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# Let T - V rightarrow W be a linear transformation from the vector spac.docx ×

# Let T - V rightarrow W be a linear transformation from the vector spac.docx

Let T : V rightarrow W be a linear transformation from the vector space V to the vector space W. Let S be a set of vectors in V and let S\' = T(S) be the set of the images of these vectors under this linear transformation. Show that if S\' is linearly independent then S was already linearly independent. Is the converse true, i.e. if S is linearly independent, does this mean that S\' is automatically linearly independent?
Solution
Let {x_1,x_2,...,x_k} be a finite subset of vectors in S. Since S is linearly independent, a_1x_1+a_2x_2+...+a_kx_k=0 implies a_1=a_2=...=a_k=0. Hence, T(a_1x_1+a_2x_2+...+a_kx_k)=0, i.e, a_1T(x_1)+a_2T(x_2)+...+a_kT(x_k)=0 implies a_1=a_2=...=a_k=0. Hence {T(x_1),T(x_2),...,T(x_k)} is linearly independent for any given T(x_1),T(x_2),...,T(x_k) in S\'. Hence S being linearly independent means S \' is linearly independent. But the converse is not true. because, T(a_1x_1+a_2x_2+...+a_kx_k)=0 doesnot necessarily mean that a_1x_1+a_2x_2+...+a_kx_k=0 (kernel of T may not be just {0})..
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Let T : V rightarrow W be a linear transformation from the vector space V to the vector space W. Let S be a set of vectors in V and let S\' = T(S) be the set of the images of these vectors under this linear transformation. Show that if S\' is linearly independent then S was already linearly independent. Is the converse true, i.e. if S is linearly independent, does this mean that S\' is automatically linearly independent?
Solution
Let {x_1,x_2,...,x_k} be a finite subset of vectors in S. Since S is linearly independent, a_1x_1+a_2x_2+...+a_kx_k=0 implies a_1=a_2=...=a_k=0. Hence, T(a_1x_1+a_2x_2+...+a_kx_k)=0, i.e, a_1T(x_1)+a_2T(x_2)+...+a_kT(x_k)=0 implies a_1=a_2=...=a_k=0. Hence {T(x_1),T(x_2),...,T(x_k)} is linearly independent for any given T(x_1),T(x_2),...,T(x_k) in S\'. Hence S being linearly independent means S \' is linearly independent. But the converse is not true. because, T(a_1x_1+a_2x_2+...+a_kx_k)=0 doesnot necessarily mean that a_1x_1+a_2x_2+...+a_kx_k=0 (kernel of T may not be just {0})..
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