# Ch-4-Prob-14-New-김준하

## 1899 days ago by matrix

A = matrix(QQ, 2, 2, [3, 0, 1, -2]) print "character polynomial of A =" print A.charpoly()
 character polynomial of A = x^2 - x - 6 character polynomial of A = x^2 - x - 6
solve(x^2 - x - 6==0,x)
 [x == 3, x == -2] [x == 3, x == -2]
A = matrix(QQ, 2, 2, [3, 0, 1, -2]) x1=3 x2=-2 print (x1*identity_matrix(2)-A).echelon_form() print (x2*identity_matrix(2)-A).echelon_form()
 [ 1 -5] [ 0 0] [1 0] [0 0] [ 1 -5] [ 0 0] [1 0] [0 0]
A = matrix(QQ, 2, 2, [3, 0, 1, -2]) print A.eigenvectors_right()
 [(3, [ (1, 1/5) ], 1), (-2, [ (0, 1) ], 1)] [(3, [ (1, 1/5) ], 1), (-2, [ (0, 1) ], 1)]
A = matrix(QQ, 4, 4, [-3, 0, -2, 8, 0, 1, 4, -2, -4, 10, -1 ,-2, 6, -4, -2, 3]) print "character polynomial of A =" print A.charpoly()
 character polynomial of A = x^4 - 118*x^2 - 168*x + 1485 character polynomial of A = x^4 - 118*x^2 - 168*x + 1485
solve(x^4 - 118*x^2 - 168*x + 1485==0,x)
 [x == 11, x == -9, x == -5, x == 3] [x == 11, x == -9, x == -5, x == 3]
A = matrix(QQ, 4, 4, [-3, 0, -2, 8, 0, 1, 4, -2, -4, 10, -1 ,-2, 6, -4, -2, 3]) x1=11 x2=-9 x3=-5 x4=3 print (x1*identity_matrix(4)-A).echelon_form() print (x2*identity_matrix(4)-A).echelon_form() print (x3*identity_matrix(4)-A).echelon_form() print (x4*identity_matrix(4)-A).echelon_form()
 [ 1 0 0 -9/13] [ 0 1 0 7/13] [ 0 0 1 11/13] [ 0 0 0 0] [ 1 0 0 2] [ 0 1 0 -1] [ 0 0 1 2] [ 0 0 0 0] [ 1 0 0 7/5] [ 0 1 0 7/5] [ 0 0 1 -13/5] [ 0 0 0 0] [ 1 0 0 -1] [ 0 1 0 -1] [ 0 0 1 -1] [ 0 0 0 0] [ 1 0 0 -9/13] [ 0 1 0 7/13] [ 0 0 1 11/13] [ 0 0 0 0] [ 1 0 0 2] [ 0 1 0 -1] [ 0 0 1 2] [ 0 0 0 0] [ 1 0 0 7/5] [ 0 1 0 7/5] [ 0 0 1 -13/5] [ 0 0 0 0] [ 1 0 0 -1] [ 0 1 0 -1] [ 0 0 1 -1] [ 0 0 0 0]
A = matrix(QQ, 4, 4, [-3, 0, -2, 8, 0, 1, 4, -2, -4, 10, -1 ,-2, 6, -4, -2, 3]) print A.eigenvectors_right()
 [(11, [ (1, -7/9, -11/9, 13/9) ], 1), (3, [ (1, 1, 1, 1) ], 1), (-5, [ (1, 1, -13/7, -5/7) ], 1), (-9, [ (1, -1/2, 1, -1/2) ], 1)] [(11, [ (1, -7/9, -11/9, 13/9) ], 1), (3, [ (1, 1, 1, 1) ], 1), (-5, [ (1, 1, -13/7, -5/7) ], 1), (-9, [ (1, -1/2, 1, -1/2) ], 1)]