Growth of a vector when projecting onto multiple (non-orthogonal) eigenspaces at once.

Question

If we have a system of ODEs $\frac{dy}{dt}=Ay$ for some matrix $A$ with eigenvalues $\lambda_j$ and corresponding projection matrices $P_j$ onto the left-eigenspaces, we know that $\frac{d}{dt}|P_jy|=\text{Re}(\lambda)|P_jy|$ and so $P_jy(t)$ grows exponentially fast with parameter $\text{Re}(\lambda_j)$. If $\overline{P}$ is the projection onto multiple eigenspaces all with $\text{Re}(\lambda_j)\in[a,b]$, can we show that $\frac{d}{dt}|P_jy|=c|P_jy|$ for some $c\in[a,b]$?

Answer 1

The answer is no. Consider \begin{equation*} A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}, \end{equation*} $c = d= 0$ and $\overline{P}=\mathrm{Id}$. Then \begin{equation*} e^{At} = \begin{bmatrix} 1 & t \\ 0 & 1 \end{bmatrix}. \end{equation*} Take $y(t) = e^{At} \, \mathrm{col}{(1,1)}$, that is, $y(t) = \mathrm{col}{(1 + t,1)}$.

Now, I do not know what norm you mean under $\lvert \cdot \rvert$. So, let us consider the three "most popular" norms:

1. The Euclidean norm: $\lvert y(t) \rvert = \sqrt{1 + (1+t)^2}$, and $\tfrac{d}{dt} \lvert y(t) \rvert = \tfrac{1 + t}{\sqrt{1 + (1+t)^2}}$ for $t \ge 0$.
2. The supremum norm: $\lvert y(t) \rvert = 1 + t$, and $\tfrac{d}{dt} \lvert y(t) \rvert \equiv 1$ for $t \ge 0$.
3. The Manhattan norm: $\lvert y(t) \rvert = 2 + t$, and $\tfrac{d}{dt} \lvert y(t) \rvert \equiv 1$ for $t \ge 0$.

Indeed, what one can hope for (in the above example) is the following result.

For each solution $\varphi(\cdot)$ and each $\epsilon > 0$ there is $C > 0$ such that $\lvert \varphi(t) \lvert \le C e^{- \epsilon t}$ for all $t \ge 0$.

But that would be a different subject.