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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!

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2 Answers

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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.

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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.

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