geduiwelh

2020-10-20

To show:

The set$\{T\left({x}_{1}\right),\text{}\dots \text{},T\left({x}_{k}\right)\}$ is a linearly independent subset of $R}^{m$

Given:

Let$T\text{}:\text{}T\text{}:\text{}{R}^{n}\to {R}^{m}$ be a linear transformation with nulity zero. If $S=\{{x}_{1},\text{}\cdots \text{}\text{},{x}_{k}\}$ is a linearly independent subset of ${R}^{n}.$

The set

Given:

Let

hosentak

Skilled2020-10-21Added 100 answers

Calculation:

Consider$a}_{1}T\left({x}_{1}\right)+\text{}\cdots \text{}\text{}{a}_{k}T\left({x}_{k}\right)={\theta}_{m$

Since, T is a linear transformation. So,

$T\left({a}_{1}{x}_{1}\right)+,\text{}\cdots \text{}\text{}+T\left({a}_{k}{x}_{k}\right)={\theta}_{m}$

$T\left({a}_{1}{x}_{1}\right)+,\text{}\cdots \text{}\text{}+\left({a}_{k}{x}_{k}\right)={\theta}_{m}$

Use the result of equation (1) in the above equation.

$T({a}_{1}{x}_{1}+\text{}\cdots \text{}\text{}+{a}_{k}{x}_{k})=T\left({\theta}_{n}\right)$

Use the results of equation (2), in the above equation.

$a}_{1}{x}_{1}+\text{}\cdots \text{}\text{}+{a}_{k}{x}_{k}={\theta}_{n$

Now,$S=\{{x}_{1},\text{}\cdots \text{}\text{}{x}_{k}\}$ is a linearly independent subset of $R}^{n$

This implies any linear combination of the elements of S will give the value of constants as 0.

This means${a}_{1}=0,\text{}\cdots \text{}\text{},{a}_{k}=0$

Therefore,$\{T\left({x}_{1}\right),\text{}\cdots \text{}\text{},T\left({x}_{k}\right)\}$ is a linearly independent subset of $R}^{m$

Conclusion:

Hence,$\{T\left({x}_{1}\right),\text{}\cdots \text{}\text{},T\left({x}_{k}\right)\}$ is linearly independent subset of $R}^{m$

Consider

Since, T is a linear transformation. So,

Use the result of equation (1) in the above equation.

Use the results of equation (2), in the above equation.

Now,

This implies any linear combination of the elements of S will give the value of constants as 0.

This means

Therefore,

Conclusion:

Hence,

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