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Question Number 159123 by BHOOPENDRA last updated on 13/Nov/21

Answered by aleks041103 last updated on 14/Nov/21

$$Def.LineartransformL:$$$$1)L(a+b)=L\left(a\right)+L\left(b\right)$$$$2)L\left(\alpha a\right)=\alpha L\left(a\right)$$$$$$$$T(\left[\begin{array}{c}a\\ b\\ c\\ d\end{array}\right]+\left[\begin{array}{c}A\\ B\\ C\\ D\end{array}\right])=T\left(\left[\begin{array}{c}a+A\\ b+B\\ c+C\\ d+D\end{array}\right]\right)=\left[\begin{array}{c}a+A+b+B\\ b+B-c-C\\ a+A+d+D\end{array}\right]=$$$$=\left[\begin{array}{c}a+b\\ b-c\\ a+d\end{array}\right]+\left[\begin{array}{c}A+B\\ B-C\\ A+D\end{array}\right]=T\left(\left[\begin{array}{c}a\\ b\\ c\\ d\end{array}\right]\right)+T\left(\left[\begin{array}{c}A\\ B\\ C\\ D\end{array}\right]\right)$$$$\Rightarrow T(\mathit{a}+\mathit{b})=T\left(\mathit{a}\right)+T\left(\mathit{b}\right)$$$$$$$$T\left(\alpha \left[\begin{array}{c}a\\ b\\ c\\ d\end{array}\right]\right)=T\left(\left[\begin{array}{c}\alpha a\\ \alpha b\\ \alpha c\\ \alpha d\end{array}\right]\right)=\left[\begin{array}{c}\alpha a+\alpha b\\ \alpha b-\alpha c\\ \alpha a+\alpha d\end{array}\right]=$$$$=\alpha \left[\begin{array}{c}a+b\\ b-c\\ a+d\end{array}\right]=\alpha T\left(\left[\begin{array}{c}a\\ b\\ c\\ d\end{array}\right]\right)$$$$\Rightarrow T\left(\alpha \mathit{a}\right)=\alpha T\left(\mathit{a}\right)$$$$$$$$\Rightarrow Tislinear$$

Answered by aleks041103 last updated on 14/Nov/21

$$c)$$$$$$$$T\left(\mathit{a}\right)=0,\mathrm{\forall }\mathit{a}\in N,N?$$$$N=\{\left[\begin{array}{c}a\\ b\\ c\\ d\end{array}\right]\mid T\left(\left[\begin{array}{c}a\\ b\\ c\\ d\end{array}\right]\right)=\left[\begin{array}{c}a-b\\ b-c\\ a+d\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]\}$$$$\left[\begin{array}{c}a-b\\ b-c\\ a+d\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]\Rightarrow \{\begin{array}{l}a-b=0\\ b-c=0\\ a+d=0\end{array}\Rightarrow a=b=c=-d=p$$$$But:$$$$\left[\begin{array}{c}a\\ b\\ c\\ d\end{array}\right]=\left[\begin{array}{c}p\\ p\\ p\\ -p\end{array}\right]=p\left[\begin{array}{c}1\\ 1\\ 1\\ -1\end{array}\right]$$$$\Rightarrow N=\{\left[\begin{array}{c}a\\ b\\ c\\ d\end{array}\right]\mid T\left(\left[\begin{array}{c}a\\ b\\ c\\ d\end{array}\right]\right)=\left[\begin{array}{c}a-b\\ b-c\\ a+d\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]\}$$$$N=Span\left(\left[\begin{array}{c}1\\ 1\\ 1\\ -1\end{array}\right]\right)\Rightarrow Dim\left(N\right)=1$$

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