Please help me with the following proof:
Suppose $\dot x=f(x(t))$ and suppose that we have:
$$ \frac{d}{dt}\left( x(t)^TPx(t) \right)\le -x(t)^TQx(t) $$
where $P$ and $Q$ are symmetric positive definite matrices and the time derivatives are taken along solutions of the system. Prove that under this condition the system is exponentially stable in the sense that solutions satisfy:
$$ \|x(t)\|\le xe^{-\mu t}\|x(0)\| $$
for some $c,\mu>0$.
My attempt:
The result is trivial for $P=Q=I$ (identity matrix). However, we have to show this for the general case. Since $P$ and $Q$ are symmetric, they are diagonalizable. So there exist transformation matrices $T_P$ and $T_Q$ such that:
$$ P=T_P\Lambda_P T_P^{-1} $$
$$ Q=T_Q\Lambda_Q T_Q^{-1} $$
where $\Lambda_P$ is a diagonal matrix with eigenvalues of $P$ along the diagonal and $\Lambda_Q$ is a diagonal matrix with eigenvalues of $Q$ along the diagonal. Since $P$ and $Q$ are positive definite, we know that these eigenvalues are all $>0$. Therefore we can write:
$$ \frac{d}{dt}\left( x(t)^TT_P\Lambda_P T_P^{-1}x(t) \right)\le -x(t)^TT_Q\Lambda_Q T_Q^{-1}x(t) $$
$$ \Rightarrow \dot x(t)^TT_P\Lambda_P T_P^{-1}x(t)+x(t)^TT_P\Lambda_P T_P^{-1}\dot x(t)\le -x(t)^TT_Q\Lambda_Q T_Q^{-1}x(t) $$
Since we are dealing with expression $\in\mathbb R$ in the above inequality, we have:
$$ \Rightarrow 2\dot x(t)^TT_P\Lambda_P T_P^{-1}x(t)\le -x(t)^TT_Q\Lambda_Q T_Q^{-1}x(t) $$
$$ \Rightarrow 2f(x(t))^TT_P\Lambda_P T_P^{-1}x(t)\le -x(t)^TT_Q\Lambda_Q T_Q^{-1}x(t) $$
However, at this point I get stuck. Since $T_Q\ne T_P$ generally, I cannot simply define a new coordinate $z=Tx$ such that only the diagonal matrices are left, where the answer will then follow directly by considering $n$ decoupled inequalities (assuming $x\in\mathbb R^n$). Could you please help me out how to continue? Or how to do the proof another way? Thanks!
Use $T=P^{1/2}$ in $z=Tx$ to get $$ \frac{d}{dt}(z^Tz)\le -z^T(P^{-1/2}QP^{-1/2})z\le-\lambda_{min}(P^{-1/2}QP^{-1/2})·z^Tz $$
Because $P$ is SPD there exists a unique SPD square root $P^{1/2}$. Then also $P^{-1/2}QP^{-1/2}$ is SPD and has a minimal positive eigenvalue $\mu=\lambda_{min}$. From the inequality above one gets trivially $$ z(t)^Tz(t)\le e^{-\mu t}z(0)^Tz(0) $$ The conclusion is reached by observing that $$ x^Tx=z^TP^{-1}z\le \frac1{λ_{min}(P)}z^Tz $$ which gives the multiplicative constant as, for instance, $$ c=\frac{λ_{max}(P)}{λ_{min}(P)}x(0)^Tx(0) $$