Question

Suppose al = (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(al) = 1, f(a2) = -1 and f(a3) = 3.

If  a = (p,q,r) belongs to R3, find f(a). 

Answer 1

Since $\left|\begin{array}{ccc} 1 & 0 & 1\\ 0 & 1 & -2\\-1 & -1 & 0 \end {array}\right|=0+0+0-(-1)-0-(-1)(-2)=-1\ne 0$ , the vectors $a_1,a_2,a_3$ are linearly independent in $\mathbb R^3$ , and thus form a basis of $\mathbb R^3$ . Therefore, each vector $a = (p,q,r)\in\mathbb R^3$ can be represent as a linear combination of $a_1,a_2,a_3$, that is $a=ta_1+sa_2+ma_3$ for some $t,s,m\in\mathbb R.$ Therefore, we have

$(p,q,r)=t(1,0,1)+s(0,1,-2)+m(-1,-1,0)=(t-m,s-m,t-2s)$. Then for each $(p,q,r)$ let us solve the following system:

$\begin{cases} p = t-m\\ q= s-m \\ r =t-2s \end{cases}$

which is equivalent to

$\begin{cases} p = t-m\\ q-p= s-t \\ r =t-2s \end{cases}$

$\begin{cases} p = t-m\\ q-p= s-t \\ r+q-p =-s \end{cases}$

$\begin{cases} p = t-m\\ r+2q-2p= -t \\ r+q-p =-s \end{cases}$

$\begin{cases} m = t-p\\ t= 2p-2q-r \\ s=p-q-r \end{cases}$

$\begin{cases} m = p-2q-r\\ t= 2p-2q-r \\ s=p-q-r \end{cases}$

Taking into account that $f:\mathbb R^3\to \mathbb R$ is a linear function, we conclude that

$f(a)=f(ta_1+sa_2+ma_3)=tf(a_1)+sf(a_2)+mf(a_3)=(2p-2q-r)\cdot 1+(p-q-r)(-1)+(p-2q-r)\cdot 3=4p-7q-3r.$

Answer: $f(a)=f(p,q,r)=4p-7q-3r.$