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Let $\mathbf{x},\mathbf{y} \in \mathbb{C}^2$ be two linearly independent vectors in two dimensional complex space. Assume that $\|\mathbf{x}\|\leq \|\mathbf{x} \pm \mathbf{y}\|$. I want to show (or understand why the following does not hold) that $\|\mathbf{x}\| \leq \|\mathbf{x} + \alpha \mathbf{y}\|$ for all $\alpha \in \mathbb{C}$ such that $|\alpha|\geq 1$. Any hints as to how to prove the inequality would be greatly appreciated, and if it doesn't necessarily hold, what other constraints need to be imposed on $\alpha$?

So far I have managed to show the trivial case that this holds where we have either $\|\mathbf{x}\|>2\|\mathbf{y}\|$ or $\|\mathbf{y}\|>2\|\mathbf{x}\|$. I am struggling to prove it for the other cases.

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This is not true in general -- for example, take $x = (10,0)$ and $y = (10i,i)$. Then:

  • $\|x\| = 10$, $\|y\| = \sqrt{101}$, and $\|x \pm y\| = \|(10 \pm 10i, \pm i)\| = \sqrt{201}$, which is bigger than $\|x\|$.

  • On the other hand, take $\alpha = i$. Then $$ x + \alpha y = (10,0) + (-10,-1) = (0,-1), $$ so $\|x + \alpha y\| = 1$ which is smaller than $\|x \| = 10$.

if it doesn't necessarily hold, what other constraints need to be imposed on $\alpha$?

It doesn't seem to me like it should be true in general. A strong constraint which you may be able to impose would be $\alpha \in \mathbb{R}$, but I guess you would probably not find that interesting.

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  • $\begingroup$ Thank you for the response. Yes, restricting $\alpha$ to $\mathbb{R}$ wouldn't be very interesting as I was wondering if you could generalise the equivalent lemma where $\mathbf{x}, \mathbf{y} \in \mathbb{R}^2$ and $\alpha \in \mathbb{R}$, for which the inequality certainly holds. $\endgroup$ – Chris Jan 12 at 0:28

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