Answer to Question #156287 in Real Analysis for wilder

Question #156287
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.
1
Expert's answer
2021-02-12T10:36:11-0500

From the question,

We'll show that f is linear if for A1,A2R22,α,βR,A_1, A_2 \in R^{2*2}, \alpha, \beta \in R, we have that f(αA1+βA2)=αf(A1)+βf(A2)f(\alpha A_1+\beta A_2)= \alpha f(A_1)+\beta f(A_2) To this end,

Let A1=(abcd),A2=(efgh)So,αA1+βA2=α(abcd)+β(efgh)    αA1+βA2=(αa+βeαb+βfαc+βgαa+βh)Now,f(αA1+βA2)=(αA1+βA2)vf(αA1+βA2)=(αa+βeαb+βfαc+βgαa+βh)(12)f(αA1+βA2)=(αaβe+2αb+2βfαcβg+2αa+2βh)f(αA1+βA2)=(αa+2αbαc+2αd)+(βe+2βfβg+2βh)f(αA1+βA2)=α(abcd)(12)+β(efgh)(12)f(αA1+βA2)=αA1v+βA2vf(αA1+βA2)=αf(A1)+βf(A2)Thus, f is a linear mappingA_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}


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