Using Newton's Method to find eigenvalues of a matrix $A \in \mathbb{R}^{n \times n}$ Let $A \in \mathbb{R}^{n \times n}$ and suppose $\exists$ a nonsingular $V \in \mathbb{R}^{n \times n}$ and a diagonal $D \in \mathbb{R}^{n \times n}$ such that $A = VDV^{-1}$. Define the map $F: \mathbb{R}^n \times \mathbb{R} \rightarrow \mathbb{R}^n \times \mathbb{R}$ by $$F(v,\lambda) = \left[\begin{array}{c} 
Av - \lambda v \\
\frac{1}{2}v^T v - \frac{1}{2}
\end{array}\right].$$ Then finding an eigenpair of $A$ with an eigenvector of unit length can be viewed as a root finding problem. One part of the question I am working on asks to show that the Jacobian $F'(v_*,\lambda_*)$ is nonsingular if $\lambda_*$ is a simple eigenvalue.
Attempt: Let $(v_*,\lambda_*)$ be an eigenpair of $A$ where $\lambda_*$ is a simple eigenvalue. Suppose the Jacobian $$F'(v_*,\lambda_*) = \left[\begin{array}{cc} 
A - \lambda_* I & -v_* \\
v_*^T & 0
\end{array}\right]$$ is singular. Then there exists a nonzero vector $(x,\mu) \in \mathbb{R}^n \times \mathbb{R}$ such that $$\left[\begin{array}{cc} 
A - \lambda_* I & -v_* \\
v_*^T & 0
\end{array}\right]\left[\begin{array}{c} 
x \\
\mu
\end{array}\right] = 0.$$ That is, \begin{align*}
Ax - \lambda_*x & = \mu v_* \\
v_*^Tx & = 0.
\end{align*} I am not sure how to derive a contradiction from here. Any help would be sincerely appreciated!
 A: Here is a solution to the original post:
Let $(v_j,\lambda_j)$ be an eigenpair of $A$ where $\lambda_j$ is the $j$-th eigenvalue of $A$ and simple. Suppose that the Jacobian $F'(v_j,\lambda_j)$ is singular. Then there exists a nonzero vector $(x,\mu) \in \mathbb{R}^n \times \mathbb{R}$ such that \begin{align}
(A-\lambda_j I)x = \mu v_j
\end{align} and \begin{align}
v_j^T x = 0.
\end{align} Using the diagonalization of $A$, we have that by the first equation, \begin{align*}
V(D - \lambda_j I)V^{-1}x = \mu v_j \Longrightarrow (D-\lambda_j I)V^{-1}x = \mu e_j
\end{align*} since $v_j = Ve_j \Longrightarrow V^{-1}v_j  = e_j$. Let $z = V^{-1}x$. Then in matrix form, this equation can be written as $$ \begin{bmatrix}
            \lambda_1-\lambda_j & & & & & & & \\
            & \ddots & & & & & & \\
            & & 0 & & & & & \\
            & & & \ddots & & & & \\
            & & & & \lambda_r-\lambda_j & & & \\
            & & & & & -\lambda_j & & \\
            & & & & & & \ddots & \\
            & & & & & & & -\lambda_j \\
            \end{bmatrix}
            \begin{bmatrix}
            z_1 \\
            \\
            \\
            \vdots \\
            \\ 
            \vdots \\
            \\ 
            \\
            z_n
            \end{bmatrix}
            =
            \begin{bmatrix}
            0 \\
            \vdots \\
            \mu \\
            \\
            \\ 
            \vdots \\
            \\ 
            \\
            0
            \end{bmatrix}$$ where the $j$-th row of $D-\lambda_j I$ is the zero vector. Hence we must have that $\mu = 0$ so $$Ax = \lambda_j x.$$ But since $\lambda_j$ is simple, $x$ must be a scalar multiple of $v_j$, i.e. $x = \alpha v_j$ for some $\alpha \in \mathbb{R}$. Since $v_j^T x = 0$ as well, we must have that $\alpha = 0$ so $x = 0$. Thus the kernel of $F'(v_j,\lambda_j)$ only contains the zero vector so it must be nonsingular.
A: There is a different solution, that doesn't require $ A $ to be diagonalisable.
Let $ x $ be an eigenvector to the simple eigenvalue $ \lambda $, so $ x\neq 0 $.
We assume $ F'(x,\lambda) $ being singular, then there exists $ \begin{pmatrix}
v\\\mu
\end{pmatrix} \in \mathbb(R)^{n+1}\backslash\{0\} $ with
$$ \begin{pmatrix}
A-\lambda I & -x\\
x^T&0
\end{pmatrix}
\begin{pmatrix}
v\\\mu
\end{pmatrix}
=
\begin{pmatrix}
0\\0
\end{pmatrix}$$
which is leading to the following equations:
\begin{align}
(A-\lambda I )v-\mu x&=0\\
x^T v&=0
\end{align}
We will now look to the following cases:
1.) If $ \mu=0 $ then $ v\neq 0 $.
From $ (1) $ it follows that $ Av=\lambda v $, which implies $ v\in E(A,\lambda) $, where $ E(a,\lambda) $ denotes the eigenspace of the eigenvalue $ \lambda $.
With $ x,v\neq 0 $ the orthogonality of $ x $ and $ v $ implied by $(2)$ shows, that $ x $ and $ v $ are linearly independent.
Both being Elements of $ E(A,\lambda) $ implies that the geometric multiplicity of $ \lambda $ is at least 2, which is a contradiction to $ \lambda $ being a simple eigenvalue, because the  geometric multiplicity can never exceed the algebraic multiplicity.
2.) Now, let $ \mu \neq 0 $. Because $ x $ is an eigenvector $ x\neq 0 $.
That leads to $ \mu x \neq 0 $, which implies $ v\neq 0 $ because otherwise $ (1) $ will give us the contradiction
$$ 0=(A-\lambda I)v=\mu x \neq 0. $$
As said before, $ v\neq 0 $ implies (with $ (2) $) that $ x $ and $ v $ are linear independent.
From $ (1) $ we get
$$ (A-\lambda I)^2 v=(A-\lambda I) \mu x =0$$
which means that $ v $ is an element of the generalized eigenspace $ H(A,\lambda) $.
From the linear independence of $ x $ and $ v $ it follows that $ \dim H(A,\lambda)\geq 2>1 $.
Because $ \dim H(A,\lambda)$ is the algebraic multiplicity of an eigenvalue this is a contradiction, to $ \lambda $ being a simple eigenvalue.
So $ F'(x,\lambda) $ must have been regular.
