
Re: Failing Linear Algebra:
Posted:
Apr 29, 2004 4:35 PM


Russell:
>Also the salient point of these variables being "free" is that they >can be set to *anything* (not just zero) and the system holds.
Right. But, once you get a system into echelon form, one free variable is set to one and the others to 0, then the system of basic variables are solved in terms of one free being 1 and the rest being 0. Then, the same thing is repeated so that each free variable equals 1 once, while holding the others at zero. This gives the basis of the system, right?
For example:
x1 + 6x5 +x7 = 5
x2 + 4x6 = 8
x3 + x5 + 8x6 = 1
x4 + 3x5 + x7 = 10
In this system, 14 are basic. 57 are free. So, first, the system is solved by setting x5 = 1, x6 = 0, and x7 = 0.
span = {(1, 8, 0, 7 1,0,0)}
Then, x5 and x7 are set at 0, while x6 = 1:
span = {(5, 4, 7, 10, 0,1,0)}
Lastly, x7 is set to 1 and x5 & x6 = 0:
span={(4, 8, 1, 9, 0,0,1)}
Right? So those three become a basis?
basis = {(1,8,0,7,1,0,0) , (5,4,7,10,0,1,0) , (4,8,1,9,0,0,1)}
Right? So, there's the basis? dim = 3, which means the dim (kernel) is three. But the original system is in R^7, so the dim of the image must be 4. Am I getting this? PLEASE correct any errors. Thanks.
Any more insight as to why exactly the dimension of the kernel would be 3 would be helpful. Is it because any of three of the variables (the free ones) can be the zero vector and the system still holds?
>That >means there are many, many column vectors satisfying such a >homogeneous matrix equation, a whole rdimensional subspace of them, >where r is the number of free variables.
Right. I've got that. The simplest way to look at it is in the following type of independence:
x+ y = 10
Y is the only free variable. But, x & y can both be 5. x can be 3 and y can be 7, or viceversa. x can be 100 and y can be 110. x can be 9.999 and y can be 0.001, etc, etc, etc. So, is the above system independent?
Also, would the basis trick work above? If y is set to 1, then x has to be 9. But if y is set to 0, then x has to be 10. So, is the basis: (9,1) or (10,0)? My guess is that it would be (10,0), which can be reduced to (1,0), right?
> >Its dimension, I think you mean. I'm not familiar with the term >"basic variables" but yes, the dimension of the image is nr using the >terminology I used above.
Yes. We call all of the variables that appear first in any of the equations, when the system is in echelon form, "basic variables". The other ones are called "free variables". So, in any system in echelon form, the number of basic variables equals the number of equations. Is that what you mean by "nr", where r is the number of free variables (dimension of the kernel)? Or, is your "r" actually the basic variables and "nr" the frees?
>Your single most damning confusion, as far as I can tell, is that you >seem not able yet to distinguish between a list (or set or space) of >vectors, and the ntuple that describes (or is) one vector.
I don't think I can even figure out what you're saying, so it must be confusing me. Could you provide an example please?
Are you trying to say that: {(0,3,4,5,1),(4,3,0,0,1), (1,1,4,9,1)} is a set of three vectors, and each one of the 3 vectors is a 5tuple? I think I've got that down.
>Take special care to notice when a list is surrounded by {}  that is >a set, usually in this context a set of vectors  and when it is >surrounded by ()  that is (usually, in this context) an ntuple, and >it (usually) refers to a *single* vector.
OK, yup. I *think* I'm getting that.
>E.g. {(1,0,0), (0,1,0), (0,0,1)} is a set of three vectors; it's one >of the possible basis sets for the vector space R^3. 1, 0, 0 are the >three components of the vector (1,0,0) in R^3.
Right. Got that, thanks. I'm good with bases when they're in that form. Otherwise, I'm having trouble, which is why the change of basis formula gets me.

