The given function-
T ( [ x y z ] ) = [ x + y β z 2 x y x + z + 1 ] T(\begin{bmatrix} x\\y\\z
\end{bmatrix})=\begin{bmatrix}x+y-z\\2xy\\x+z+1\end{bmatrix} T ( β£ β‘ β x y z β β¦ β€ β ) = β£ β‘ β x + y β z 2 x y x + z + 1 β β¦ β€ β
LetΞ± = [ Ξ± 1 Ξ± 2 Ξ± 3 ] , Ξ² = [ Ξ² 1 Ξ² 2 Ξ² 3 ] β R 3 \alpha=\begin{bmatrix} \alpha_1\\\alpha_2\\\alpha_3
\end{bmatrix},\beta=\begin{bmatrix} \beta_1\\\beta_2\\\beta_3
\end{bmatrix}\in R^3 Ξ± = β£ β‘ β Ξ± 1 β Ξ± 2 β Ξ± 3 β β β¦ β€ β , Ξ² = β£ β‘ β Ξ² 1 β Ξ² 2 β Ξ² 3 β β β¦ β€ β β R 3
T ( Ξ± + Ξ² ) = T ( [ Ξ± 1 Ξ± 2 Ξ± 3 ] + [ Ξ² 1 Ξ² 2 Ξ² 3 ] ) = T ( [ Ξ± 1 + Ξ² 1 Ξ± 2 + Ξ² 2 Ξ± 3 + Ξ² 3 ] ) = [ Ξ± 1 + Ξ² 1 + Ξ± 2 + Ξ² 2 β ( Ξ± 3 + Ξ² 3 ) 2 ( Ξ± 1 + Ξ² 1 ) ( Ξ± 2 + Ξ² 2 ) Ξ± 1 + Ξ² 1 + Ξ± 3 + Ξ² 3 + 1 ] = [ Ξ± 1 + Ξ± 2 β Ξ± 3 2 Ξ± 1 Ξ± 2 Ξ± 1 + Ξ± 3 + 1 ] + [ Ξ² 1 + Ξ² 2 β Ξ² 3 2 Ξ² 1 Ξ² 2 + 2 ( Ξ± 1 Ξ² 2 + Ξ² 1 Ξ± 2 ) Ξ² 1 + Ξ² 3 ] β T ( Ξ± ) + T ( Ξ² ) T(\alpha+\beta)=T(\begin{bmatrix} \alpha_1\\\alpha_2\\\alpha_3
\end{bmatrix}+\begin{bmatrix} \beta_1\\\beta_2\\\beta_3
\end{bmatrix})
\\[9pt]
=T(\begin{bmatrix} \alpha_1+\beta_1\\\alpha_2+\beta_2\\\alpha_3+\beta_3
\end{bmatrix})
\\[9pt]
=\begin{bmatrix} \alpha_1+\beta_1+\alpha_2+\beta_2-(\alpha_3+\beta_3)\\2(\alpha_1+\beta_1)(\alpha_2+\beta_2)\\\alpha_1+\beta_1+\alpha_3+\beta_3+1
\end{bmatrix}\\[9pt]
=\begin{bmatrix} \alpha_1+\alpha_2-\alpha_3\\2\alpha_1 \alpha_2\\\alpha_1+\alpha_3+1
\end{bmatrix}+\begin{bmatrix} \beta_1+\beta_2-\beta_3\\2\beta_1 \beta_2+2(\alpha_1 \beta_2+\beta_1 \alpha_2)\\\beta_1+\beta_3
\end{bmatrix}\\[9pt]\neq T(\alpha)+T(\beta) T ( Ξ± + Ξ² ) = T ( β£ β‘ β Ξ± 1 β Ξ± 2 β Ξ± 3 β β β¦ β€ β + β£ β‘ β Ξ² 1 β Ξ² 2 β Ξ² 3 β β β¦ β€ β ) = T ( β£ β‘ β Ξ± 1 β + Ξ² 1 β Ξ± 2 β + Ξ² 2 β Ξ± 3 β + Ξ² 3 β β β¦ β€ β ) = β£ β‘ β Ξ± 1 β + Ξ² 1 β + Ξ± 2 β + Ξ² 2 β β ( Ξ± 3 β + Ξ² 3 β ) 2 ( Ξ± 1 β + Ξ² 1 β ) ( Ξ± 2 β + Ξ² 2 β ) Ξ± 1 β + Ξ² 1 β + Ξ± 3 β + Ξ² 3 β + 1 β β¦ β€ β = β£ β‘ β Ξ± 1 β + Ξ± 2 β β Ξ± 3 β 2 Ξ± 1 β Ξ± 2 β Ξ± 1 β + Ξ± 3 β + 1 β β¦ β€ β + β£ β‘ β Ξ² 1 β + Ξ² 2 β β Ξ² 3 β 2 Ξ² 1 β Ξ² 2 β + 2 ( Ξ± 1 β Ξ² 2 β + Ξ² 1 β Ξ± 2 β ) Ξ² 1 β + Ξ² 3 β β β¦ β€ β ξ = T ( Ξ± ) + T ( Ξ² )
Since, T ( Ξ² ) = [ Ξ± 1 + Ξ± 2 β Ξ± 3 2 Ξ± 1 a l p h a 2 Ξ± 1 + a l p h a 3 + 1 ] β [ Ξ² 1 + Ξ² 2 β Ξ² 3 2 Ξ² 1 Ξ² 2 + 2 ( Ξ± 1 Ξ² 2 + Ξ² 1 Ξ± 2 ) Ξ² 1 + Ξ² 3 ] T(\beta)=\begin{bmatrix} \alpha_1+\alpha_2-\alpha_3\\2\alpha_1 alpha_2\\\alpha_1+alpha_3+1
\end{bmatrix}\neq\begin{bmatrix} \beta_1+\beta_2-\beta_3\\2\beta_1 \beta_2+2(\alpha_1 \beta_2+\beta_1 \alpha_2)\\\beta_1+\beta_3
\end{bmatrix} T ( Ξ² ) = β£ β‘ β Ξ± 1 β + Ξ± 2 β β Ξ± 3 β 2 Ξ± 1 β a lp h a 2 β Ξ± 1 β + a lp h a 3 β + 1 β β¦ β€ β ξ = β£ β‘ β Ξ² 1 β + Ξ² 2 β β Ξ² 3 β 2 Ξ² 1 β Ξ² 2 β + 2 ( Ξ± 1 β Ξ² 2 β + Ξ² 1 β Ξ± 2 β ) Ξ² 1 β + Ξ² 3 β β β¦ β€ β
Therefore T is not liner transformation between vector space.
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