
Re: Failing Linear Algebra:
Posted:
Apr 29, 2004 3:24 AM


On 28 Apr 2004 22:32:04 GMT, Anonymous wrote:
>Russell: > >> If the kernel >>is {0}, i.e. the set of rows (or the set of columns) of the matrix is >>linearly independent, and the matrix is square, then the matrix is >>invertible and you have an isomorphism. Otherwise, one or more rows >>or columns is a linear combination of the others; and that means there >>are nonzero vectors which, when multiplied by your matrix, give zero. >>In other words, those vectors are members of the kernel of your linear >>transformation. See how it all fits together? > >Right. So, if a matrix is homogenous and in echelon form, the number of free >variables is the same as the kernel, because if any free variable is set equal >to 0, then the system holds.
To my knowledge, the term "homogeneous" is applied to a certain matrix equation, not to the matrix itself. And the kernel is a set, not a number; you should be talking about the *dimension* of the kernel in this context. Also you should be aware that the "free variables" you speak of here, as far as I understand you, are not themselves vectors in your space; they are *components* of the nx1 column vector _x_ that you are multiplying by your matrix to get the nx1 _0_ column vector. Also the salient point of these variables being "free" is that they can be set to *anything* (not just zero) and the system 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.
Therefore, the number of basic variables is the >same as the image, 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. These are the components of _x_ that *must* be set to zero for the homogeneous equation to hold, assuming that the matrix is in rowechelon form.
Btw you are again not thinking abstractly enough for my taste; this matrix equation (or system of equations if you prefer to call it that) is one very nice application of matrix theory, but by no means is it the only one.
> >>Btw you asked elsewhere what an eigenvalue is. The definition is >>simplicity itself, learn it! > >Yeah, I know the definition. I think I stated it in a post already. > >>Btw I found nothing objectionable >>about Lipschutz from an abstract point of view; if he made it any >>simpler he would mislead, but he treads that fine line very nicely >>IMHO. > >OK, thanks. I'll be sure to use it to study.
Of course, your mileage may vary. If you do use it, be sure not to neglect the exercises that are proofs. Indeed if triage is needed, you probably should concentrate mainly on proofs over the next couple weeks.
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. Of course every component of a vector does also happen to be a vector in R, but in most cases R is a *different* vector space than the one you are considering, i.e. the components are *not* vectors in the context of the problem or proof you are working on. So to avoid confusion, you shouldn't call them vectors.
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. Perhaps it is an unknown or arbitrary vector, but it is *one* nonetheless. And nonethemore.
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.

