- #1
Philmac
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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 annihilators. The book I'm reading explains everything in extremely general terms and doesn't provide any numerical examples, so I can't wrap my head around any of this. I'd really appreciate it if my loose understanding of these concepts could be critiqued/corrected and, if possible, some simple numerical examples could be provided. I really can't make heads or tails of some of this.
Note: I've never seen this bracket notation before, so I'll briefly introduce it in case it isn't something standard:
[x,y][itex]\equiv[/itex]y(x)
First, dual spaces. My understanding of a linear functional is that it's a black box where vectors go in and scalars come out (e.g. dot product). The dual space V' of a vector space V, is the set of all linear functionals that can be applied to that vector space. So, why is this called a "space"? How can things like integration and dot products (i.e. operations) form a space? The author also refers to the elements of V' as "vectors" -- how can an operation be a vector? My understanding of a vector is that it is a value with both magnitude and direction. Obviously, operations produce values, but V' is the set of operations, not values.
Second, dual bases. I just don't understand this at all, so I'll just provide the definition in this book:
So, what I think this means is that there is one operation in V' for each V, for which yj(xi)=1 for j=i and 0 for all j≠i. But does V' necessarily have dimension n?
Third, reflexivity. I just don't understand this at all. Here's the definition in the book:
Fourth, annihilators. I think I understand this concept somewhat, but the proofs presented don't make sense to me. My understanding of an annihilator is that it is any subset of V' which evaluates to 0 for all x in V. The thing I'm confused about is the annihilator of an annihilator.
Note: I've never seen this bracket notation before, so I'll briefly introduce it in case it isn't something standard:
[x,y][itex]\equiv[/itex]y(x)
First, dual spaces. My understanding of a linear functional is that it's a black box where vectors go in and scalars come out (e.g. dot product). The dual space V' of a vector space V, is the set of all linear functionals that can be applied to that vector space. So, why is this called a "space"? How can things like integration and dot products (i.e. operations) form a space? The author also refers to the elements of V' as "vectors" -- how can an operation be a vector? My understanding of a vector is that it is a value with both magnitude and direction. Obviously, operations produce values, but V' is the set of operations, not values.
Second, dual bases. I just don't understand this at all, so I'll just provide the definition in this book:
I think ∂ij is the Kronecker delta, but I'm not 100% sure.If V is an n-dimensional vector space and if X={x1,...,xn} is a basis in V, then there is a uniquely determined basis X' in V', X'={y1,...,yn}, with the property that [xi,yi]=∂ij. Consequently the dual space of an n-dimensional space is n-dimensional.
The basis X' is called the dual basis of X.
So, what I think this means is that there is one operation in V' for each V, for which yj(xi)=1 for j=i and 0 for all j≠i. But does V' necessarily have dimension n?
Third, reflexivity. I just don't understand this at all. Here's the definition in the book:
If V is a finite-dimensional vector space, then corresponding to every linear functional z0 on V' there is a vector x0 in V such that z0(y)=[x0,y](x0) for every y in V'; the correspondence z0[itex]\leftrightarrow[/itex]x0 between V'' and V is an isomorphism.
The correspondence described in this statement is called the natural correspondence between V'' and V.
It is important to observe that the theorem shows not only that V and V'' are isomorphic -- this much is trivial from the fact that they have the same dimension -- but that the natural correspondence is an isomorphism. This property of vector spaces is called reflexivity; every finite-dimensional vector space is reflexive.
Fourth, annihilators. I think I understand this concept somewhat, but the proofs presented don't make sense to me. My understanding of an annihilator is that it is any subset of V' which evaluates to 0 for all x in V. The thing I'm confused about is the annihilator of an annihilator.
Now, I'm willing to accept this proposition, but the proof is relatively short yet I cannot make sense of it. The proof is:If M is a subspace in a finite-dimensional vector space V, then M00 (=(M0)0) = M.
The problem I have here is "By definition, M00 is the set of all vectors x such that [x,y]=0 for all y in M0". Shouldn't it be the set of all vectors z (or whatever letter you like) in V'?By definition, M00 is the set of all vectors x such that [x,y]=0 for all y in M0. Since, by the definition of M0, [x,y] = 0 for all x in M and all y in M0, it follows that M[itex]\subset[/itex]M00. The desired conclusion now follows from a dimension argument. Let M be m-dimensional; then the dimension of M0 is n-m, and that of M00 is n-(n-m)=m. Hence M = M00, as was to be proved.