When does a rectangular matrix A $\in$ $\mathbb C^{\times }$ have the property such that $\|Ax\|_2=\|x\|_2$, where $x \in \mathbb C^ $ This is one of my homework questions which I am trying to solve I started with the claim that $\|Ax\|_2$ will be less than equals to $\|A\|_2$ $\|x\|_2$ and from here I can only get the condition for $\|Ax\|_2 = \|x\|_2$, that $\|A\|_2 \ge 1$ should hold.
pls suggest me approach for solving this question
 A: short way
$$
\|Ax\|_2^2 = (Ax)^\dagger(Ax) = x^\dagger A^\dagger A x,\qquad \text{and} \qquad \|x\|_2^2 =  x^\dagger x
$$
where $\dagger$ denotes the conjugate transpose. Now, since $\forall x \in \mathbb C^n : \|Ax\|_2 = \|x\|_2$ we get
$$
x^\dagger (A^\dagger A) x = x^\dagger x \iff  x^\dagger (A^\dagger A -E_n) x = 0 \\
$$
Consider a quadratic form $Q_A:\mathbb C^n \to \mathbb C$ such that $ x \mapsto x^\dagger(A^\dagger A -E_n)x$, now if this is the zero quadratic form, i.e. $\forall x \in \mathbb C^n :Q_A(x) = x^\dagger(A^\dagger A -E_n)x = 0$ then its matrix is zero $A^\dagger A - E_n = 0$, and consequently
$$
A^\dagger A = E_n \in \mathbb C^{n\times n}
$$

More explicitly:
$$
y_i = (Ax)_i = \sum_{j=1}^n A_{ij} x_j \\\implies \|y\|_2^2 = \sum_{i=1}^m (\sum_{j=1}^n A_{ij} x_j)^*(\sum_{j=1}^n A_{ij} x_j) = \sum_{i=1}^m \sum_{j,k=1}^n A_{ij}^* A_{ik} x_j^* x_k
$$
which should be equal to
$$
\|x\|_2^2 = \sum_{i=1}^n \sum_{j,k=1}^n \delta_{ij} \delta_{ik} x_j^* x_k
$$
So we get
$$
\sum_{i=1}^n \sum_{j,k=1}^n \delta_{ij} \delta_{ik} x_j^* x_k = \sum_{i=1}^m \sum_{j,k=1}^n A_{ij}^* A_{ik} x_j^* x_k \\
\sum_{i=1}^n \sum_{j,k=1}^n \delta_{ij} \delta_{ik} x_j^* x_k - \sum_{i=1}^m \sum_{j,k=1}^n A_{ij}^* A_{ik} x_j^* x_k = 0 \\
\sum_{j,k=1}^n \left(\sum_{i=1}^n \delta_{ij} \delta_{ik} - \sum_{i=1}^m A_{ij}^* A_{ik}\right) x_j^* x_k = 0 \\
\overset{\forall x \in \mathbb C^n}{\implies} \sum_{i=1}^n \delta_{ij} \delta_{ik} = \sum_{i=1}^m A_{ij}^* A_{ik}
$$
from the last equation
$$
\delta_{jk} = \sum_{i=1}^m A_{ij}^*A_{ik}, \quad j,k \in \{1,\cdots,n\}
$$
A: Below follows an analysis that hinges on the polarization identity:
$$ \forall x, y \in \mathbb{C}^n \: : \: \langle x, y \rangle_n = \frac{1}{4} \left( \|x + y\|_n^2 + \|x - y\|_n^2 - i \|x + iy\|_n^2 + i \|x - iy\|_n^2\right).$$
Here $\langle x, y \rangle_n$ denotes the inner product between $x, y \in \mathbb{C}^n$ and $\|x\|_n$ denotes the norm given by $$\|x\|_n^2 = \langle x, x \rangle_n.$$ The proof of the polarization identity will be omitted. I use the subscript $n$ to emphasize the dimension of the vector space.
Returning to the problem at hand. We have $A \in \mathbb{C}^{m \times n}$ such that
$$\forall x \in \mathbb{C}^n \: : \: \|Ax\|_m = \|x\|_n.$$
We say that $A$ preserves norms. We mention in passing $A$ is necessarily one-to-one. Specifically if $Ax=0$, then $$\|x\|_n = \|Ax\|_m = 0,$$ so $x=0$ and we must have $m \ge n$. Now from the polarization identity we have
$$ \forall x, y \in \mathbb{C}^n \: : \: \langle Ax,Ay \rangle_m = \langle x,y \rangle_n.$$
We see that $A$ also preserves inner products.
It follows that $$ \forall x, y \in \mathbb{C}^n \: : \: \langle x,A^*Ay \rangle_n = \langle Ax, Ay \rangle_m =\langle x, y \rangle_n.$$
Now let $B = [b_{ij}] \in \mathbb{C}^{n \times n}$ be given by $B = A^*A$ and let $e_k$ denote the $k$ column of the identity matrix $I_n$ of dimension $n$. Then
$$ b_{ij} = \langle e_i, Be_j \rangle_n = \langle e_i, e_j \rangle_n = \begin{cases} 1 & i=j \\ 0 & i \not =j \end{cases}.$$
We conclude that $B = A^*A = I_n$. It follows that $A$ is an orthogonal matrix.
A: Let $.^H$ be the Hermetian (conjugate) transpose, and let $\langle \cdot,\cdot\rangle$ be the standard inner product on $\mathbb{C}^n$.
In the forward direction, assume that $\lVert Ax \rVert = \lVert x\rVert$ for $x$ nonzero. Then this implies that:
$$
\langle Ax, Ax \rangle = \langle x,A^H Ax \rangle = \langle x,x\rangle \implies \langle x, A^H Ax - x\rangle = 0.
$$
Note that:
$$
\lVert A^H x\rVert^2 = \langle x,AA^H x\rangle\leq \lVert x\rVert\lVert AA^H x\rVert = \lVert x\rVert\lVert A^H x\rVert \implies \lVert A^Hx\rVert\leq \lVert x\rVert.
$$
But using Cauchy-Schwarz:
$$
\begin{split}
0 = \langle x,A^H Ax - x\rangle^2  \leq \lVert x\rVert^2\lVert A^H Ax - x\rVert^2 =\\ \lVert x\rVert^2(\lVert A^H Ax\rVert^2 - \langle A^H Ax,x\rangle - \langle x,A^H Ax\rangle + \lVert x\rVert^2) \leq\\
 \lVert x\rVert^2(\lVert Ax\rVert^2 - \langle A^H Ax,x\rangle- \langle x,A^H Ax\rangle +\lVert x\rVert^2) =\\ -\lVert x\rVert^2(\langle x,A^H Ax - x\rangle - \overline{\langle x,A^H Ax - x\rangle }) = 0,
\end{split}
$$
which implies that $\lVert A^H Ax - x\rVert = 0$, which in turn implies that $A^H A x = x\implies A^H A = I_n$
In the backward direction, assume $A^H A = I_n$. But then we trivially have
$$
\lVert x\rVert^2= \langle x,x\rangle = \langle x,A^H A x\rangle = \langle Ax,Ax\rangle= \lVert Ax\rVert^2.
$$
