I do. 0 and 1 are the additive and multiplicative identities, respectively.that is unless you have your own definition of the numbers 0 and 1.
But I found on http://mathworld.wolfram.com/MatrixExponential.html that exp(A) (where A is a square matrix) is defined asOriginally posted by HallsofIvy
No, O^{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.
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.Btw my algebra book mentions that A0 = I for any square matrix A. No mention that O0 is undefined.
Sorry, I gave you the wrong image. Please look at this for the sigma definition of exp(A):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 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.
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)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.
We start with a definition: The real numbers constitute a complete simply ordered field.