The question is : Prove that (A, B) is controllable if and only if (A + BK, B) is controllable for all K.
My proof thus far:
- Let $u=kx +v$
- Consider the Im(Qc) = Im(B) + (A+BK)*Im(B) + ... + (A+BK)^n-1 * Im(B)
- I know that it can be shown that for any vector 'v' contained in R^n that Im(B) + (A+BK)v = Im(B) + Av...however I am not sure how to show this
My idea is to show that Im(B) + (A+BK)v is contained in Im(B) + Av and vice versa. Any help would be greatly appreciated
Take $y,z \in \operatorname{Im} B$. We need to show that $y + (A+BK)z \in \operatorname{Im} B + A \operatorname{Im} B$, i.e. there exist $u,v \in \operatorname{Im} B$ such that $y+ (A+BK)z = u + Av$. Indeed, take $v=z$ and $u = y+BKz \in \operatorname{Im} B$.
For the other direction let $u,v \in \operatorname{Im} B$ arbitrary and show that there exist $y,z \in \operatorname{Im} B$ such that $u + Av = y + (A+BK)z$. Taking $z=v$ and $y=u-BKv \in \operatorname{Im} B$ we can conclude the result.