# Since w is a subspace we know that g t v is always an

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Chapter 6 / Exercise 14
Elementary Linear Algebra
Larson
Expert Verified
Since W is a subspace, we know that g ( T )( v ) is always an element is W . 5. Let { W i } i I be the collection of T -invatirant subspaces and W be the intersection of them. For every v W , we have T ( v ) W i for every i I , since v is an element is each W i . This means T ( v ) is also an element in W . 6. Follow the prove of Theorem 5.22. And we know that the dimension is the maximum number k such that { z,T ( z ) ,T 2 ( z ) ,...,T k - 1 ( z )} is independent and the set is a basis of the subspace. (a) Calculate that z = ( 1 , 0 , 0 , 0 ) ,T ( z ) = ( 1 , 0 , 1 , 1 ) , T 2 ( z ) = ( 1 , - 1 , 2 , 2 ) ,T 3 ( z ) = ( 0 , - 3 , 3 , 3 ) . So we know the dimension is 3 and the set { z,T ( z ) ,T 2 ( z )} is a basis. 133
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Chapter 6 / Exercise 14
Elementary Linear Algebra
Larson
Expert Verified
(b) Calculate that z = x 3 ,T ( z ) = 6 x,T 2 ( z ) = 0 . So we know that the dimension is 2 and the set { z,T ( z )} is a basis. (c) Calculate that T ( z ) = z . So we know that the dimension is 1 and { z } is a basis. (d) Calculate that z = 0 1 1 0 ,T ( z ) = 1 1 2 2 , T 2 ( z ) = 3 3 6 6 . So we know that the dimension is 2 and the set { z,T ( z )} is a basis. 7. Let W be a T -invariant subspace and T W be the restricted operator on W . We have that R ( T W ) = T W ( W ) = T ( W ) W. So at least it’s a well-defined mapping. And we also have T W ( x ) + T W ( y ) = T ( x ) + T ( y ) = T ( x + y ) = T W ( x + y ) and T W ( cx ) = T ( cx ) = cT ( x ) = cT W ( x ) . So the restriction of T on W is also a lineaer operator. 8. If v is an eigenvector of T W corresponding eigenvalue λ , this means that T ( v ) = T W ( v ) = λv . So the same is true for T . 9. See Example 5.4.6. (a) For the first method, we may calculate T 3 ( z ) = ( 3 , - 3 , 3 , 3 ) and rep- resent it as a linear combination of the basis T 3 ( z ) = 0 z - 3 T ( z ) + 3 T 2 ( z ) . So the characteristic polynomial is - t 3 + 3 t 2 - 3 t . For the second method, denote β to be the ordered basis { z,T ( z ) ,T 2 ( z ) ,T 3 ( z )} . 134
And we may calculate the matrix representation [ T W ] β = 0 0 0 1 0 - 3 0 1 3 and directly find the characteristic polynomial of it to get the same result. (b) For the first method, we may calculate T 3 ( z ) = 0 and represent it as a linear combination of the basis T 2 ( z ) = 0 z + 0 T ( z ) . So the characteristic polynomial is t 2 . For the second method, denote β to be the ordered basis { z,T ( z )} . And we may calculate the matrix representation [ T W ] β = 0 0 1 0 and directly find the characteristic polynomial of it to get the same result. (c) For the first method, we may calculate T ( z ) = z . So the characteristic polynomial is - t + 1. For the second method, denote β to be the ordered basis { z } . And we may calculate the matrix representation [ T W ] β = 1 and directly find the characteristic polynomial of it to get the same result. (d) For the first method, we may calculate T 2 ( z ) = 3 T ( z ) . So the char- acteristic polynomial is t 2 - 3 t . For the second method, denote β to be the ordered basis { z,T ( z )} . And we may calculate the matrix representation [ T W ] β = 0 0 1 3 and directly find the characteristic polynomial of it to get the same result.