Constructing a specific Rank-One Matrix

Question

Given u $\in \mathbb{R}^{n}$ and v $\in \mathbb{R}^{m}$ with unit $L^{2}$ norm, i.e. $\|u\|_{2}$ = $\|v\|_{2}$ = 1. Construct a rank-one matrix B $\in \mathbb{R}^{mxn}$ such that $Bu = v$ and $\|B\|_{2}$ = 1.

Answer 1

Let us take $\|B\|_{2}=sup \dfrac{\|BX\|_{2}}{\|X\|_{2}} \ \ (\forall X \neq 0, X \in \mathbb{R}^n)$ or equivalently:

$\|B\|_{2}=sup \|BX\|_{2}\ \ \ (\forall X \neq 0 \ s.t. \|X\|_{2}=1$) (induced norm).

Remark: $\|B\|_{2}$ can have a different meaning (Frobenius norm).

Let us consider the rank-one matrix $B=vu^T$ (its image space is generated by $v$).

Indeed, $Bu=vu^Tu=v.1=v$. Moreover

$\|B\|_{2}=sup \ \|(vu^t)X\|_{2}=sup \ \|v(u^tX)\|_{2} \ \ \ $ (supremum taken for all$ X \neq 0 \ s.t. \|X\|_{2}=1$).

$\|B\|_{2}=sup \ |u^tX|\|v\|_{2} = sup \ |u^tX| \ \ \ (1)$.

But, using the equality case of Cauchy-Schwarz, being given that both vector $u$ and vectors $X$ have unit norm, the sup in (1) is attained for $X=u$ and thus has value $|u^tu|=1$.