Are eigenvalues (resp. unit eigenvectors) dependent continuously on elements $a_{ij}$ of a symmetric matrix $A$? Let $A(t)=(a_{ij}(t)),~(t\in \mathbb R)$ is a symmetric matrix such that $a_{ij}(t)=a_{ji}(t)$ is a real-valued continuous function. Let $\lambda_1(t) \ge \cdots \ge \lambda_n(t)$ is all of the eigenvalues of $A(t)$.


*

*show that each $\lambda_i(t)$ is continuous in $t$. 


(Note: if $i=1$ or $n$, it may be easy thanks to the well-known relation $\lambda_\max = \max_{x\not= 0} \frac{x^T A x}{x^T x}$, but it becomes very hard for me even in the case $i=2$)


*Moreover, for a given $i$, if $v_i(t)$ is the eigenvector corresponding
$\lambda_i(t)$ with unit length in the standard eudlidean metric, is the function $t\mapsto v_i(t)\in \mathbb R^n$ continuous as
well?

*What if we assume $A(t)$ is skew-symmetric rather than symmetric?

*What if we consider, in analogy, the smooth version of the same problems, replacing all "continuous" by "smooth"?
 A: I outlines of my argument for key question 1) For 3) it is clear
that the $ \gamma_j (t) = - i \lambda_j$ are also continuous real
functions where $ i ^ 2 = -1 $.
My remarks 1) The first point is that the
concept of continuous functions is local. So all my reasoning is
locally at one point.
2) The fact that $ \lambda $ are defined and real  ($ A
(t) $ symmetrical).
3) By a method analogue in the prove of Gershgorin
theorem (localization of the eigenvalues of matrix), and the fact
that the coefficients $ a_ {ij} $  of the matrix $A (t) $ are
continuous, we can assume that $ \lambda_{ij} $ are localy bounded
by the same constant.
4) Using some functions continuous proprieties   in a
neighborhood of a  given point follows:
a) if $ f $ is bounded and $ g $ continues,   then $ fg$
continues involves  $ f $ continuous
b) If $f$ and $g$ are defined and positive, then $f + g$
 continuous involves $f$ and $g$ continuous.
c) if $f$ well defined and
bounded with $f ^ 2$ continues, then $ f$ continues.
5) the coefficients $f_j(t)$ of the characteristic
polynomial of $A (t) $  are  continuous (because in the $ det (A
(t)) $ intervenes only algebraic operations on continuous
functions $  a_ {ij}$ that retain continues propriety.
6) Relations between roots $\Lambda_{i}$ and the
coefficients of the characteristic polynomial $ f_j$ of $ A (t) $
(more precisely the $  f_j$ are function symmetric polynomials in
the $ \lambda_i$).
A last remark  sorry for my English,  and why you suppose that the
functions $ \lambda_i(t)$ are ordered?.
Prove of key question: if $n-1$ of $\lambda_j(t)$ are continues,
then by relation $\sum \lambda_j=\pm f_{n-1}$ we tack the other.
if $n-2$ of $\lambda_j(t)$ are continues, suppose are
$\lambda_3,\lambda_4,\cdot\cdot\cdot,\lambda_n$ and we show that
$\lambda_1, \lambda_2$ are continues. So the $det(A(t))$ and
remarque a) and 5) we have $\lambda_1,\lambda_2$
continues. also by 5) $\lambda_1+\lambda_2$ is continue and so
$\lambda_1^2+\lambda_2^2+2\lambda_1\lambda_2$ is continue, so
$\lambda_1^2+\lambda_2^2$ is continue so $\lambda_1^2$ and
$\lambda_2^2 $ are continues. By c)  we conclude that
$\lambda_1$ and $\lambda_2 $ are continues.
