- #1

agro

- 46

- 0

^{0}defined as I

_{n*n}???

How about if O is not a square matix (i.e. O has m*n dimension where m is not equal to n).

Thanks a lot...

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter agro
- Start date

- #1

agro

- 46

- 0

How about if O is not a square matix (i.e. O has m*n dimension where m is not equal to n).

Thanks a lot...

- #2

HallsofIvy

Science Advisor

Homework Helper

- 43,021

- 970

0

If A is not a square matrix, then A

- #3

phoenixthoth

- 1,605

- 2

a lesson on exponentiation.

let's assume that A and B are finite nonnegative integers.

A^B is DEFINED to mean the following: the number of FUNCTIONS from a set of B elements to a set of A elements.

then 0^0 is the number of functions from the empty set to itself. there is ONE: the empty function.

therefore, 0^0 = 1.

however, it would be committing a fallacy to assume this has philosophical implications such as something coming from nothing.

cheers,

phoenix

let's assume that A and B are finite nonnegative integers.

A^B is DEFINED to mean the following: the number of FUNCTIONS from a set of B elements to a set of A elements.

then 0^0 is the number of functions from the empty set to itself. there is ONE: the empty function.

therefore, 0^0 = 1.

however, it would be committing a fallacy to assume this has philosophical implications such as something coming from nothing.

cheers,

phoenix

Last edited:

- #4

Hurkyl

Staff Emeritus

Science Advisor

Gold Member

- 14,967

- 19

When your domain is the nonnegative integers, 0^0 is undefined.

- #5

phoenixthoth

- 1,605

- 2

nonnegative integers are their own cardinal numbers, essentially.

for example, the cardinal number of {a} is 1 and the cardinal number of the empty set is 0. if the empty set is denoted *, then the very definition of one is {*). there is no difference between 1 and {a}, by definition. that is unless you have your own definition of the numbers 0 and 1.

by the way, since you're a pf mentor, perhaps you can answer this question: what is meant by the label "radio wave" that appears underneath my name?

cheers,

phoenix

for example, the cardinal number of {a} is 1 and the cardinal number of the empty set is 0. if the empty set is denoted *, then the very definition of one is {*). there is no difference between 1 and {a}, by definition. that is unless you have your own definition of the numbers 0 and 1.

by the way, since you're a pf mentor, perhaps you can answer this question: what is meant by the label "radio wave" that appears underneath my name?

cheers,

phoenix

Last edited:

- #6

Hurkyl

Staff Emeritus

Science Advisor

Gold Member

- 14,967

- 19

that is unless you have your own definition of the numbers 0 and 1.

I do. 0 and 1 are the additive and multiplicative identities, respectively.

One thing that seems to be neglected, with good reason, in the construction of the nonnegative integers is that the construction yields a

(the above point isn't the main point I'm trying to make, but it

The main thing is that mathematical structures include more than simply objects; they include the

As for the radio wave, it's your forum title! The forum titles here are light waves, so as you post more, your frequency goes up. (and I'm a lowly advisor, not mentor!)

- #7

agro

- 46

- 0

Originally posted by HallsofIvy

^{0}is NOT I when O is the "0" matrix anymore than

0^{0}= 1 in numbers. They are both "undefined".

If A is not a square matrix, then A^{n}is not defined for any n.

But I found on http://mathworld.wolfram.com/MatrixExponential.html that exp(A) (where A is a square matrix) is defined as

http://mathworld.wolfram.com/mimg1040.gif [Broken]

But that page also mentions that:

for a zero matrix O, exp(0) = 1

Is that a definition of "exp(O)" or a result from the general exp(A) formula? If it isn't a definition, that means O

Can you give a link that mentions that O

Thanks a lot...

Last edited by a moderator:

- #8

Hurkyl

Staff Emeritus

Science Advisor

Gold Member

- 14,967

- 19

Btw my algebra book mentions that A0 = I for any square matrix A. No mention that O0 is undefined.

It is a common

Incidentally, I forgot to mention a major difference between cardinal exponentiation and integer exponentiation; for cardinal numbers, 0^x = 1 for

- #9

phoenixthoth

- 1,605

- 2

i agree that x^x has a nasty discontinuity when considered as a function from R to R and, in that sense, from that perspective, 0^0 is undefined.

neither approach seems any more valid than the other; i guess it's just that set theory satisfies me more and starting from the bottom up rather than assuming all that stuff works and starting from the middle or top and going down to foundational number theory.

cheers,

phoenix

- #10

agro

- 46

- 0

Originally posted by Hurkyl

It's true that exp(O) = I, but I don't see how you are extrapolating from this that O^{0}= I...

It is a commonconventionto define the notation x^0 to be synonymous with 1 because it makes writing many formulae and theorems simpler, and it is a subtlety that is often neglected, or even overlooked.

Incidentally, I forgot to mention a major difference between cardinal exponentiation and integer exponentiation; for cardinal numbers, 0^x = 1 forallx, while for integers, 0^x = 0 for all positive x.

Sorry, I gave you the wrong image. Please look at this for the sigma definition of exp(A):

http://mathworld.wolfram.com/mimg1039.gif [Broken]

In the summation, there is "A^n". Then if exp(0) = I is not by definition, O^0 must be defined as I.

Last edited by a moderator:

- #11

Hurkyl

Staff Emeritus

Science Advisor

Gold Member

- 14,967

- 19

i guess it's just that set theory satisfies me more and starting from the bottom up rather than assuming all that stuff works and starting from the middle or top and going down to foundational number theory.

The "bottom up" approach has no advantage in this respect; that this particular set of nonnegative integers exists is one of the

And besides, the "top down" approach makes use of such models for the purposes of

Back to agro: In my browsing of mathworld, they seem to leave out a lot of these minor details... it feels to me that the site is intended more as a reference to people who already know the material rather than for those still learning it.

But anyways, this edge case about x^0 when x = 0 arises so often when using power series that, by convention, it is understood that x^0 is supposed to

- #12

phoenixthoth

- 1,605

- 2

what, in your way of looking at things, is the DEFINITION of a real number?

cheers,

phoenix

- #13

Hurkyl

Staff Emeritus

Science Advisor

Gold Member

- 14,967

- 19

We start with a definition:The real numbers constitute a complete simply ordered field.

- #14

phoenixthoth

- 1,605

- 2

(i know the answers)

what is complete?

what is ordered?

what is a field?

in set theory, all you have to do is define "set." and then develop approximately twelve axioms of ZF theory. this is basically the same as the number of axioms in the definition of a complete ordered field, i guess; so maybe both ways require the same amount of work.

btw, they leave the word set undefined.

- #15

Hurkyl

Staff Emeritus

Science Advisor

Gold Member

- 14,967

- 19

Simply writing down the axioms of the real numbers is far less work, easier to generalize, and has less baggage attached. The only advantage your approach has is that it is trivial to prove the real numbers are consistent relative to ZFC (and, TMK, the only reason this construction even exists is because people were interested in relative consistency).

- #16

Matrix algebra can be defined as an operator algebra, so matrices are operators that operate on vector spaces. Let A be an operator operating on the object ψ

so, A ψ = ψ'

In operator algebra A^n is defined as operating on ψ, n times with the operator A e.g. A^2 ψ = A A ψ. So A^0 means operating on ψ zero times with the operator A, hence ψ is not operated on, and is unchanged

so, A^0 ψ = ψ

It naturally follows from this that A^0 = I, where I is the identity operator. This operator algebra generalizes exactly to matrix algebra.

so, A ψ = ψ'

In operator algebra A^n is defined as operating on ψ, n times with the operator A e.g. A^2 ψ = A A ψ. So A^0 means operating on ψ zero times with the operator A, hence ψ is not operated on, and is unchanged

so, A^0 ψ = ψ

It naturally follows from this that A^0 = I, where I is the identity operator. This operator algebra generalizes exactly to matrix algebra.

Last edited by a moderator:

Share:

- Last Post

- Replies
- 8

- Views
- 418

- Last Post

- Replies
- 3

- Views
- 239

- Replies
- 44

- Views
- 2K

- Last Post

- Replies
- 5

- Views
- 455

- Last Post

- Replies
- 2

- Views
- 395

- Last Post

- Replies
- 23

- Views
- 730

- Replies
- 8

- Views
- 428

- Replies
- 3

- Views
- 229

- Replies
- 12

- Views
- 322

- Last Post

- Replies
- 5

- Views
- 418