91Ó°ÊÓ

$\mathrm{T}\left(\mathrm{q}\right({\mathrm{x}}_1,{\mathrm{x}}_2\left)\right)=\mathrm{q}({\mathrm{x}}_1,0)$

Consider ${\mathrm{q}}_1\left({\mathrm{x}}_1,{\mathrm{x}}_2\right)={\mathrm{a}}_1{\mathrm{x}}_1^2+{\mathrm{b}}_1{\mathrm{x}}_1{\mathrm{x}}_2+{\mathrm{c}}_1{\mathrm{x}}_2^2$

$q_2(x_1,x_2)=a_2{x}_1^2+b_2{x}_1{x}_2+c_2{x}_2^2∈Q_2\mathrm{and}\mathrm{Î\pm }∈\mathrm{R}$then9

$$\begin{gathered} T\left(a_1\left(x_1,x_2\right)+a_2\left(x_1,x_2\right)\right)=T\left(a\left(a_1{x}_1^2+b_1{x}_1{x}_2+c_1{x}_2^2\right)+a_2{x}_1^2+b_2{x}_1{x}_2+c_2{x}_2^2\right) \\ =T\left(\left(a{a}_1+a_2\right)x_1^2+\left(a{b}_1+b_2\right)x_1{x}_2+\left(a_1+c_2\right)x_2^2\right) \\ =\left(a{a}_1+a_2\right)x_1^2 \\ =\left(a{a}_1+a_2\right)x_1^2 \\ =a{q}_1\left(x_1,0\right)+a_2\left(x_1,0\right) \\ =\mathrm{Î\pm }T\left(a_1\left(x_1,x_2\right)\right)+T\left(a_2\left(x_1,x_2\right)\right) \end{gathered}$$

Since T satisfy

$\mathrm{T}({\mathrm{Î\pm \pm ç}}_1+{\mathrm{q}}_2)=\mathrm{Î\pm °Õ}\left({\mathrm{q}}_1\right)+\mathrm{T}\left({\mathrm{q}}_2\right)\mathrm{´Ú´Ç°ùÎ\pm }∈\mathrm{R}\mathrm{and}{\mathrm{q}}_1,{\mathrm{q}}_2∈{\mathrm{Q}}_2$

$\mathrm{T}:{\mathrm{Q}}_2→{\mathrm{P}}_2,\mathrm{T}\left(\mathrm{q}\right({\mathrm{x}}_1,{\mathrm{x}}_2\left)\right)=\mathrm{q}({\mathrm{x}}_1,0)$is a linear transformation.

Kernel of T:

$$\begin{gathered} \ker \left(T\right)=\left\{q\left(x_1,x_2\right)∈Q_2:T\left(q\left(x_1,x_2\right)\right)=0∈P_2\right\} \\ =\left\{q\left(x_1,x_2\right)∈Q_2:q\left(x_1,0\right)=0∈P_2\right\} \\ =\left\{q\left(x_1,x_2\right)=a{x}_1^2+b{x}_1{x}_2+c{x}_2^2∈Q_2:a{x}_1^2=0\right\} \\ =\left\{q\left(x_1,x_2\right)=a{x}_1^2+b{x}_1{x}_2+c{x}_2^2∈Q_2:a=0\right\} \\ =\left\{q\left(x_1,x_2\right)=b{x}_1{x}_2+c{x}_2^2∈Q_2:b,c∈R\right\} \end{gathered}$$

Image of T:

$$\begin{gathered} \mathrm{Im}\left(T\right)=\left\{T\left(q\left(x_1,x_2\right)\right)∈P_2:q\left(x_1,x_2\right)∈Q_2\right\} \\ =\left\{q\left(x_1,0\right)∈P_2:q\left(x_1,x_2\right)=a{x}_1^2+b{x}_1{x}_2+c{x}_2^2∈Q_2\right\} \\ =\left\{q\left(x_1,0\right)=a{x}_1^2∈P_2:a∈R\right\} \\ =\left\{p\left(x\right)=a{x}^2∈P_2:a∈R\right\} \end{gathered}$$

$$\begin{gathered} \mathrm{rank}\left(\mathrm{T}\right)=\dim \left(\mathrm{Img}\right(\mathrm{T}\left)\right)=1 \\ \mathrm{nullity}\left(\mathrm{T}\right)=\dim \left(\ker \right(\mathrm{T}\left)\right)=2 \end{gathered}$$

Yes, is a linear transformation.

$$\begin{gathered} \ker \left(\mathrm{T}\right)=\mathrm{q}({\mathrm{x}}_1,{\mathrm{x}}_2)={\mathrm{bx}}_1{\mathrm{x}}_2+{\mathrm{cx}}_2^2∈{\mathrm{Q}}_2:\mathrm{b},\mathrm{c}∈\mathrm{R} \\ \mathrm{Im}\left(\mathrm{T}\right)=\mathrm{p}\left(\mathrm{x}\right)={\mathrm{ax}}^2∈{\mathrm{P}}_2:\mathrm{a}∈\mathrm{R} \\ \mathrm{rank}\left(\mathrm{T}\right)=1\mathrm{and}\mathrm{nullity}\left(\mathrm{T}\right)=2 \end{gathered}$$