# What is matrix^0

1. Sep 11, 2003

### agro

If O is a zero/null matrix with the dimension n*n, is O0 defined as In*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...

2. Sep 11, 2003

### HallsofIvy

Staff Emeritus
No, O0 is NOT I when O is the "0" matrix anymore than
00= 1 in numbers. They are both "undefined".

If A is not a square matrix, then An is not defined for any n.

3. Sep 11, 2003

### phoenixthoth

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

Last edited: Sep 11, 2003
4. Sep 11, 2003

### Hurkyl

Staff Emeritus
That definition of exponentiation is only used when your domain is the cardinal numbers.

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

5. Sep 11, 2003

### phoenixthoth

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

Last edited: Sep 11, 2003
6. Sep 12, 2003

### Hurkyl

Staff Emeritus
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 model of the nonnegative integers, not the nonnegative integers.

(the above point isn't the main point I'm trying to make, but it is one of my bigger pet peeves... probably because I believed as you do at one time as well)

The main thing is that mathematical structures include more than simply objects; they include the operations on the objects as well. The cardinal numbers and the integers are different mathematical structures. For the cardinal numbers, 0^0 = 1 makes sense... but it does not make sense for 0^0 to exist for the integers because the integers are (isomorphic to) a subset of the real numbers, and the function x^y has a nasty discontinuity at 0^0. (a very nasty discontinuity in the complex case!)

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. Sep 12, 2003

### agro

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 O0 must be defined as I.

Can you give a link that mentions that O0 is undefined (or defined as I)? (I failed to find one) Btw my algebra book mentions that A0 = I for any square matrix A. No mention that O0 is undefined.

Thanks a lot...

Last edited by a moderator: May 1, 2017
8. Sep 12, 2003

### Hurkyl

Staff Emeritus
It's true that exp(O) = I, but I don't see how you are extrapolating from this that O0 = I...

It is a common convention to 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 for all x, while for integers, 0^x = 0 for all positive x.

9. Sep 12, 2003

### phoenixthoth

we're bumping into two approaches to foundational number theory. my approach is set theoretic and bottom up and yours is starting from abstract algebra (rings and such) and essentially from the middle down.

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. Sep 12, 2003

### agro

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: May 1, 2017
11. Sep 12, 2003

### Hurkyl

Staff Emeritus
The "bottom up" approach has no advantage in this respect; that this particular set of nonnegative integers exists is one of the axioms of Zermelo set theory rather than any sort of deduction. (the axiom of infinity, in particular)

And besides, the "top down" approach makes use of such models for the purposes of relative consistency. For example, one writes down the properties that a structure should have, such as defining the real numbers to be a complete ordered field, and one of the things for which one subsequently searches is a model of this definition to prove that the axioms are consistent (relative to the theory in which the model exists).

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 mean 1 rather than to mean x^0... so if you plug in 0 for x, you never have to evaluate 0^0 becaue the notation x^0 means 1.

12. Sep 12, 2003

### phoenixthoth

set theory does have the advantage because there are fewer terms undefinied/definied and fewer axioms to consider.

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

cheers,
phoenix

13. Sep 12, 2003

### Hurkyl

Staff Emeritus
Quoting directly from Advanced Calculus by Buck:

14. Sep 12, 2003

### phoenixthoth

then you've opened a can of worms.

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. Sep 12, 2003

### Hurkyl

Staff Emeritus
How could your approach be any less work than the axiomatic definition? You have to develop three number systems before you can even define this model of the real numbers, and even then you still have to prove the model really is a complete ordered field.

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. Sep 12, 2003

### MathNerd

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 &psi;

so, A &psi; = &psi;'

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

so, A^0 &psi; = &psi;

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: Sep 12, 2003