# Simultaneously bounding stable and unstable components

I am reading a passage from Perko's book about the Stable Manifold Theorem. Here is the problem:

Let $$\dot x = f(x)$$ be a system where $$f: E \subset \mathbb{R}^{n} \rightarrow \mathbb{R}^n$$ (with $$E$$ open and containing the origin). The linearization of this system at $$0$$ is $$\dot x = Ax$$ where $$A = Df(0)$$. Suppose $$A$$ has $$k$$ eigenvalues $$\lambda_{j}$$ with negative real part and $$n - k$$ eigenvalues $$\lambda_{j}$$ with positive real part. Then we can write the eigendecomposition as $$A = C\Lambda C^{-1}$$ with

$$\Lambda = \begin{bmatrix} P & 0 \\ 0 & Q \end{bmatrix}$$, where $$P$$ is the matrix for the first $$k$$ eigenvalues and $$Q$$ is the matrix for the remaining $$n - k$$ eigenvalues. For the eigenvalues in $$P$$, choose $$\alpha > 0$$ sufficiently small so that $$Re(\lambda_{j}) < -\alpha < 0$$.

If we define $$U(t) = \begin{bmatrix} e^{Pt} & 0 \\ 0 & 0 \end{bmatrix}$$ and $$V(t) = \begin{bmatrix} 0 & 0 \\ 0 & e^{Qt} \end{bmatrix}$$, then $$\dot U = \Lambda U$$ and $$\dot V = \Lambda V$$.

Now we get to something that Perko claims but does not prove: There exists $$K > 0$$ sufficiently large and $$\sigma > 0$$ sufficiently small that

$$\|U(t)\| \leq Ke^{-(\alpha + \sigma)t}$$ for $$t \geq 0$$ and $$\|V(t)\| \leq Ke^{\sigma t}$$ for $$t \leq 0$$.

There is a discrete version of the first part discussed here, but that problem only addresses a matrix $$A$$ with no unstable subspace. Can someone explain how to prove the full claim here? Thanks.

• Why is $A$ necessarily disgonalisable? – copper.hat Apr 20 at 16:59
• For sure, Perko does not assume diagonalizability. A general Jordan form could be used for $\Lambda$, $P$, and $Q$. I don't know how to prove the claim even for $A$ diagonalizable, though. – user231 Apr 20 at 17:01

## 1 Answer

Note that if $$\operatorname{re} \lambda < \beta$$ for all $$\lambda \in \sigma(B)$$, then there is some $$K$$ such that $$\|e^{Bt}\| \le K e^{\beta t}$$ for all $$t \ge 0$$. (This is true even if $$B$$ is not diagonalisable.)

Let $$q = \min_{\lambda \in \sigma(Q)} \lambda$$, $$p = \max_{\lambda \in \sigma(P)} \lambda$$. Note that $$p < -\alpha < 0 < q$$.

Pick any $$\sigma$$ such that $$p < -(\alpha+\sigma) < 0 < \sigma < q$$.

Then there is a $$K$$ such that $$\|e^{Pt} \| \le K e^{-(\alpha+\sigma)t}$$ and $$\|e^{-Qt}\| \le K e^{-\sigma t}$$ for all $$t \ge 0$$.