Dynamical changing of an eigenvector Consider a matrix $A\in\mathbb{R}^{n\times n}$. One of the eigenvalues of $A$ is zero and all the others are positive. Suppose $w\in\mathbb{R}^n$ is an eigenvector with the zero eigenvalue, i.e,
$$Aw=0$$
Suppose $A$ is a function of time $t$. Hence the time derivative of $A$ and $w$ satisfy
$$(Aw)'=\dot{A}w+ A\dot{w}=0$$
Has anybody encountered similar problems as below: if $\|\dot{A}\|$ is sufficiently small, under what kind of conditions $\|\dot{w}\|$ is also very small? Or in other words, if $\|\dot{A}\|$ is bounded from upper, when is $\|\dot{w}\|$ also bounded from upper?
PS: I omit some specifics of $A$ above. In case you may be familiar with graph theory, I am considering a transpose of a Laplacian matrix $A=L^T$. From graph theory, if the underlying graph is strongly connected, the Laplacian has a positive left eigenvector with the zero eigenvalue. By positive eigenvector, I mean all elements of the eigenvvector are positive.
 A: I have solved this problem. It is mainly based on the paper "1988 - Derivatives and Perturbations of Eigenvectors". If you are interested in the eigenvector perturbation problem, you may refer to that paper. Here I would like to highlight several points:


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*The eigenvalue has to satisfy some constraints. For example, in the above mentioned paper, the eigenvalue should be simple. For symmetric matrices and their eigenvalue perturbation, you may refer to "1970 - The Rotation of Eigenvectors by a Perturbation. III".

*The main ideas to solve eigenvector perturbation are:
a) using generalized matrix inverse such as Moore-Penrose inverse or group inverse. If the eigenvalue is changing, we need to use group inverse as in the above mentioned paper. If the eigenvalue is constant, we may use more common Moore-Penrose inverse.
b) constrain the length of the eigenvector. Eigenvector only represents a direction. We need to pose a constraint such as $w^Tw=1$ or $w^Ty=1$ for some vector $y$.
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Come back to the original problem. If the eigenvalue is zero, constant and simple. From $A\dot{w}+\dot{A}w=0$, we have
$$\dot{w}=-A^\dagger \dot{A}w+x_0$$
where $A^\dagger$ is the Moore-Penrose inverse and $x_0$ is in the null space of $A$. Because $Aw=0$ and the zero eigenvalue is simple, we know $x_0=kw$ with $k\in\mathbb{R}$.
Furthermore, consider the length constraint $w^Tw=1$, we have
$$k=w^TA^\dagger\dot{A}w$$
Therefore,
$$\dot{w}=-(I-ww^T)A^\dagger \dot{A}w$$
Next is simple:
$$\|\dot{w}\|\le\|A^\dagger\|\|\dot{A}\|$$
Note $I-ww^T$ is an orthogonal projection matrix.
