subseteq V$$
At this point we use the fact that
$$A^n \operatorname{Im} B \subseteq \operatorname{Im} B + A\operatorname{Im} B + A^2 \operatorname{Im} B + \dots + A^{n-1} \operatorname{Im} B$$
which follows from the Cayley-Hamilton Theorem. Therefore
$$\langle A | \operatorname{Im} B \rangle = \operatorname{Im} B + A\operatorname{Im} B + A^2 \operatorname{Im} B + \dots + A^{n-1} \operatorname{Im} B \subseteq V$$
This means any $A$-invariant subspace that contains $\operatorname{Im} B$ should also contain $\langle A | \operatorname{Im} B \rangle$.
We know that $\operatorname{Im} (B X) \subseteq \operatorname{Im} B$ for any $X$ with compatible size. Therefore, your calculations show that $\langle A +BK | \operatorname{Im} B \rangle \subseteq \langle A | \operatorname{Im} B \rangle$. It is also easy to see that $\langle A +BK | \operatorname{Im} B \rangle$ is $A$-invariant. So using the previous fact, we can conclude that $\langle A | \operatorname{Im} B \rangle \subseteq \langle A +BK | \operatorname{Im} B \rangle$
