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What is the difference between two equivalent definitions and a characterization?

To me it seems like the same thing.

For example an invertible matrix $A$. Def1 is $A^{-1}$ exists and Def2 determinant non zero.

Letting Def2 be a property, then it characterises the invertible matrecies.

And vise versa.

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  • $\begingroup$ could you give an example of what you mean? $\endgroup$ – mdave16 Jun 18 '17 at 16:32
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    $\begingroup$ @mdave16 ill add one $\endgroup$ – user415535 Jun 18 '17 at 16:34
  • $\begingroup$ "characterization"? Well, a definition is a definition and a characterization is a property the object may have. If the object must have the characterization and if nothing except the object has the characterization and if the characterization can only apply to the object for logical reasons and having the characterization "forces" anything with it to be the object then the characterization can be seen as an alternative definition, but in general they adjectives (characteristics) and nouns (definitions) are completely different concepts. $\endgroup$ – fleablood Jun 18 '17 at 16:46
  • $\begingroup$ @fleablood en.wikipedia.org/wiki/Characterization_(mathematics). This was what I had in mind. $\endgroup$ – user415535 Jun 18 '17 at 17:10
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I think, you can think of them interchangeably.

Imagine having a really big box. Then you can label a box with something like "$A^{-1}$ exists" and you define the box to be the box of invertibles (the box obviously living in the "matrix" floor/category), so it's the box of invertible matrices. Now, you could label a second box with "$\det(A) \neq 0$" and you notice this box is exactly the same as the one before, then it's a characterisation/2nd definition.

However having a second definition for something is not a well-defined concept (sometimes), so you call it a characterisation rather than a second definition.

But out there somewhere, some other mathematician has done this in the reverse order, they have defined invertible matrices as ones with non zero determinant and have found the characterisation of $A^{-1}$ existing. This is perfectly fine as they are equivalent properties.

Does this answer your question? I think it's more a linguistic question than a logical one

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  • $\begingroup$ You have an example when it is not a well defined concept? $\endgroup$ – user415535 Jun 18 '17 at 20:19
  • $\begingroup$ speaking linguistically, once you have defined something, you cannot "re-define" it for convenience. So defining a linear functions and then re-labelling this box "differentiable" functions is not well defined, because i don't know that linear is equivalent to differentiable. And indeed, they aren't equivalent. $\endgroup$ – mdave16 Jun 18 '17 at 21:55
  • $\begingroup$ yes ofc, but if they are equivalent? $\endgroup$ – user415535 Jun 19 '17 at 3:33
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    $\begingroup$ then it's perfectly well defined and nice. Again, this is really a linguistic distinction, and this is a distinction is almost too subtle to care about. $\endgroup$ – mdave16 Jun 19 '17 at 12:07
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If you have a property $D$ and can show that $D\iff E,$ you could take either $D$ or $E$ as the def'n of the property. Usually the one that gets called the def'n is usually felt to be simpler or more fundamental.

Applying some basic some calculus and trigonometry we can prove that $\pi=4\sum_{n=0}^{\infty}(-1)^n/(2n+1)$ but it is generally not preferred to define $\pi$ by this formula and work back from it to prove that $\pi$ is the area of a circle of radius $1$.

In topology there are many equivalent ways to define continuity of a function. Most of them are usually regarded as useful corollaries of continuity rather than the def'n.

In algebra, when we have a structure with a binary operation written as multiplication, with a unique two-sided identity element $I,$ it is standard terminology to say that $A$ is invertible iff $AB=BA=I$ for some $B.$ It is an important theorem in matrix theory that $A$ is invertible iff $det(A)\ne 0.$

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