Is the norm of a non-negative vector always smaller than the norm of the sum of two non-negative vectors

normed-spaces

I am not a mathematician, so this question may be trivial, but I do not know whether it is correct or not. So any help is appreciated

Let $u$ and $v$ be two non-negative vectors, i.e., all elements in both vectors are non-negative.
Intuitively, I would say that the following inequality holds:$$\|u\| \leq \|u+v\|,$$
i.e., the norm over $u$ is smaller then the norm over the sum of $u$ and $v$. As the norm basically defines the 'length', the sum of non-negative vectors $u$ and $v$ should give a larger 'length'.

My question is, does this statement hold? And if so, what would be a proof?

Edit: I actually ment non-negative vectors instead of positive vectors. Thanks Rob Arthan for pointing this out.

Best Answer

First, let me assume you are working with the usual Euclidean norm on the real $n$-dimensional vector space, $\Bbb{R}^n$. sLet $u = (u_1, \ldots, u_n)$ and $v = (v_1, \ldots, v_n)$ where the $u_i$ and $v_i$ are positive real numbers. Then for each $i$, $u_i < u_i+v_i$, so $$ \|u\| = \sqrt{u_1^2 + \ldots u_n^2} < \sqrt{(u_1+v_1)^2 + \ldots (u_n+v_n)^2} = \|u + v\| $$ If you want the analogous result with "non-negative" in place of "positive", just replace each $<$ above by $\le$.

You can ignore the rest of this answer if you are not interested in more general norms.

In general, a norm on $\Bbb{R}^n$ is completely determined by its unit disc, $D$, which can be an arbitrary convex body that is symmetric about the origin, i.e., $D = -D$. In $\Bbb{R}^2$ you can define a norm whose unit disc is the parallelogram with vertices at $(-1, 0)$, $(1, 1)$, $(1, 0)$ and $(-1, -1$). Under this norm $\|(0, 1/2)\| = 1 > \|(3/4, 3/4)\| = 3/4$.

Related Solutions

[Math] Finding bounds on the norm of the difference between two vectors.

Your argument more or less goes through in general. The cases $p=1$ and $p=+\infty$ are special and may easily be done by hand. We assume here that $1<p<+\infty$. On the compact set $T\times T$ the maximum is attained for some couple $(u,v)$. We fix this extremal couple in the following. Observe that with your notation for the index sets: $$ 1=\|u\|_p^p = \sum_{i\in C}|u_i|^p + \sum_{i\in D} |u_i|^p $$ (and similarly for $v$) so that we may rearrange to get: $$ D(u,v)^p = \sum_{i\in C} u_i^p + \sum_{i\in B} v_i^p + \sum_{i\in D} |u_i-v_i|^p = 2 + \sum_{i\in D} \left(|u_i-v_i|^p - u_i^p -v_i^p\right) $$ For $a\geq b\geq 0$ clearly $|a-b|^p-a^p-b^p \leq -|b|^p$ and more generally for any $a,b\geq 0$ one has: $|a-b|^p-a^p-b^p \leq - \min\{a^p,b^p\}$. Therefore, $$ D(u,v)^p \leq 2 - \sum_{i\in D} \min\{u_i,v_i\}^p$$ which is strictly less than 2 unless $D$ was empty in the first place. QED

This method of proof is not restricted to finite dimensions and works in a general measure space.

EDIT: A solution using Lagrange multipliers: Suppose that the index set $D$ is non-empty. Now, $u_i$ and $v_i$ are both non-zero for $i\in D$ so we may use Lagrange multipliers to study the extremal value for this functional, the constraints being that $g(u)=\sum_{i\in C∪D} u_i^p −1$ and $h(v)=\sum_{i\in B∪D} v_i^p−1$ both should vanish. Writing $F(u,v,a,b)=D(u,v)p−ag(u)−bh(v)$ we see that at our extremal point, all partial derivatives of $F$ should vanish (we use here $p>1$ so that $D(u,v)$ is differentiable also when $u_i=v_i$ and only consider non-vanishing components of $u_i$ and $v_i$). In particular, taking the derivative w.r.t. $u_i$ with $i\in D$: $$p|u_i−v_i|^{p−1}{\rm sign}(u_i−v_i)−(1+a)p|u_i|^{p−1}=0 $$ This implies that ${\rm sign}(u_i−v_i)$ must be constant for $i\in D$ (we may assume $≥0$) and then that $v_i=t u_i$, $i\in D$ for some constant $0<t≤1$. Thus, $$D(u,v)^p=2+((1−t)^p−1−t^p)\sum_{i \in D}|u_i|^p$$ Finally, the t-dependent factor is strictly negative unless t=0, implying that the index set $D$ must have been empty in the first place. So your conclusion goes through and $\max_{T×T}D(u,v)=2^{1/p}$. A formula which is also valid for $p=1$ and $p=\infty$.

[Math] Calculating the angle between two vectors

Since the norm is just the magnitude or modulus or whatever you call it, this code should do the trick:

from numpy import arccos, array
from numpy.linalg import norm

# Note: returns angle in radians
def theta(v, w): return arccos(v.dot(w)/(norm(v)*norm(w)))

sulfur = array([0.0, 0.0, 0.102249])
hydrogen_1 = array([0.0, 0.968059, -0.817992])
hydrogen_2 = array([0.0, -0.968059, -0.817992])
print(theta(hydrogen_1-sulfur, hydrogen_2-sulfur))

The numpy function is documented here.

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Related Solutions

[Math] Finding bounds on the norm of the difference between two vectors.

[Math] Calculating the angle between two vectors

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