I encountered the following function in Numerical Methods, where y is 2 by 1 Vector:
You can see the Jacobian of y here:
You can also see the eigenvalues of jacobian Function where |y|=1:
I tried to take the derivative and find the eigenvectors myself but my lack of analysis knowledge maybe made it hard for me to do it, I would really appreciate it when somebody explain this in detail.
Considering the lines of $f(y)$ we have
$$ \frac{d f(y)}{d y_k}=\frac{d}{dy_k}\left(A_{ij}y_j+\lambda\left(1-\sum_{l=1}^3y_l^2\right)y_i\right) $$
and now
$$ \frac{d}{dy_k}A_{ij}y_j = A_{ik}e_k\\ \frac{d}{dy_k}\left(1-\sum_{l=1}^3y_l^2\right)y_i = \left(1-\sum_{l=1}^3y_l^2\right)e_k-2y_k y_i e_k $$
here $e_k$ represents the $k-$row unit. So we have
$$ \frac{d}{dy_k}\left(A_{ij}y_j+\lambda\left(1-\sum_{l=1}^3y_l^2\right)y_i\right) = A_{ik}e_k+\left(1-\sum_{l=1}^3y_l^2\right)e_k-2y_k y_i e_k $$
or in a more compact form
$$ Df(y) = A+\lambda \left(1-||y||^2\right)I_3-2\lambda y y^{\top} $$
with $I_3$ the identity rank 3 matrix.
Now regarding the eigenvalues
substituting into $Df(y)$ the relationship $y_1 = \sqrt{1-y_2}$ we have
$$ \left( \begin{array}{cc} 2 \lambda \left(y_2^2-1\right) & -2 \lambda \sqrt{1-y_2^2} y_2-1 \\ 1-2 \lambda y_2 \sqrt{1-y_2^2} & -2 \lambda y_2^2 \\ \end{array} \right) $$
with eigenvalues
$$ \left\{-\sqrt{\lambda ^2-1}-\lambda ,\sqrt{\lambda ^2-1}-\lambda \right\} $$
