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}.

geduiwelh 2020-10-20 Answered
To show:
The set {T(x1),  ,T(xk)} is a linearly independent subset of Rm
Given:
Let T : T : RnRm be a linear transformation with nulity zero. If S={x1,   ,xk} is a linearly independent subset of Rn.
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Expert Answer

hosentak
Answered 2020-10-21 Author has 100 answers
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
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