Answer to Question #144440 in Linear Algebra for Sourav Mondal

Question

Suppose a1 = (1, 0, 1), a2 = (0, 1, -2) and
a3 = (-1, -1, 0) are vectors in R³ and
f : R³ -> R is a linear functional such that
f(a1) = 1, f(a2) = -1 and f(a3) = 3. If
a = (p,q,r) belongs to R3, find f(a).

Answer 1

A mapping $f:V\to \Bbb{F}$ from the vector space $V$ to scalar field $\Bbb{F}$ is called a linear functional iff:

$f(\alpha x+\beta y)=\alpha f(x)+\beta f(y)$

holds for all $x,y$ in $V$ and for all $\alpha, \beta$ in $\Bbb{F}$

Given

$a=(p,q,r)=\lambda_1(1,0,1)+\lambda_2(0,1,-2)+\lambda_3(-1,-1,0)$

$\begin{matrix} \lambda_1-\lambda_3=p \\ \lambda_2-\lambda_3=q \\ \lambda_1-2\lambda_2=r \\ \end{matrix}$

$\begin{matrix} \lambda_1=\lambda_3+p \\ \lambda_2=\lambda_3+q \\ \lambda_3+p-2\lambda_3-2q=r \\ \end{matrix}$

$\begin{matrix} \lambda_1=2p-2q-r \\ \lambda_2=p-q-r \\ \lambda_3=p-2q-r \\ \end{matrix}$

$a=(2p-2q-r)a_1+(p-q-r)a_2+(p-2q-r)a_3$

$f(a)=(2p-2q-r)f(a_1)+(p-q-r)f(a_2)+$

$+(p-2q-r)f(a_3)=$

$=(2p-2q-r)(1)+(p-q-r)(-1)+$

$+(p-2q-r)(3)=$

$=2p-2q-r-p+q+r+3p-6q-3r=$

$=4p-7q-3r$

$f(a)=4p-7q-3r$