I came across a basic lemma of L-2 norm space in a book (without proof), the statement is as follows:

Assuming $x \in \mathbb{F}^{n}$ and $y \in \mathbb{F}^{m}$ ($x,y \ne 0$), if $||x||_{2} \leq ||y||_{2}$ then there is a matrix $\Delta \in \mathbb{F}^{n \times m}$ with $||\Delta||_{2} \leq 1$ such that $x = \Delta y$.

Could anybody give me an idea how does the above claim come? Thank you in advance!


It is rather trivial. Rank-1 matrix, $\Delta = \frac{xy^T}{\|y\|_2^2}$ satisfies both the conditions.

  • $\begingroup$ Thank you very much. One more question is: can we extend the above claim for any two matrix $x \in \mathbb{F}^{n \times n}$ and $y \in \mathbb{F}^{n \times n}$? In my opinion, it cannot be. $\endgroup$
    – lyhuehue01
    Mar 31 '17 at 7:27
  • $\begingroup$ @lyhuehue01 You are correct, the result would not hold. This is because, $rank(\Delta y) \leq rank(y)$. Therefore if, $rank(x) > rank(y)$, your proposition could not hold. But it will always hold if $rank(x) \leq rank(y)$ $\endgroup$
    – GraspIt
    Apr 12 '17 at 4:17

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