
Re: Failing Linear Algebra:
Posted:
Apr 28, 2004 8:38 AM


On 27 Apr 2004 22:49:33 GMT, Anonymous wrote:
>Toni: > >>Define "vector". You can't really, since you haven't properly defined >>a vector space. Hint: axioms. > >I don't understand. A vector is any collection: (x1, ...., x^n) of anything. >In math, the vectors are numbers.
Absolutely not. (1,2,3) is an example of a vector in a certain vector space. But what "vector" actually means is "element of a vector space", and what "vector space" actually means is [insert longish definition here].
An example I always give to emphasize that a vector space is exactly what the definition says it is, not what we think it is:
Let V = {the movie "Kill Bill"}. Let's say M is that movie, to save typing; now V = {M}.
For x, y in V define x + y = M. For x in V and a real number r, define fx = M.
Now it's easy to verify (_if_ you _know_ the definition of "vector space" it's easy, anyway) that V, together with the addition and scalar multiplication defined above, is a vector space. So now the movie "Kill Bill" has become a vector.
(Note it's the movie itself that's a vector, not (the movie) or the sequence of frames of the movie or whatever.)
>>Is span({0,0},{0,2}) a plane? > >No, because a1 (0,0) + a2 (0,2) = 2a2. Multiplied by any scalar, this gives >you a line through the origin and 2a2. So, it's just a line. I guess I should >have said any vector space in R^2 is at MOST a plane?? Or, maybe, any basis >for R^2 is a plane. Right? Since the (0,0) part of the span above is not >needed. So, the above span is not independent and is therefore not a basis. >That's why you just get a line. > >> Is span({0,0,0},{0,0,1},{0,0,2}) a solid >>area? > >Again, this would be a line (????): 3a3. Only either (0,0,1) or (0,0,2) is >needed to produce a basis.
************************
David C. Ullrich

