Let V and W be vector spaces and T:V→W be linear. Let {y1,…,yk} be a linearly independent subset of R(T). If S={x1,…,xk} is chosen so that T(ξ)=yi for i=1,…,k, prove that S is linearly independent.

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Let V and W be vector spaces and T:V→W be linear. Let {y1,…,yk} be a linearly independent subset of R(T). If S={x1,…,xk} is chosen so that  T(ξ)=yi for i=1,…,k, prove that S is linearly independent.

Thomas Lynn · Accepted Answer

Let us take {w1,w2,w3…wk} be the linear independent subset of R(t) instead of {y1…yk} and S={v1,v2,…vk} is subset of V such that T(vi)=Wi...(1).  The transformation is linear, where V and W are two vector spaces.  To show S={v1,v2,…vk} is LI.  For any scalars a1,a2,…ak consider a1v1+a2v1+…+akvk=0...(2)  Since T is a function, the above equation can be written as follows,  T(a1v1+a2v2+…+akvk)=T(0).  Since T is linear transformation,  T(a1v1)+T(a2v2)+…+T(akvk)=0  a1T(v1)+a2T(v2)+…+akT(vk)=0  a1w1+a2w2+…+akwk=0 (From (1))  Since, the set {w1,w2,w3…wk} is LI, the only choice is a1=a2…=ak=0...(3)  So, from equation (2) and (3), the set S={v1,v2,…vk} is LI.  Hence, proved.

Let V and W be vector spaces and T:V \rightarrow

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