matrices of eigen values - Rajiv Kumar Math II 1 Matrices...

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Unformatted text preview: Rajiv Kumar Math II 1 Matrices of linear transformation & eigen values eigen vectors Rajiv Kumar Math II 2 Definition : Natural matrix of a linear transformation T : V n V m Rajiv Kumar Math II 3 Example : Let T: V 3 V 3 be a linear map, defined by ( 29 ( 29 1 3 1 2 3 2 1 3 2 1 , , , , x x x x x x x x x x T-- +- = Find the matrix [T: S , S ] if S ={(1,0,0), (0,1,0),(-1,1,1)} S ={(0,1,0),(0,0,1),(3,1,0)} Rajiv Kumar Math II 4 S ={(1,0,0), (0,1,0),(-1,1,1)} S ={(0,1,0),(0,0,1),(3,1,0)} ( 29 ( 29 1 3 1 2 3 2 1 3 2 1 , , , , x x x x x x x x x x T-- +- = T(1,0,0) = (1,-1,-1) T(0,1,0) = (-1,1,0) T(-1,1,1) = (-1,2,2) [(1,-1,-1)] S =(- 4/3,-1,1/3) [(-1,1,0)] S =(4/3,0,-1/3) [(-1,2,0)] S =(7/3,2,-1/3) Rajiv Kumar Math II 5 ---- 3 1 3 1 3 1 2 1 3 7 3 4 3 4 A= Rajiv Kumar Math II 6 Let T: V 3 V 3 be a linear map. ( 29 ( 29 ( 29 { } 1 , , , , 1 , , , , 1 = S and ( 29 ( 29 ( 29 { } , , 1 , , 1 , 1 , 1 , 1 , 1 = S are ordered bases of V 3 . If Matrix of Transformation is Rajiv Kumar Math II 7 ( 29 V V : T th e F in d 3 3 = 1 1 1 , : S S T As per definition of (T:S ,S ) [Tu 1 ] S = (1,0,0) [Tu 2 ] S = (0,1,0) [Tu 3 ] S = (0,0,1) Rajiv Kumar Math II 8 ( 29 ( 29 ( 29 { } 1 , , , , 1 , , , , 1 = S ( 29 ( 29 ( 29 { } , , 1 , , 1 , 1 , 1 , 1 , 1 = S [T(1,0,0)] S = (1,0,0) T(1,0,0)= 1(1,1,1)+ 0(1,1,0)+ 0(1,0,0) T(1,0,0)= (1,1,1) T(0,1,0)= 0(1,1,1)+ 1(1,1,0)+ 0(1,0,0) =(1,1,0) T(0,0,1)= 0(1,1,1)+ 0(1,1,0)+ 1(1,0,0) =(1,0,0) Rajiv Kumar Math II 9 T(x 1 ,x 2 , x 3 )= T(x 1 (1,0,0) +x 2 (0,1,0) +x 3 (0,0,1)) T(x 1 ,x 2 , x 3 )= x 1 T (1,0,0) +x 2 T(0,1,0) +x 3 T(0,0,1) T(x 1 ,x 2 , x 3 )= x 1 (1,1,1) +x 2 (1,1,0) +x 3 (1,0,0) T(x 1 ,x 2 , x 3 )= (x 1 +x 2 +x 3 , x 1 +x 2 , x 1 ) Rajiv Kumar Math II 10 Definition : A transformation T:V V is said to be nilpotent if T n is zero transformation for some (+ve) integer n>1 , smallest such integer is called degree of nilpotence. Rajiv Kumar Math II 11 A= - 3 1 1 2 1 Q. 4 Page 157 If S ={(1,1), (-1,1)} S ={(1,1,1),(1,-1,1),(0,0,1)}, Find T Such that [T: S ,S ] = A Rajiv Kumar Math II 12 S ={(1,1), (-1,1)} ={u 1 , u 2 } S ={(1,1,1),(1,-1,1),(0,0,1)}, [Tu 1 ] S = (1,0,-1) [Tu 2 ] S = (2,1,3) T(1,1)= 1(1,1,1)+ 0(1,-1,1)- 1(0,0,1) T(1,1)= (1,1,0) T(-1,1)= 2(1,1,1)+ 1(1,-1,1)+ 3(0,0,1) =(3,1,6) Rajiv Kumar...
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matrices of eigen values - Rajiv Kumar Math II 1 Matrices...

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