A first, crude explanation

我對 "rank A = rank AT" 的第一版解釋。設 A 為 m x n 矩陣。

考慮 independent columns 的個數 c。我們知道 c 加上 A 的 null space 的維度恆為 n。直覺上,A 把愈多 domain 裡的向量銷為 0,在 codomain 所能展開的空間就愈小。這是大家熟悉的 rank-nullity equation。

次考慮 independent rows 的個數 r。視每個 row 為一個 homogeneous linear equation 的係數,把這些 equations 聯立起來,解空間 ─ 恰好仍是 A 的 null space ─ 的維度加上 r 亦恆為 n。直覺上,有愈多 independent equations 的限制,解空間就愈小。這是關於 annihilators 的維度式。

因為 null space 的維度和 n 固定,c 就等於 r 嘍。


把 rows 和 columns 互換一下,可以得到星星比較少的同構式:

Im t ≅ (Im t)* ≅ V' * / Annih Im t = V' * / Ker t* ≅ Im t*



Blogger XOO11/11/2009 6:25 pm 說:

In general, dual space often represents "adjunction" between two structures. (See p.88, MacLane's category textbook)

The isomorphism between V and V* holds only if they are finite-dimensional. It may explain that it is not completely natural to say V is isomorphic to V*, right?

Blogger Josh Ko11/13/2009 7:58 am 說:

For now I just regard V* as "transposed V", i.e., a element of V* is a row which is the transpose of an element (column) of V. This correspondence between dual spaces and row spaces does not seem to be perfect, though. For example, we have (A^T)^T = A but only V** \cong V. If you happen to have some free time, perhaps you can elaborate on "it is not completely natural to say V is isomorphic to V*"?

Blogger XOO11/18/2009 2:37 pm 說:

Given a dualization operator D, we could form an adjunction from Vec to Vec, and by the adjunction we have the unit which is a natural transformation from V to V**. However it is not a natural isomorphism in general.

Adjunction is a weaker notion of equivalence of categories, and equivalene is weaker than isomorphism of categories.

Currently, you may think adjunction as Galois connection.

Do you have a copy of MacLane's category textbook?

Blogger Josh Ko11/19/2009 10:37 am 說:

I do, but I do not have time to read it before I complete the research proposal. Hopefully I'll be able to acquaint myself with (pure) category theory next year. In fact I am now quite curious about exactly how successful category theory is (in terms of pure mathematics), and to have an answer it seems necessary to understand the theory to a certain extent. That requires much time and energy which I cannot devote to the theory now.


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