Recent content by Philmac

Define f:V\rightarrow V'' such that for each x in V, f(x) is the member of V'' such that for all y in V', f(x)(y)=y(x) Injectivity: The definition of f implies that for every x in V, there is an f(x) in V'' (I think this is all that is required) or Show that if for arbitrary x1,x2 in...

Sorry, a bunch of other issues came up and I haven't had time to properly sit down and think about it yet; I probably should have said something before. I'll finally have some time tomorrow afternoon, so I'll post what I come up with then :)

I'm exceptionally terrible at proving things, but I'll try. First, proving that f is an isomorphism: V is an n-dimensional vector space V' is the dual space of V and is therefore n-dimensional V''(=(V')') is the dual space of V' and is therefore n-dimensional Since dimV=dimV'', V is...

Are you saying that the textbook is wrong? The symbol in the book is =, not \approx or \cong. (Although, equivalency implies isometry) Not quite sure I follow here. Does this have something to do with reflexivity? That is to say, for every z in M00, z(y)=0, so y(x)=0? Which means that M00...

Here it is:

Ohh, I get it now (I think)! The claim isn't that z0(y) is always the same value, the claim is simply that z0(y) will always be the same as y(x0). Which should be obvious just by looking at the equation provided, but, oh well. I think I'm ready for annihilators now.

I like your q, 1+1=2 analogy. That, in combination with what you said below, triggered a bit of an epiphany. V' is the set of functionals that take x as their argument and V'' is the set of functionals that take y as their argument. I find this notation to be slightly misleading (I'm probably...

Thank you! I still don't quite understand yet, but I think I'm a bit closer. If I'm not mistaken, it seems that we define the elements of V'' such that regardless of the value of y (or omega) in V', there is a bijective map (an isomorphism, in this context, I believe) between V and V'', in...

Hm, alright. Thank you. I think I'm ready for reflexivity.

But how do you know that the appropriate operations will be available? And how do you know that V' is always n-dimensional? Take R1 for example, I can think of at least two linear operations: multiplication and integration. So wouldn't V' have a dimension of at least 2 (even though n=1)?

Is it always possible to do this?

My understanding of the Kronecker delta ∂ij is that it evaluates to 1 when i=j and to 0 when i≠j. So, doesn't this mean that yj(xi) must evaluate to 1 for all i=j and 0 for all i≠j for the Kronecker delta to make sense here? What guarantee is there that this will happen? In R≤3 this makes...

Thanks again! The book I'm reading is "Finite-Dimensional Vector Spaces" by Paul R. Halmos. I've read your post many times but I'm having a great deal of trouble understanding it. Math notation has always been extremely confusing for me (I really need concrete examples with numbers), so I...

Thank you very much! There is a list of axioms defining both fields and vector spaces at the beginning of the book I'm reading. I do agree that a dual space satisfies these axioms. However, I errantly believed that "vectors" had to have some geometric interpretation. I thought that perhaps...

My background in linear algebra is pretty basic: high school math and a first year course about matrix math. Now I'm reading a book about finite-dimensional vector spaces and there are a few concepts that are just absolutely bewildering to me: dual spaces, dual bases, reflexivity and...