Why is there always a matrix with norm 1 that maps a prespecified unit vector to another one? Let $x,y \in {\mathbb{C}^n}$ and $\left\| x \right\| = \left\| y \right\| = 1$.Why is there a matrix $H \in {\mathbb{C}^{n \times n}}$ with $\left\| H \right\| = 1$ such that $Hx=y$?
 A: I will answer in terms of linear maps rather than matrices because thinking in terms of linear maps seems cleaner to me, although it is easy to translate back and forth between linear maps and matrices.
Suppose $x, y \in \mathbf{C}^n$ and $\|x\| = \|y\| = 1$. Define a linear map $H \colon \mathbf{C}^n \to \mathbf{C}^n$ by
$$
H(\lambda x + z) = \lambda y
$$
for each $\lambda \in \mathbf{C}$ and each $z \in \{x\}^\perp$. Then $Hx = y$. Also, for each $\lambda \in \mathbf{C}$ and each $z \in \{x\}^\perp$ we have
$$
\|H(\lambda x + z)\| = \|\lambda y\| = |\lambda| \le \sqrt{|\lambda|^2 + \|z\|^2} = \|\lambda x + z\|.
$$
Thus $\|H\| \le 1$.
A: Let $e_1=[1,0,\ldots,0]^T$, $\|x\|_2=\|y\|_2=:\rho$. Let $H_x$ be a unitary matrix such that the first column of $H_x$ is $x/\rho$. Then we have $H_x(\rho e_1)=x$. This matrix can be constructed by extending $x/\rho$ to an orthonormal basis. Similarly, let $H_y$ be a unitary matrix such that $H_y(\rho e_1)=y$. Then
$$
\rho e_1=H_x^*x=H_y^*y
$$
and hence
$$
Hx:=H_yH_x^*x=y.
$$
Since $H_x$ and $H_y$ are unitary, $H=H_yH_x^*$ is unitary as well and therefore $\|H\|_2=1$.
You can also construct $H$ using the Householder transformation, but the derivation is quite tedious in the complex case. If everything is real, setting $H=I-2vv^T$ with $v:=(x-y)/\|x-y\|_2$ does the job.
A: As the OP does not specify what norms are used, I assume that the vector norm is arbitrary and the matrix norm is induced by it.
For every vector norm on $\mathbb C^n$, the open unit ball $B=\{x: \|x\|<1\}$ is non-empty and convex. By Hahn-Banach separation theorem, there exists a hyperplane $x+P$ (where $P$ is some $(n-1)$-dimensional space) such that the open unit ball lies on one side of the hyperplane. It follows that the closed unit ball $\overline{B}$ is sandwiched between $-x+P$ and $x+P$. Clearly $x\notin P$, or else some nonzero vector outside $P$ will have zero norm.
It follows that if a vector inside $\overline{B}$ is written as a sum in the form of $ax+p$, where $a\in\mathbb C$ and $p\in P$, we must have $|a|\le1$. So, if $H$ is the matrix that maps $x$ to $y$ and maps every vector in $P$ to zero, then $Hx=y,\ \|Hx\|=1$ and $\|H(ax+p)\|=|a|\le1$. Hence $\|H\|=1$.
