Question

Consider the transformation $\mathbf{T}\mathbf{\left(}\mathbf{q}\left(x_1,x_2,x_3\right)\mathbf{\right)}\mathbf{=}\mathbf{q}\mathbf{(}{\mathbf{x}}_{\mathbf{1}}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{1}\mathbf{)}\mathbf{}\mathbf{from}\mathbf{}{\mathbf{Q}}_{\mathbf{3}}\mathbf{}\mathbf{to}\mathbf{}{\mathbf{P}}_{\mathbf{2}}$. Is T a linear transformation? If so, find the image, rank, kernel, and nullity of T

Short answer

the solution is

Yes,T is a linear transformation

$$\begin{gathered} \ker \left(\mathrm{T}\right)=\left\{\mathrm{q}\left({\mathrm{x}}_1,{\mathrm{x}}_2,{\mathrm{x}}_3\right)={\mathrm{bx}}_2^2+{\mathrm{cx}}_3^2+{\mathrm{dx}}_1{\mathrm{x}}_2-\left(\mathrm{b}+\mathrm{c}\right){\mathrm{x}}_2{\mathrm{x}}_3-{\mathrm{dx}}_3{\mathrm{x}}_1\in {\mathrm{Q}}_3\right\} \\ \mathrm{lm}\left(\mathrm{T}\right)=\left\{\mathrm{p}\left(\mathrm{x}\right)={\mathrm{\alpha x}}^2+\mathrm{\beta x}+\mathrm{\lambda }\in {\mathrm{P}}_2\right\} \\ \mathrm{rank}\left(\mathrm{T}\right)=3\mathrm{and}\mathrm{nullity}\left(\mathrm{T}\right)=3 \end{gathered}$$

Step-by-step solution

given information

$\mathrm{T}\left(\mathrm{q}({\mathrm{x}}_1,{\mathrm{x}}_2,{\mathrm{x}}_3)\right)=\mathrm{q}({\mathrm{x}}_1,1,1)$

linear transformation

$$\begin{gathered} \mathrm{Consider},{\mathrm{q}}_1\left({\mathrm{x}}_1,{\mathrm{x}}_2,{\mathrm{x}}_3\right)={\mathrm{a}}_1{\mathrm{x}}_1^2+{\mathrm{b}}_1{\mathrm{x}}_2^2+{\mathrm{c}}_1{\mathrm{x}}_3^2+{\mathrm{d}}_1{\mathrm{x}}_1{\mathrm{x}}_2+{\mathrm{e}}_1{\mathrm{x}}_2{\mathrm{x}}_3+{\mathrm{f}}_1{\mathrm{x}}_3{\mathrm{x}}_1, \\ {\mathrm{q}}_1\left({\mathrm{x}}_1,{\mathrm{x}}_2,{\mathrm{x}}_3\right)={\mathrm{a}}_1{\mathrm{x}}_1^2+{\mathrm{b}}_1{\mathrm{x}}_2^2+{\mathrm{c}}_1{\mathrm{x}}_3^2+{\mathrm{d}}_1{\mathrm{x}}_1{\mathrm{x}}_2+{\mathrm{e}}_1{\mathrm{x}}_2{\mathrm{x}}_3+{\mathrm{f}}_1{\mathrm{x}}_3{\mathrm{x}}_1,\in {\mathrm{Q}}_3\mathrm{and}\mathrm{\alpha }\in \mathrm{R},\mathrm{then} \\ {\mathrm{a}}_2{\mathrm{x}}_1^2+{\mathrm{b}}_2{\mathrm{x}}_2^2+{\mathrm{c}}_2{\mathrm{x}}_3^2+{\mathrm{d}}_2{\mathrm{x}}_1{\mathrm{x}}_2+{\mathrm{e}}_1{\mathrm{x}}_2{\mathrm{x}}_3+{\mathrm{f}}_1{\mathrm{x}}_3{\mathrm{x}}_1) \\ =\mathrm{T}\left(\left({\mathrm{\alpha d}}_1+{\mathrm{d}}_2\right){\mathrm{x}}_1^2+\left({\mathrm{\alpha b}}_1+{\mathrm{b}}_2\right){\mathrm{x}}_2^2+({\mathrm{\alpha c}}_1+{\mathrm{c}}_2\right){\mathrm{x}}_3^2+ \\ \left({\mathrm{\alpha d}}_1+{\mathrm{d}}_2\right){\mathrm{x}}_1{\mathrm{x}}_2+\left({\mathrm{\alpha e}}_1+{\mathrm{e}}_2\right){\mathrm{x}}_2{\mathrm{x}}_3+({\mathrm{\alpha f}}_1+{\mathrm{f}}_2){\mathrm{x}}_3{\mathrm{x}}_1 \\ =\left({\mathrm{\alpha d}}_1+{\mathrm{a}}_2\right){\mathrm{x}}_1^2+({\mathrm{\alpha b}}_1+{\mathrm{b}}_2+({\mathrm{\alpha c}}_1+{\mathrm{c}}_2)+ \\ \left({\mathrm{\alpha d}}_1+{\mathrm{d}}_2\right){\mathrm{x}}_1+\left({\mathrm{\alpha e}}_1+{\mathrm{e}}_2\right)+({\mathrm{\alpha f}}_1+{\mathrm{f}}_2) \\ =\mathrm{\alpha }\left({\mathrm{a}}_1{\mathrm{x}}_1^2+{\mathrm{b}}_1+{\mathrm{c}}_1+{\mathrm{d}}_1{\mathrm{x}}_1+{\mathrm{e}}_1{\mathrm{f}}_1{\mathrm{x}}_1\right)+\left({\mathrm{a}}_2{\mathrm{x}}_1^2+{\mathrm{b}}_2+{\mathrm{c}}_2+{\mathrm{d}}_2{\mathrm{x}}_1+{\mathrm{e}}_2{\mathrm{f}}_2{\mathrm{x}}_1\right) \\ ={\mathrm{\alpha q}}_1\left({\mathrm{x}}_1,1,1\right)+{\mathrm{q}}_2\left({\mathrm{x}}_1,1,1\right) \\ =\mathrm{\alpha T}\left({\mathrm{q}}_1\left({\mathrm{x}}_1,{\mathrm{x}}_2,{\mathrm{x}}_3\right)\right)+\mathrm{T}\left({\mathrm{q}}_2\left({\mathrm{x}}_1,{\mathrm{x}}_2,{\mathrm{x}}_3\right)\right) \end{gathered}$$,

Since satisfy

$$\begin{gathered} \mathrm{T}\left({\mathrm{\alpha q}}_1+{\mathrm{q}}_2\right)=\mathrm{\alpha T}\left({\mathrm{q}}_1\right)+\mathrm{T}\left({\mathrm{q}}_2\right)\mathrm{for}\mathrm{\alpha }\in \mathrm{R}\mathrm{and}{\mathrm{q}}_1,{\mathrm{q}}_2{\mathrm{iQ}}_{3,} \\ \mathrm{so}\mathrm{T}:{\mathrm{Q}}_3\to {\mathrm{P}}_2,\mathrm{T}\left(\mathrm{q}\left({\mathrm{x}}_1,{\mathrm{x}}_2,{\mathrm{x}}_3\right)\right)=\mathrm{q}\left({\mathrm{x}}_{1}1,1\right) \\ \mathrm{Is}\mathrm{a}\mathrm{linear}\mathrm{transformation}. \end{gathered}$$

find kernel of T and image of T

$$\begin{gathered} \ker \left(\mathrm{T}\right)=\left\{\mathrm{q}\left({\mathrm{x}}_1,{\mathrm{x}}_2,{\mathrm{x}}_3\right)\in {\mathrm{Q}}_3:\mathrm{T}\left(\mathrm{q}\left({\mathrm{x}}_1,{\mathrm{x}}_2,{\mathrm{x}}_3\right)\right)=0\in {\mathrm{P}}_2\right\} \\ =\left\{\mathrm{q}\left({\mathrm{x}}_1,{\mathrm{x}}_2,{\mathrm{x}}_3\right)\in {\mathrm{Q}}_2:\mathrm{q}\left({\mathrm{x}}_1,1,1\right)=0\in {\mathrm{P}}_2\right\} \\ =\left\{{\mathrm{ax}}_1^2+{\mathrm{bx}}_2^2+{\mathrm{dx}}_1{\mathrm{x}}_2+{\mathrm{ex}}_2{\mathrm{x}}_3+{\mathrm{fx}}_3{\mathrm{x}}_1\in {\mathrm{Q}}_3:\mathrm{q}\left({\mathrm{x}}_1,1,1\right)=0\right\} \\ =\left\{\mathrm{q}\left({\mathrm{x}}_1,{\mathrm{x}}_2,{\mathrm{x}}_3\right)\in {\mathrm{Q}}_3:{\mathrm{ax}}_1^2\left(\mathrm{d}+\mathrm{f}\right){\mathrm{x}}_1+\left(\mathrm{b}+\mathrm{c}+\mathrm{e}\right)=0\in {\mathrm{P}}_2\right\} \\ =\left\{{\mathrm{ax}}_1^2+{\mathrm{bx}}_2^2+{\mathrm{cx}}_3^2+{\mathrm{dx}}_1{\mathrm{x}}_2+{\mathrm{ex}}_2{\mathrm{x}}_3+{\mathrm{fx}}_3{\mathrm{x}}_1\in {\mathrm{Q}}_3:\mathrm{a}=0,\mathrm{f}=-\mathrm{b}-\mathrm{c}\right\} \\ =\left\{\mathrm{q}\left({\mathrm{x}}_1,{\mathrm{x}}_2,{\mathrm{x}}_3\right)={\mathrm{bx}}_2^2+{\mathrm{cx}}_3^2+{\mathrm{dx}}_1{\mathrm{x}}_2-\left(\mathrm{b}+\mathrm{c}\right){\mathrm{x}}_2{\mathrm{x}}_3{\mathrm{dx}}_3{\mathrm{x}}_1:\mathrm{b},\mathrm{c},\mathrm{d}\in \mathrm{R}\right\} \\ \mathrm{lm}\left(\mathrm{T}\right)=\left\{\mathrm{T}\left(\mathrm{q}\left({\mathrm{x}}_1,{\mathrm{x}}_2,{\mathrm{x}}_3\right)\right)\in {\mathrm{P}}_2:\mathrm{q}\left({\mathrm{x}}_1,{\mathrm{x}}_2,{\mathrm{x}}_3\right)\in {\mathrm{Q}}_3\right\} \\ =\left\{\mathrm{q}\left({\mathrm{x}}_1,1,1\right)\in {\mathrm{P}}_2:\mathrm{q}\left({\mathrm{x}}_1,{\mathrm{x}}_2,{\mathrm{x}}_3\right)={\mathrm{ax}}_1^2+{\mathrm{bx}}_3^2+{\mathrm{dx}}_1{\mathrm{x}}_2+{\mathrm{ex}}_1{\mathrm{x}}_2+{\mathrm{fx}}_3{\mathrm{x}}_1\in {\mathrm{Q}}_3\right\} \\ =\left\{\mathrm{q}\left({\mathrm{x}}_1,1,1\right)={\mathrm{a}}_1{\mathrm{x}}_1^2+\left(\mathrm{d}+\mathrm{f}\right){\mathrm{x}}_1+\left(\mathrm{b}+\mathrm{c}+\mathrm{e}\right)\in {\mathrm{P}}_2\right\} \\ \left\{\mathrm{p}\left(\mathrm{x}\right)={\mathrm{\alpha x}}^2+\mathrm{\beta x}+\mathrm{y}\in {\mathrm{P}}_2:\mathrm{\alpha }=\mathrm{a},\mathrm{\beta }=\mathrm{d}+\mathrm{f},\mathrm{y}=\mathrm{b}+\mathrm{c}+\mathrm{e}\in \mathrm{R}\right\} \\ \mathrm{rank}\left(\mathrm{T}\right)=\dim \left(\mathrm{lm}\left(\mathrm{T}\right)\right)=3 \\ \mathrm{nullity}\left(\mathrm{T}\right)=\dim \left(\ker \left(\mathrm{T}\right)\right)=4 \end{gathered}$$

conclusion

Yes,T is a linear transformation

$$\begin{gathered} \ker \left(\mathrm{T}\right)=\left\{\mathrm{q}\left({\mathrm{x}}_1,{\mathrm{x}}_2,{\mathrm{x}}_3\right)={\mathrm{bx}}_2^2+{\mathrm{cx}}_3^2+{\mathrm{dx}}_1{\mathrm{x}}_2-\left(\mathrm{b}+\mathrm{c}\right){\mathrm{x}}_2{\mathrm{x}}_3-{\mathrm{dx}}_3{\mathrm{x}}_1\in {\mathrm{Q}}_3\right\} \\ \mathrm{lm}\left(\mathrm{T}\right)=\left\{\mathrm{p}\left(\mathrm{x}\right)={\mathrm{\alpha x}}^2+\mathrm{\beta x}+\mathrm{\lambda }\in {\mathrm{P}}_2\right\} \\ \mathrm{rank}\left(\mathrm{T}\right)=3\mathrm{and}\mathrm{unllity}\left(\mathrm{T}\right)=3 \end{gathered}$$