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4. Feb 2023•0 gefällt mir•3 views

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Let V and W be vector spaces and T: V rightarrow a one-to-one linear transformation. Prove or give a counter example to each of the following If {v1,,vk} is a basis of V then {T(vk)} is a basis of W. If {T(v1),, T(vk)} is a basis of W then {v1,, vk} is a basis of V. Solution a.) False. Let V=R 2 , W=R 3 . v 1 = (1,0), v 2 = (0,1) [a basis for V] T(x,y) = (x,y,0) [clearly one-to-one] T(v 1 ) = (1,0,0), T(v 2 ) = (0,1,0) [clearly NOT a basis for W] What caused this to fail? dim(V) does not equal dim(W). Since it is 1-1, we knew that dim(V) <= dim(W), but not that they were equal, which is needed for the desired result. b.) True. Suppose c 1 v 1 + ... + c k v k = 0. Apply T to both sides: T(c 1 v 1 + ... + c k v k ) = T(0) c 1 T(v 1 ) + ... + c k T(v k ) = 0 [since T is linear] c 1 = c 2 = ... = c k = 0 [since the T(v i ) are linearly independent] Therefore the v i are linearly independent. Let v be any element of V. Consider T(v) in W. Since the T(v i ) form a basis, there exist d 1 ,...,d k such that: d 1 T(v 1 ) + ... + d k T(v k ) = T(v) T(d 1 v 1 + ... + d k v k ) = T(v) [since T is linear] d 1 v 1 + ... + d k v k = v [since T is one-to-one!] Therefore v is a linear combination of the v i . Hence the v i span V. Since the v i span V and are linearly independent, they from a basis for V. .

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- 1. Let V and W be vector spaces and T: V rightarrow a one-to-one linear transformation. Prove or give a counter example to each of the following If {v1,,vk} is a basis of V then {T(vk)} is a basis of W. If {T(v1),, T(vk)} is a basis of W then {v1,, vk} is a basis of V. Solution a.) False. Let V=R 2 , W=R 3 . v 1 = (1,0), v 2 = (0,1) [a basis for V] T(x,y) = (x,y,0) [clearly one-to-one] T(v 1 ) = (1,0,0), T(v 2 ) = (0,1,0) [clearly NOT a basis for W] What caused this to fail? dim(V) does not equal dim(W). Since it is 1-1, we knew that dim(V) <= dim(W), but not that they were equal, which is needed for the desired result. b.) True. Suppose c 1 v 1 + ... + c k v k = 0. Apply T to both sides: T(c 1 v 1 + ... + c k v k ) = T(0) c 1 T(v 1 ) + ... + c k T(v k ) = 0 [since T is linear] c 1 = c 2 = ... = c k = 0 [since the T(v i ) are linearly independent] Therefore the v i are linearly independent. Let v be any element of V. Consider T(v) in W. Since the T(v i ) form a basis, there exist d 1 ,...,d k such that: d 1 T(v 1 ) + ... + d k T(v k ) = T(v) T(d 1 v 1 + ... + d k v k ) = T(v) [since T is linear] d 1 v 1 + ... + d k v k = v [since T is one-to-one!] Therefore v is a linear combination of the v i . Hence the v i span V.
- 2. Since the v i span V and are linearly independent, they from a basis for V.