矩阵分析答案03

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6, Find algm(A) and geom(A), where A =
1 1 0 0 2 1 0 0 2
)
.
solution: PA (t) = (t − 1)(t − 2)2 , so eigenvalues of A are {1, 2}, and algm(1) = 1, algm(2) = ( 2. ) 0 −1 0 0 −1 −1 , so rank (1 · I − A) = 2. 1 · I − A= 0 0 −1 By definition and property of null space, geom(1) = dim(null(1 · I − A)) = n − rank (1 · I − A) = 3 − 2 = 1, where n is the number or rows/columns of square) matrix A. ( 1 −1 0 0 0 −1 , so rank (2 · I − A) = 2 2 · I − A= 0 0 0 Then geom(2) = 3 − 2 = 1.
( λ ... 1
. . . 0
0 ) . . . . . . . . . λn





then: Aq x = Aq−1 Ax = Aq−1 λx = λAq−1 x = · · · = λq x Since Aq = 0, so λq x = Aq x = 0x = 0 ⇒ λ = 0. λ is an arbitrary eigenvalue of A, so all eigenvalues of A are zeros.
3
Homework 3 1, D is a diagonal matrix, compute the characteristic polynomial PD (t), and prove PD (D) = 0. proof: Let D = Then t − λ1 0 ... 0 t − λ2 . . . 0 0 PD (t) = det(tI − D) = det( . . . . ) . . . . . . . . 0 ... . . . t − λn = (t − λ1 )(t − λ2 ) . . . (t − λn ) Plug in D to PD (t), we have 0 λ1 − λ2 0 . . . 0 ... 0 0 ... ... . . . . . . . . . . . . . . . λn − λ 1 0 0 . . . λn − λ2 λ1 − λn 0 ... 0 0 λ2 − λn . . . . . . ... . . . . . . . . . . . . 0 0 ... 0 Basically we multiply the i-th diagonal entry to the i-th diagonal, i = 1, ..., n.So 0 0 ... 0 0 0 ... ... PD (D) = . . . . . . . . . . . . 0 0 ... 0 0 0 ... 0 λ2 − λ1 . . . PD (D) = . . . . . . . . . 0 0 ... 2, A ∈ Mn is called a nilpotent matrix if Aq = 0 for some positive integer q . The minimum of such q is called the index of nilpotence. Show all eigenvalues of a nilpotent matrix are all zeros. Give an example that a nonzero matrix all of whose eigenvalues are equal to 0. proof: Let λ be an eigenvalue of A, x be the associated eigenvector for λ, 1
(
)Βιβλιοθήκη Baidu
2
AB is an diagonal matrix so it’s diagonalizable. The eigenvalues of BA are {0, 0}, if BA is also diagonalizable, ) then ( 0 0 there exists an invertible matrix P s.t. BA = P −1 0 0 P , but the right hand side is always equal to 0 while the left hand side is nonzero matrix. Contradiction. So BA is not diagonalizable.
(
0 2 0 0
)
is a nonzero matrix all of whose eigenvalues are zeros.
3, If B is similar to A, then rank (B ) = rank (A). solution: By (0.4.6) (c) in the textbook, if B = XAY , where X, Y are both nonsingular, then rank (B ) = rank (A). Here we have B = P −1 AP , directly we get rank (B ) = rank (A). 4, A matrix A is square root of B , if A2 = B . Show that every diagonalizable matrix has a square root. proof: √ √ Let B = P −1 diag {b1 , ..., bn }P , let A = P −1 diag { (b1 ), ..., (bn )}P , then A2 = B . So A is a square root of B . 5, Let A, B ∈ Mn , suppose either A or B is nonsingular. If AB is diagonalizable, is also diagonalizable. Consider ( ) show that ( BA ) 0 1 1 1 A = 0 0 and B = 0 0 to show that this need not be true if both A and B are singular. proof: By theorem (3.2.8) in the slides (lecture 3), if at least one of A and B is nonsingular, then AB is similar to BA. So there exists an invertible matrix P , such that BA = P −1 ABP If AB is diagonalizable, then exists an invertible matrix Q, s.t. AB = Q−1 ΛQ, where Λ is a diagonal matrix. −1 −1 −1 Λ(QP ), is also diagonalizable. Then BA = (QP )) ( = P ) Q ΛQP ( 0 1 1 1 If A = 0 0 and B = 0 0 , both are singular because the determents of are ( both matrices )( ) 0.( ) ( )( ) 0 1 1 1 0 0 1 1 0 1 Then AB = 0 0 = , BA = 0 0 0 0 0 0 0 0 = 0 1 0 0