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I have proved the following convergence, but I'm not convinced with my answer. Suppose we have the following $$\lim_{t \rightarrow \infty} \text{tr}\int_{0}^{t}e^{sQ^T}Pe^{sQ} ds$$ where tr is trace. I know $P$ is positive semi-definite, $Q$ is negative-definite. Let the Jordan normal form of $Q$ be $Q = SJS^{-1}$.

Step 1: bring trace inside and use cyclic permutation property $$\lim_{t \rightarrow \infty} \int_{0}^{t}\text{tr}(Pe^{sQ}e^{sQ^T}) ds$$ Step 2: $P$ is positive semi-definite, $Q$ negative-definite, so $e^{sQ}$ is positive semi-definite. Space of PSD Matrices has an inner product, and tr can be one, so apply Cauchy-Schwarz inequality and get $\text{tr}(Pe^{sQ}e^{sQ^T}) \le \sqrt{\text{tr}(P^2)\text{tr}((e^{sQ}e^{sQ^T})^2)}$. But since they are PSD, we can take out the squares and get $$\text{tr}(Pe^{sQ}e^{sQ^T}) \le \sqrt{\text{tr}(P)^2\text{tr}((e^{sQ}e^{sQ^T}))^2} = {\text{tr}(P)\text{tr}((e^{sQ}e^{sQ^T}))}$$

Step 3: Take out $\text{tr}(P)$ from the integral $$\lim_{t \rightarrow \infty} \int_{0}^{t}\text{tr}(Pe^{sQ}e^{sQ^T})ds \le \lim_{t \rightarrow \infty} \text{tr}(P)\int_{0}^{t}\text{tr}(e^{sQ}e^{sQ^T})ds$$

Step 4: Apply Step 2 to term inside integral $$\lim_{t \rightarrow \infty} \text{tr}(P)\int_{0}^{t}\text{tr}(e^{sQ}e^{sQ^T}) ds \le \lim_{t \rightarrow \infty} \text{tr}(P)\int_{0}^{t}\text{tr}(e^{sQ})\text{tr}(e^{sQ^T})ds$$ Step 5: Trace is sum of eigenvalues $$\lim_{t \rightarrow \infty} \text{tr}(P)\int_{0}^{t}\text{tr}(e^{sQ})\text{tr}(e^{sQ^T})ds = \lim_{t \rightarrow \infty} \text{tr}(P)\int_{0}^{t} 2\sum_{i=0}^N e^{s\lambda_i} $$ Step 6: Swap integral and summation, and integrate $$\lim_{t \rightarrow \infty} \text{tr}(P) 2\sum_{i=0}^N \int_{0}^{t} e^{s\lambda_i} = \lim_{t \rightarrow \infty} \text{tr}(P) 2\sum_{i=0}^N \frac{e^{t\lambda_i} - 1}{\lambda_i}$$ Step 7 All $\lambda_i < 0$ $\Rightarrow \lim_{t \rightarrow \infty} e^{t\lambda_i} \rightarrow 0$. Take out $-1$ from all $\lambda$ and we get $$\le 2*\frac{tr(P)}{|tr(Q)|}$$ Is there anything wrong with this proof that I'm not seeing?

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  • $\begingroup$ You say that your matrices are positive/negative etc... But then are you refering to symmetric matrices ? $\endgroup$
    – Bertrand R
    Aug 16 '13 at 15:38
  • $\begingroup$ @BertrandR I only know $P$ is symmetric, $Q$ is not symmetric but all eigenvalues of $Q$ are negative $\endgroup$
    – user90275
    Aug 16 '13 at 15:41
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    $\begingroup$ Ok let me write an answer, but your step 2 is wrong : why can you "take out the squares?". $\endgroup$
    – Bertrand R
    Aug 16 '13 at 15:43
  • $\begingroup$ I see. It should be sum of the squares of the eigenvalues which is greater than square of the sum. I can't keep the bound. $\endgroup$
    – user90275
    Aug 16 '13 at 15:50
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So first of all : $P$ being symmetric Possitive semi-definite, for all matrix $A$, $A^T P A$ is still symmetric positive semi-definite.

$e^{sQ^T}=(e^{sQ})^T$ shows that $tr(e^{sQ^T}Pe^{sQ})\geq 0$. We just need to have domination for the limit.

Then use Cauchy-Schwarz inequality in a different way (denote $\sqrt{P}$ a standard square root of $P$) : $$tr(e^{sQ}Pe^{sQ^T})=tr((\sqrt{P}e^{sQ})^T(\sqrt{P}e^{sQ})$$

Now we need to work on $tr((\sqrt{P}e^{sQ})^T(\sqrt{P}e^{sQ})=||\sqrt{P}e^{sQ}||^2$ : Froebenius norm verify : $||AB||\leq||A|| \times||B||$ (Use Cauchy Schwarz)

This way : $$tr((\sqrt{P}e^{sQ})^T(\sqrt{P}e^{sQ})\leq ||\sqrt{P}||^2||e^{sQ}||^2$$

Now we only have to worry about $||e^{sQ}||^2$ : This may be an overkill but I don't see how to do this otherwise if Q is not symmetric (in this case it is trivial because you can compute the norm) : $Q=D+N$ where $DN=ND$ and D diagonalizable with same eigenvalues than Q and N nilpotent (Dunford decomposition). Same reasoning shows (not obvious, $DN=ND$ crucial) : $$||e^{sQ}||^2\leq ||e^{sD}||^2 ||e^{sN}||^2$$ Roughly, the idea is that $||e^{sN}||^2$ is polynomial in s and $||e^{sD}||^2$ approximately $e^{2\lambda s}$ with $\lambda\in sp(Q)$.

$||e^{sN}||^2\in O(t^m)$ for some m is a polynomial in s (write the exponential serie to convince yourself).

$||e^{sD}||^2=||\Gamma^{-1}e^{s\Delta}\Gamma||^2$ where $\Delta$ is the diagonal with Q's eigenvalues. Then : $$||e^{sD}||^2\leq ||\Gamma^{-1}||^2||\Gamma||^2 ||e^{s\Delta}||^2$$ Of course $$||e^{s\Delta}||^2=\sum_{\lambda\in sp(Q)} e^{2s\lambda}$$

Finally :

$$tr((\sqrt{P}e^{sQ})^T(\sqrt{P}e^{sQ})\leq C R(s) \sum_{\lambda\in sp(Q)} e^{2s\lambda}$$ where C is a constant and R(s) is a polynomial in s.

Because $\lambda<0$ for all $\lambda\in sp(Q)$, you have a proper domination for the initial term, and then convergence of the quantity you were interested into

EDIT : There is a slight imprecision about the Dunford decomposition which can only be done in the complex case (or when you can make Q triangular in $\mathbb{R}$, but we can assume that this is the case given what the OP said, and otherwise work on $Re(\lambda)$

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