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How to prove this without using algebraism?

Prove that $\lVert\vec{u}\rVert=0 \iff \vec{u}=\vec{0}$.

Question 1.8 of Geometria Analítica Third Ed., Ivan Camargo & Paulo Boulos

ObliviousCapybara
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user113581321
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1 Answers1

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Recalling that, by definition, norms are both:

  • positive definite (i.e. for all $\vec{x}$, $\lVert \vec{x}\rVert = 0 \implies \vec{x} = \vec{0}$)
  • homogeneous (i.e. for all scalars $s$ and vectors $\vec{x}$, $\lVert s\vec{x} \rVert = s\lVert \vec{x} \rVert$)
  • subadditive (not needed here)

We prove the two implications, starting with $\implies$:

$\lVert\vec{u}\rVert = 0$ implies $\vec{u} = \vec{0}$ by positive definiteness.

Moving on to $\impliedby$:

$\vec{u} = \vec{0}$ implies $\vec{u} = 0\vec{u}$ and thus $\lVert\vec{u}\rVert = \lVert 0\vec{u}\rVert$ and $\lVert 0\vec{u}\rVert = 0\lVert \vec{u}\rVert = 0$ by homogeneity.

Note that sometimes in the definition of norm positive definitness is, in fact, replaced by $\lVert \vec{u} \rVert = 0 \iff \vec{u} = \vec{0}$, since it is equivalent and a very commonly used property of norms.

Hope this helps!

ObliviousCapybara
  • 487
  • Thanks man! Out of curiosity aside, do you know any good books that explains these proofs about analytic geometry and linear algebra to indicate me? – user113581321 Apr 06 '22 at 17:14
  • @user113581321 sadly no, i have a computer science background and I've never studied this stuff in depth outside of what's needed for numerical methods stuff ;( – ObliviousCapybara Apr 06 '22 at 20:02
  • @ObliviousCapybara: How did you know what the OP meant by $\Vert \vec{u} \Vert$? – Jesse Madnick Apr 06 '22 at 21:00
  • @JesseMadnick what do you mean? I've taken a few courses about linear algebra and matrix computations, but what i mean is that the main focus of those classes wasn't on pure mathematics but more about computational implications and such, therefore i never read any book which is purely focused on geometry... If you are interested in those topics i've read "Numerical Linear Algebra" by Trefethen, Bau; and (partially) "Matrix Analysis" by Horn, Johnson. The latter is definitely more in-depth and involved than the first, while the first one is application-oriented. – ObliviousCapybara Apr 06 '22 at 21:40
  • @JesseMadnick Also, i've read a few books about linear algebra during my bachelor but they weren't that good tbh, and they weren't written in english. – ObliviousCapybara Apr 06 '22 at 21:42
  • @JesseMadnick I just realized you probably meant "how did you know that OP's norm definition didn't include that property, as it often happens", and the answer is that since he states that the question is taken from a book, I assumed the book doesn't directly include that property in the definition. And the other usual definition of norm that I've encountered is the one I put in the answer. – ObliviousCapybara Apr 06 '22 at 21:52
  • @ObliviousCapybara: Right. My point was that the OP didn't provide their definition of $\Vert \vec{u} \Vert$, so it's hard to know how to answer. It could be, say, that the context is $ \mathbb{R}^n$, that $\Vert \vec{u} \Vert$ is defined by $( \sum |u_j|^2)^{1/2}$, and the problem wants the reader to show that $\Vert \cdot \Vert$ is a norm. Anyway, I just looked at the OP's textbook, and it appears that the definition of $\Vert \vec{u} \Vert$ is something else entirely (i.e., not what either of us have said so far), though my Portuguese isn't strong enough to know exactly what's meant. – Jesse Madnick Apr 07 '22 at 02:16