Answer to Question #156287 in Real Analysis for wilder

Question

Let A = {(x, y, z) E R^3ㅣx - y + z = 0} where E represents "an element of" and R the set of real numbers
A mapping f is defined in R^(2*2) , the set of all square matrices of order 2 with entries in R by f: R^(2*2) → R^2
A → Av where v = (-1 2).
(a) Show that f is a linear mapping.

Answer 1

From the question,

We'll show that f is linear if for $A_1, A_2 \in R^{2*2}, \alpha, \beta \in R,$ we have that $f(\alpha A_1+\beta A_2)= \alpha f(A_1)+\beta f(A_2)$ To this end,

Let $A_1= \begin{pmatrix} a & b \\ c & d \end{pmatrix}, A_2=\begin{pmatrix} e & f \\ g & h \end{pmatrix}\\ So, \\ \alpha A_1+ \beta A_2= \alpha \begin{pmatrix} a & b \\ c & d \end{pmatrix}+ \beta \begin{pmatrix} e & f \\ g & h \end{pmatrix}\\ \implies\\ \alpha A_1+ \beta A_2=\begin{pmatrix} \alpha a+ \beta e & \alpha b+ \beta f\\ \alpha c +\beta g & \alpha a+ \beta h \end{pmatrix}\\ Now, \\ f(\alpha A_1+ \beta A_2)=(\alpha A_1+ \beta A_2)v\\ f(\alpha A_1+ \beta A_2)= \begin{pmatrix} \alpha a+ \beta e & \alpha b+ \beta f\\ \alpha c +\beta g & \alpha a+ \beta h \end{pmatrix}\begin{pmatrix} -1 \\ 2 \end{pmatrix}\\ f(\alpha A_1+ \beta A_2)=\begin{pmatrix} -\alpha a- \beta e +2 \alpha b+2 \beta f\\ -\alpha c -\beta g +2 \alpha a+2 \beta h \end{pmatrix}\\ f(\alpha A_1+ \beta A_2)=\begin{pmatrix} -\alpha a +2 \alpha b \\ \alpha c +2 \alpha d \end{pmatrix}+\begin{pmatrix} -\beta e +2 \beta f \\ -\beta g +2 \beta h \end{pmatrix}\\ f(\alpha A_1+ \beta A_2)= \alpha\begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} -1 \\ 2 \end{pmatrix}+\beta \begin{pmatrix} e & f \\ g & h \end{pmatrix}\begin{pmatrix} -1 \\ 2 \end{pmatrix}\\ f(\alpha A_1+ \beta A_2)= \alpha A_1v+ \beta A_2v\\ f(\alpha A_1+ \beta A_2)=\alpha f(A_1)+\beta f(A_2)\\ \text{Thus, f is a linear mapping}$