A condition on controllability based on invariant subspace

Question

Let A be a $n \times n$ matrix, we say that a linear subspace of $\mathbb{R}^n$ is invariant under $A$ if every vector $x$ in that subspace has the property that $Ax$ also belongs to that subspace. Show that the system $$\dot{x}(t)=Ax(t)+bu(t)~,~b \in \mathbb{R}^n$$is controllable if and only if $b$ doesnot belong to invariant subspace of $A$.

Answer 1

If we can work from the condition regarding the rank of the controllability matrix, then it suffices to note that the column-space of the controllability matrix $[b,Ab,\dots,A^{n-1}b]$ is the smallest $A$-invariant subspace that contains $b$. This invariant subspace will be a proper subspace of $\Bbb R^n$ if and only if the dimension of this subspace is less than $n$.

We can show that the column-space is the smallest invariant subspace that contains $b$ as follows: to show that the column-space is invariant, note that every element of the column space can be written in the form $$ x = \sum_{i=0}^{n-1} a_i A^ib $$ for some coefficients $a_i \in \Bbb R^n$. Thus, it suffices to show that for $0 \leq i \leq n-1$, $A(A^ib)$ is an element of this column space. It's clear that when $i < n- 1$, $A(A^ib) = A^{i+1}b$ will also be an element of this column space. For the case of $i = n$, note that by the Cayley-Hamilton theorem we must have $$ A^n = c_0I + c_1 A + \cdots + c_{n-1}A^{n-1} $$ for some coefficients $c_i \in \Bbb R$. It follows that $A(A^{n-1}b)$ is an element of this column space since $$ A(A^{n-1}b) = A^n b = c_0b + c_1 Ab + \cdots + c_{n-1}A^{n-1}b $$ which is an element of the column space.

To show that the column-space is the smallest such invariant space, it suffices to note that an invariant space that contains $b$ must (by definition) contain $Ab$, and must therefore contain $A(Ab) = A^2b$, and so on.