By Kenneth Hoffman and Ray Kunze Editorial
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Post 12 months be aware: First released January fifteenth 1998
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Additional resources for Algebra Lineal (Spanish Edition)
Thus if A : V → V is a linear mapping, the induced pull-back mapping: A∗ : Λn (V ∗ ) → Λn (V ∗ ) , is just “multiplication by a constant”. 5) for all ω in Λn (V ∗ ). 5) it’s easy to derive a number of basic facts about determinants. 3. 6) det(AB) = det(A) det(B) . Proof. 4) and (AB)∗ ω = det(AB)ω = B ∗ (A∗ ω) = det(B)A∗ ω = det(B) det(A)ω , so, det(AB) = det(A) det(B). 4. If I : V → V is the identity map, Iv = v for all v ∈ V , det(I) = 1. We’ll leave the proof as an exercise. Hint: I ∗ is the identity map on Λn (V ∗ ).
In particular, if f1 and f2 are diffeomorphisms and v = v1 (f2 )∗ (f1 )∗ v = (f2 ◦ f1 )∗ v . The results we described above have “dual” analogues for oneforms. Namely, let U and V be open subsets of R n and Rm , respectively, and let f : U → V be a C ∞ -map. Given a one-form, µ, on V one can define a “pull-back” one-form, f ∗ µ, on U by the following method. For p ∈ U let q = f (p). , an element µq ◦ dfp of Tp∗ Rn . 16. The one-form f ∗ µ is the one-form defined by the map p ∈ U → (µq ◦ dfp ) ∈ Tp∗ Rn where q = f (p).
Hint: You are required to show that the sums (−1)σ a1,σ(1) . . an,σ(n) σ ∈ Sn (−1)σ aσ(1),1 . . aσ(n),n σ ∈ Sn and are the same. Show that the second sum is identical with (−1)τ aτ (1),1 . . aτ (n),n summed over τ = σ −1 ∈ Sn . 7. Let A be an n × n matrix of the form A= B ∗ 0 C where B is a k × k matrix and C the × k block is zero. Show that × matrix and the bottom det A = det B det C . 8) every non-zero term is of the form (−1)στ b1,σ(1) . . bk,σ(k) c1,τ (1) . . c ,τ ( ) where σ ∈ Sk and τ ∈ S .
Algebra Lineal (Spanish Edition) by Kenneth Hoffman and Ray Kunze Editorial