The general question has been treated [here](https://mathoverflow.net/questions/116123/how-to-find-define-eigenvectors-as-a-continuous-function-of-matrix), and the response was negative. My question is about more particular perturbations. The counterexamples given in the previous question have variations not only on the diagonal, and moreover, in the simple example given by A. Quas, the discontinuous change in eigenvectors take place where the eigenvalue is double.



> 
> Suppose $A$ is a symmetric, positive definite matrix, with simple lowest eigenvalue, and $D$ is a diagonal matrix. Does the first eigenvalue and the first eigenvector of $A+D$ vary continuously for $\|D\|\_1 <\varepsilon$ ($\varepsilon$ can be as small as possible, and $\|\cdot \|\_1$ is the $L^1$ norm)?
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> In the case this is true, it is possible to bound the variation of the eigenvector in terms of $\|D\|\_1$?
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In response to the comments, I add a few details:


* $A$ is constant, $D$ is a diagonal matrix which is variable with $N$ parameters (the matrices are of size $N \times N$)
* $D$ is just assumed to vary continuously with norm smaller than $\varepsilon$.



Eigenvectors as continuous functions of matrix - diagonal perturbations
As @Joe Silverman said in his comment, the key here is that $\lambda$ is simple.


I copy here a few Theorems on the subject that might be helpful. 



> 
> From the book [Linear Algebra and Its Applications](http://rads.stackoverflow.com/amzn/click/0471751561) (by Peter D. Lax):
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**Theorem 7, p.130:**
Let $A(t)$ be a differentiable square matrix-valued function of the real variable $t$. Suppose that $A(0)$ has an eigenvalue $a\_0$ of multiplicity one, in the sense that $a\_0$ is a simple root of the characteristic polynomial of $A(0)$. Then for $t$ small enough, $A(t)$ has an eigenvalue $a(t)$ that depends differentiably on $t$, and which equals $a\_0$ at zero, that is $a(0)=a\_0$.


**Theorem 8, p.130:**
Let $A(t)$ be a differentiable matrix-valued function of $t$, $a(t)$ an eigenvalue of $A(t)$ of multiplicity one. Then we can choose an eigenvector $h(t)$ of $A(t)$ pertaining to the eigenvalue $a(t)$ to depend differentiably on $t$.



> 
> From the Book [Matrix Analysis](http://rads.stackoverflow.com/amzn/click/0521548233) (by Roger A. Horn & Charles R. Johnson)
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**Corollary 6.3.8, p.407:** Let $A,E\in \Bbb C^{n\times n}$. Assume that $A$ is Hermitian and $A+E$ is normal, let $\lambda\_1,\ldots,\lambda\_n$ be the eigenvalues of $A$ arranged in increasing order $\lambda\_1\leq \ldots\leq \lambda\_n$ and let $\hat \lambda\_1,\ldots,\hat \lambda\_n$ be the eigenvalues of $A+E$, ordered so that $\Re(\hat\lambda\_1)\leq\ldots\leq\Re(\hat\lambda\_n) $. Then
$$ \sum\_{i=1}^n |\hat \lambda\_i-\lambda\_i|^2 \leq \|E\|\_F^2,$$
where $\|\cdot\|\_F$ is the Frobenius norm.


This is a somehow refined version of Theorem 7 above:


**Theorem 6.3.12, p.409:**
Let $A,E\in \Bbb C^{n\times n}$ and suppose that $\lambda$ is a simple eigenvalue of $A$. Let $x$ and $y$ be, respectively, right and left eigenvectors of $A$ corresponding to $\lambda$. Then  

 a) for each given $\epsilon>0$ there exists a $\delta>0$ such that, for all $t\in\Bbb C$ such that $|t|<\delta$, there is a unique eigenvalue $\lambda(t)$ of $A+tE$ such that $|\lambda(t)-\lambda-ty^\*Ex/y^\*x|\leq |t| \epsilon$  

b) $\lambda(t)$ is continuous at $t=0$, and $\lim\_{t\to 0}\lambda(t)=\lambda$  

c) $\lambda(t)$ is differentiable at $t=0$, and
$$ \left.\frac{\operatorname{d}\lambda(t)}{\operatorname{d}t}\right|\_{t=0}=\frac{y^\*Ex}{y^\*x}$$


