To show: The set {T(x1), … ,T(xk)} is a linearly independent subset of Rm Given: Let T : T : Rn→Rm be a linear transformation with nulity zero. If S={x1, ⋯ ,xk} is a linearly independent subset of Rn.

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To show:  The set {T(x1), … ,T(xk)} is a linearly independent subset of Rm  Given:  Let T : T : Rn→Rm be a linear transformation with nulity zero. If S={x1, ⋯  ,xk} is a linearly independent subset of Rn.

hosentak · Accepted Answer

Calculation:  Consider a1T(x1)+ ⋯  akT(xk)=θm  Since, T is a linear transformation. So,  T(a1x1)+, ⋯  +T(akxk)=θm  T(a1x1)+, ⋯  +(akxk)=θm  Use the result of equation (1) in the above equation.  T(a1x1+ ⋯  +akxk)=T(θn)  Use the results of equation (2), in the above equation.  a1x1+ ⋯  +akxk=θn  Now, S={x1, ⋯  xk} is a linearly independent subset of  Rn  This implies any linear combination of the elements of S will give the value of constants as 0.  This means a1=0, ⋯  ,ak=0  Therefore, {T(x1), ⋯  ,T(xk)} is a linearly independent subset of Rm  Conclusion:  Hence, {T(x1), ⋯  ,T(xk)} is linearly independent subset of Rm

To show: The set { T(x_1), ... ,T(x_k)} is a linearly independent subset of R^{m} Given: Let T : T : R^{n} rightarrow R^{m} be a linear transformation with nulity zero. If S={x_{1}, cdots , x_{k}} is a linearly independent subset of R^{n}.

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