Find all $x\in\mathbb{C^n}$ such that $||Ax||_2=1$ and $||x||_2=1$

Question

Given $A\in\mathbb{C}^{n\times n}$, find all $x\in\mathbb{C^n}$ such that $||Ax||_2=1$ and $||x||_2=1$.

Lets do SVD: $A=U\Sigma V^*$, where $\Sigma=\mathrm{diag}\{\sigma_1,\ldots,\sigma_n\}$.

We do change of a coordinates $x=Vy,$ then we need to find $y\in\mathbb{C^n}$ such that $||\Sigma y||_2=1$ and $||y||_2=1$.

Answer 1

Consider the system

$\left\{ \begin{matrix} \sigma_1^2|y_1|^2+\sigma_2^2|y_2|^2+\ldots+\sigma_n^2|y_n|^2=1 \\ |y_1|^2+|y_2|^2+\ldots+|y_n|^2=1 \end{matrix}\right.$

Let $\sigma_i=\max\{\sigma_1,\ldots,\sigma_n\}$ and $\sigma_j=\min\{\sigma_1,\ldots,\sigma_n\}$.

We are assuming that $\sigma_i^2>1>\sigma_j^2\geq0$.

We can rewrite this system as

$\left\{ \begin{matrix} \sigma_i^2|y_i|^2+\sigma_j^2|y_j|^2=1-\sum_{l\neq i,j}\sigma_l^2|y_l|^2\\ |y_i|^2+|y_j|^2=1-\sum_{l\neq i,j}|y_l|^2 \end{matrix}\right.$

By Cramer's rule, we obtain

$|y_j|^2=\det\begin{pmatrix}\sigma_i^2 & 1-\sum_{l\neq i,j}\sigma_l^2 |y_l|^2\\ 1 & 1-\sum_{l\neq i,j}|y_l|^2\end{pmatrix}/\det\begin{pmatrix}\sigma_i^2 & \sigma_j^2\\ 1 & 1\end{pmatrix}$ and $|y_i|^2=\det\begin{pmatrix}1-\sum_{l\neq i,j}\sigma_l|y_l|^2 & \sigma_j^2\\ 1-\sum_{l\neq i,j}|y_l|^2 & 1\end{pmatrix}/\det\begin{pmatrix}\sigma_i^2 & \sigma_j^2\\ 1 & 1\end{pmatrix}$.

Notice that

1. $\det\begin{pmatrix}\sigma_i^2 & \sigma_j^2\\ 1 & 1\end{pmatrix}=\sigma_i^2-\sigma_j^2>0$
2. $\det\begin{pmatrix}\sigma_i^2 & 1-\sum_{l\neq i,j}\sigma_l^2|y_l|^2\\ 1 & 1-\sum_{l\neq i,j}x_l^2\end{pmatrix}=\sigma_i^2-1+\sum_{l\neq i,j}(\sigma_l^2-\sigma_i^2)|y_l|^2$.

Since $\sigma_l^2-\sigma_i^2\leq 0$ e $0<\sigma_i^2-1$, we can find a solution when $\sum_{l\neq i,j}(\sigma_i^2-\sigma_l^2)x_l^2\leq \sigma_i^2-1.$

3. $\det\begin{pmatrix}1-\sum_{l\neq i,j}\sigma_l^2|y_l|^2 & \sigma_j^2\\ 1-\sum_{l\neq i,j}|y_l|^2 & 1\end{pmatrix}=1-\sigma_j^2+\sum_{l\neq i,j}(\sigma_j^2-\sigma_l^2)|y_l|^2$.

Since $\sigma_j^2-\sigma_l^2\leq 0$ e $0<1-\sigma_j^2$, we can find a solution when $\sum_{l\neq i,j}(\sigma_l^2-\sigma_j^2)|y_l|^2\leq 1-\sigma_j^2.$