請問這題答案是 -1/8 嗎 我算是這樣 還有二題不知道怎麼做 謝謝
Let A be a m×n matrix.
If B is a nonsingular m×m matrix, use the Rank-Nullity Theorem to show that BA
and A have the same nullspace and hence the same rank.
If C is a nonsingular n×n matrix, use the conclusion above to show that AC and A
have the same rank.
An n×n matrix A is called nilpotent if Am=0 for some positive integer m. Consider an n×n
nilpotent matrix A, and choose the smallest number m such that Am=0. Pick a vector v in Rn such
that Am-1v≠0.
(10 pts) Show that the vectors v, Av, A2v, …, Am-1v are linearly independent.
3 則留言:
你應該要給det(A)..
1. Let C = [0 2 0; 1 -1 -1; 0 0 1] => B = C(A^T)
=> B^-1 = ((A^T)^-1)(C^-1)
det(AB^-1) = det(A)det(B^-1)
= det(A)(det(A^T)^-1)(det(C)^-1)
= det(A)(det(A)^-1)(det(C)^-1)
= det(C)^-1 = -1/2
2. 老師上課時有證過 ker(A) = ker(BA),
也就是他一直強調的列運算保 kernel
所以 nullity(A) = nullity(BA)
=> rank(A) = n-nullity(A) = n-nullity(BA) = rank(BA),
AC 的想法和 BA 意思差不多
3. 這在上課時應該也有提過 (chap 6), 假設
(a_0)v + (a_1)Av + (a_2)(A^2)v + … + (a_m-1)A^(m-1)v = 0, 先拿 A^(m-1) 乘以上面這個式子, 可得 a_0=0, 後面以此類推可證出 a_i = 0, for all i
這題不用 最後會消掉 答案的確是-1/2
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