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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 $A(t)$.

  1. 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 $i=2$)

  1. 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?

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

  3. What if we consider, in analogy, the smooth version of the same problems, replacing all "continuous" by "smooth"?

Hang
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    What do you mean by "smooth version" of the problem? – BigbearZzz May 27 '16 at 08:48
  • See https://www.math.upenn.edu/~kazdan/509S07/eigenv5b.pdf – user7530 May 27 '16 at 14:57
  • Remove the catastrophic condition $\lambda_1(t) \ge \cdots \ge \lambda_n(t)$. –  May 27 '16 at 18:45
  • The condition $\lambda_1\ge\cdots\ge\lambda_n$ gives some basic issues for points (3) and (4). For (3) consider $$\begin{pmatrix} 0 & t \ -t & 0 \end{pmatrix}$$ whose eigenvalues are $ti$ and $-ti$ (who's the bigger one?); for (4) consider $$A(t)=\begin{pmatrix}t&0\0&-t\end{pmatrix}$$ where $\lambda_1(t)=\lvert t\rvert,\ \lambda_2(t)=-\lvert t\rvert$. –  May 27 '16 at 22:58

1 Answers1

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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.

m.idaya
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