Rank of the matrix product $C e^{At} B$

Question

Let $A \in \mathbb R^{n \times n}$. Fix $m<n$ and let $B \in \mathbb R^{n \times m}$, $C \in \mathbb R^{(m-1) \times n}$ be two matrices with full rank. What I am interested in is the matrix product of the three matrices $C e^{At} B$, which is $(m-1)$ by $m$.

I have observed in examples that if we have (cf. controllability matrix)

\begin{align*} \text{rank } C \begin{pmatrix} B & AB & \dots & A^{n-1}B \end{pmatrix} = m-1 \end{align*}

then we also have

\begin{align*} \text{rank } C e^{At} B = m-1. \end{align*} Is this somehow obvious? How could one prove it? Is it even possible that the rank of the two matrices $C \begin{pmatrix} B & AB & \dots & A^{n-1}B \end{pmatrix}$ and $C e^{At} B$ is always the same? How is this proved?

Answer 1

It follows from the Cayley-Hamilton Theorem, which states that every matrix satisfies its characteristic polynomial. This means

$$A^n + a_{n-1} A^{n-1} + \dots + a_1 A + a_0 I = 0$$

where $A$ is an $n \times n$ matrix. The most important implication of this is that every power of a matrix can be written as the sums of the powers less than $n$. This means

$$e^{At} = \alpha_{n-1}(t) A^{n-1} + \dots \alpha_1(t) A + \alpha_0(t)$$

where $\alpha_i : \mathbb{R} \to \mathbb{R}$ are some scalar functions. Since scalars do not affect the rank your question is somewhat obvious.

Answer 2

Okay, let me also briefly summarize, what I have gotten so far:

So, first of all, like obareyy correctly explains, is that we can write

\begin{align*} Ce^{At}B = C \left( \sum_{k=0}^\infty A^k \frac{t^k}{k!} \right) B. \end{align*}

Since, we want to investigate the image space of $Ce^{At}B$ let us observe $x \in \text{Image}(Ce^{At}B)$, i.e. $x = Ce^{At}B \xi$. By definition of $e^{At}$ we have that $x = \lim_{N \to \infty} C (\sum_{k=0}^N A^k \frac{t^k}{k!})B) \xi =: \lim_{N \to \infty} x_N$.

Clearly, $x_N$ is a linear combination of the columns of $CB, CAB, \dots, CA^{N}B$. By virtue of the Cayley-Hamilton theorem, we can further conclude that $x_N$ is a linear combination of the columns of $CB, CAB, \dots, CA^{n-1}B$. In other words $x_N \in \text{Image} \begin{pmatrix} CB & CAB & \dots CA^{n-1}B \end{pmatrix}$ and by taking the limit we have:

$\text{Image } Ce^{At}B$ is a subspace of $\text{Image} \begin{pmatrix} CB & CAB & \dots CA^{n-1}B \end{pmatrix}$.

But what I want is this question:

Question: For given $A, B, C$ such that $\begin{pmatrix} CB & CAB & \dots CA^{n-1}B \end{pmatrix}$ has full row rank, for which $t \ge 0$ can we guarantee that $Ce^{At}B$ is also full row rank?

Answer 3

Your conjecture is not correct in general. Here is a counterexample. Let $C=(I_{m-1},0)$, $B$ be the matrix with a $1$ at the top left corner and $0$ elsewhere, and $A$ be the circulant permutation matrix $$ \pmatrix{0&&&1\\ 1&\ddots\\ &\ddots&\ddots\\ &&1&0}. $$ Then $C(B, AB, A^2B, \ldots, A^{n-1}B)$ contains $I_{m-1}$ as a submatrix and hence it has full rank. However, as $B$ has rank 1, $Ce^{At}B$ has at most rank 1 for every $t$. So, when $m-1\ge2$, $Ce^{At}B$ always has deficient rank.