If $||x||_2 = 3$ and $||y||_2 = 7$ find the minimum of $\langle x, y\rangle$

Question

If $||x||_2 = 3$ and $||y||_2 = 7$, find the minimum of $\langle x, y\rangle$

My thought process is that it would be 3 since if $||x||_2 = 3$, we can have $|x_1| = 1$, $|x_2| = 1$, $|x_3| = 1$ and every other term can equal $0$. Similarly, if $||y||_2 = 7$ then we can have $||y_1|| = 1$, $||y_2|| = 1$, ... , $||y_7|| = 1$ and every other term can equal $0$. So, we would have that $$\langle x, y\rangle = |x_1| |y|_1 + |x_2| |y|_2 + |x_3| |y|_3 + |x_4| |y|_4 + |x_5| |y|_5 + |x_6| |y|_6 + |x_7| |y|_7 + ... + |x_k| |y|_k\\=(1\times1) + (1\times1) + (1\times1) + (0\times1) + (0\times1) + (0\times1) + (0\times1) + (0\times0) + ... + (0\times0) = 3$$. So our minimum is 3.

Is is train of logic correct?

mjw · Answer 1 · 2020-11-18T02:04:33.860

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Take $x$ and $y$ to be in opposite directions. For example $x=(3,0)$ and $y=(-7,0)$.

The minimum is $$-3\cdot 7 = -21.$$

Or are $x$ and $y$ $n-$dimensional?:

$x=(3,0,0,\cdots,0)$ and $y=(-7,0,0,\cdots,0)$ and again the minimum is $-21.$

UPDATE: To derive this result, we can use Lagrange multipliers.

$$ L = \langle x,y \rangle - \lambda ( \langle x,x \rangle-9) - \mu( \langle y,y \rangle-49)$$

Computing $0=\frac{\partial L}{\partial x}=y-2\lambda x$ and $0=\frac{\partial L}{\partial y}=x-2\mu y$ leads to the condition that $y= \alpha x$, for some $\alpha$, so $x$ and $y$ are in the same direction (maximizing the inner product) or in opposite directions (minimizing the inner product).

edited Nov 18 '20 at 02:04

answered Nov 18 '20 at 01:21

mjw

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x and y are n-dimensional – Nuemann12 Nov 18 '20 at 01:26
That's fine. Let $y= -\frac{7}{3} x$, then $<x,y>=-21.$ – mjw Nov 18 '20 at 01:28
I assumed that the absolute values forced the inner product to be positive. How can it be -21? After all isn't ||x||$_2^2$ = <x,x>? – Nuemann12 Nov 18 '20 at 01:30
$x$ and $y$ are free to point in whatever direction they want. – mjw Nov 18 '20 at 01:32
Ah. So the norm simply gets the distance between them in a way that disregards direction? – Nuemann12 Nov 18 '20 at 01:33
As you said, the norm is non-negative but the inner product could be negative. – mjw Nov 18 '20 at 01:34
Let us continue this discussion in chat. – Nuemann12 Nov 18 '20 at 01:35
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Inner product could be defined as $\langle x,y\rangle = \sum_{j=1}^n x_j y_j$, then the norm, $||x||_2^2 = \langle x,x\rangle \ge0.$ – mjw Nov 18 '20 at 01:35
$\frac{\partial \langle x,y \rangle }{\partial x} = y$ and $\frac{\partial \langle x,x \rangle }{\partial x} = 2x$, etc. – mjw Nov 18 '20 at 01:53

If $||x||_2 = 3$ and $||y||_2 = 7$ find the minimum of $\langle x, y\rangle$

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