# Linear combination of two vectors in complex space

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.

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.
• 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. Jan 12, 2019 at 0:28