Say What Now?
Right now, I'm diving into that wonderful tome known as "Linear Algebra Made Easy". I'm trying to prep myself for what's going to happen to me in August when my Ph. D. classes start.
So far, it's been tough, but it's straightforward and I don't feel like I'm lost. However, the problem is that the problems do not have solutions given in the back of the book. So I don't know the answer to: "does every subspace have a subspace which acts as an additive inverse under subspace addition?" (If you know, please put the answer in the comments.)
I wrote an e-mail to Dr. S, who
a) was glad I was studying (I haven't done that much; I'm only just getting over my illness), and
b) recommended several books to read. Furthermore, he told me to think about a lot of subjects such as:
* how many ways can you interpret a matrix being symmetric or positive definite?
* what is an economic or probabilistic interpretation of the adjoint to a matrix?
* what are separating hyperplanes and linear functionals?
* How is a probability measure like a vector?
* How is a markov chain like a matrix?
* How does the concept of correlation relate to the eigen-structure of a covariance matrix....?
And I don't know what any of the above means! AAAAGGGGGHHHHH!!
posted by James at 10:17 PM
3 Comments:
I'm assuming that by subspace addition, you mean where S and T are subspaces of a vector space V, their sum is S + T = {s + t: s ∈ S, t ∈ T}
I don't think that every subspace has an additive inverse under subspace addition; that would mean for every subspace S of a vector space V there would be a subspace -S such that S + -S = {0}, where {0} is the subspace consisting of just the zero vector, 0. Now suppose that S has is not the subspace containing the zero vector and that it has such an inverse, T = -S. That means for any s ∈ S all t ∈ T are such that s + t = 0 This will only be true where t = -s; otherwise s + t ≠ 0 So if T is the additive inverse of S, then it is not the additive inverse of S; the nonexistence of additive inverses for arbitrary subspace follows by contradiction.
There is probably a briefer, clearer proof of the same, but that's what I come up with. I'd venture the only subspace with an additive inverse is the zero subspace itself.
Dr. Sinisterra,
You got it. I first came up with the idea that T would have to be the -S subspace, which means that T = -S. But -S = S, and you would have S + S = 0, which would mean S = O (the zero subspace).
Dr. S (the Dr. S who is my professor) concurred, but his proof was much more like yours than it was like me. At least, we're all in agreement: the only subpace with an additive inverse is the zero subspace, which consists only of the zero vector.
Hrm. I like your proof better than mine—much cleaner and shorter.
Any progress with the others?
—Frank S. (to avoid ambiguity)
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