High School

If \( T_1(x) \) and \( T_2(x) \) are onto linear transformations from \(\mathbb{R}^n\) to \(\mathbb{R}^m\), then is \( w(x) = T_1(x) + T_2(x) \) also onto?

A. True, by the definition of linear transformation.
B. True, by the definition of linear transformation and the definition of onto.
C. False. Consider \( T_2(x) = T_1(x) \), where \( T_1 \) is onto.
D. False. Consider \( T_2(x) = -T_1(x) \), where \( T_1 \) is one-to-one.
E. False. Consider \( T_2(x) = -T_1(x) \), where \( T_1 \) is onto.

Answer :

[tex]$W(\mathbf{x})$[/tex]is a linear transformation. it is not onto .The given statement is false.

The objective is to determine whether the following statement is true or false.

"If [tex]$$T_1(\mathbf{x})$ and $T_2(\mathbf{x})$[/tex] are onto linear transformations from [tex]$\mathbf{R}^n$[/tex] to [tex]$\mathbf{R}^m$[/tex] then [tex]$W(\mathbf{x})=T_1(\mathbf{x})+T_2(\mathbf{x})$[/tex] is an onto linear transformation."

To check whether [tex]$W(\mathbf{x})$[/tex] is a linear transformation or not:

For a, b ∈ R and for x, y ∈ [tex]$\mathbf{R}^n$[/tex],

[tex]$$\begin{aligned}W(a \mathbf{x}+b \mathbf{y})= & T_1(a \mathbf{x}+b \mathbf{y})+T_2(a \mathbf{x}+b \mathbf{y}) \quad\left(\text { Since } W(\mathbf{x})=T_1(\mathbf{x})+T_2(\mathbf{x})\right) \\= & {\left[a T_1(\mathbf{x})+b T_1(\mathbf{y})\right]+\left[a T_2(\mathbf{x})+b T_2(\mathbf{y})\right] } \\& \quad\left(\text { Since } T_1(\mathbf{x}) \text { and } T_2(\mathbf{x}) \text { are onto linear transformations }\right) \\\end[/tex]

[tex]= & a\left[T_1(\mathbf{x})+T_2(\mathbf{x})\right]+b\left[T_1(\mathbf{y})+T_2(\mathbf{y})\right] \\\\= & a W(\mathbf{x})+b W(\mathbf{y})\end{aligned}$$[/tex]

Therefore, [tex]$W(\mathbf{x})$[/tex] is a linear transformation.

To check whether [tex]$W(\mathbf{x})$[/tex] is onto:

Define the linear transformations:

[tex]$$T_1: \mathbf{R} \rightarrow \mathbf{R}$ such that $T_1(\mathbf{x})=\mathbf{x}$[/tex]

[tex]$$T_2: \mathbf{R} \rightarrow \mathbf{R}$ such that $T_2(\mathbf{x})=-\mathbf{x}$[/tex]

Then, [tex]$T_2(\mathbf{x})=-\mathbf{x}=-T_1(\mathbf{x}) \Rightarrow T_2(\mathbf{x})=-T_1(\mathbf{x})$[/tex]

The range of both functions [tex]$T_1, T_2$ is $\mathbf{R}$[/tex] This is same as the co-domain of [tex]$T_1, T_2$[/tex]. So, they are onto functions.

If A and B are the two sets, if for every element of B, there is at least one or more element matching with set A, it is called the onto function.

Therefore, [tex]$T_1, T_2: \mathbf{R} \rightarrow \mathbf{R}$[/tex]are onto linear transformations.

And, [tex]$W=T_1+T_2: \mathbf{R} \rightarrow \mathbf{R}$[/tex] is a mapping.

Now, [tex]$W(\mathbf{x})=T_1(\mathbf{x})+T_2(\mathbf{x})$[/tex]

[tex]$$\begin{aligned}& =T_1(\mathbf{x})-T_1(\mathbf{x}) \\& =0\end{aligned}$$[/tex]

So, the range of [tex]$$W(\mathbf{x})$ is $\{0\} \neq \mathbf{R}($[/tex] co-domain of [tex]$W(\mathbf{x}))$[/tex]

Then, is [tex]$W(\mathbf{x})$[/tex] not onto.

So, the given statement is false.

Hence, the correct answer is 5th option

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