Suppose $x_i > 0$ and $y_j -x_j$, then $c_{ij} = 1/(x_i+x_j)$. These matrices are infinitely divisible, i.e., $[c_{ij}^r]$ is also positive definite for all $r > 0$.
Spectral properties of Cauchy-like matrices and kernels are studied here.
For additional information and an easier read, on the general case ($1/(x_i+y_j)$), you might enjoy looking at the recent book by Pinkus.
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 (by Peter D. Lax):
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 (by Roger A. Horn & Charles R. Johnson)
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
$$ \left.\frac{\operatorname{d}\lambda(t)}{\operatorname{d}t}\right|_{t=0}=\frac{y^*Ex}{y^*x}$$
Best Answer
The $2^n\times 2^n$ dimensional Hadamard matrices $H_{2^n}$ are also called Sylvester matrices or Walsh matrices. There are only two distinct eigenvalues $\pm 2^{n/2}$, so the eigenvectors are not in general orthogonal. An orthogonal basis of eigenvectors is constructed recursively in A note on the eigenvectors of Hadamard matrices of order $2^n$ (1982) and in Some observations on eigenvectors of Hadamard matrices of order $2^n$ (1984). See also Chapter 5 of Hadamard Matrix Analysis and Synthesis (2012).