Question #143171

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). 


1
Expert's answer
2020-11-09T15:45:24-0500

Since 101012110=0+0+0(1)0(1)(2)=10\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 a1,a2,a3a_1,a_2,a_3 are linearly independent in R3\mathbb R^3 , and thus form a basis of R3\mathbb R^3 . Therefore, each vector a=(p,q,r)R3a = (p,q,r)\in\mathbb R^3 can be represent as a linear combination of a1,a2,a3a_1,a_2,a_3, that is a=ta1+sa2+ma3a=ta_1+sa_2+ma_3 for some t,s,mR.t,s,m\in\mathbb R. Therefore, we have


(p,q,r)=t(1,0,1)+s(0,1,2)+m(1,1,0)=(tm,sm,t2s)(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)(p,q,r) let us solve the following system:


{p=tmq=smr=t2s\begin{cases} p = t-m\\ q= s-m \\ r =t-2s \end{cases}


which is equivalent to


{p=tmqp=str=t2s\begin{cases} p = t-m\\ q-p= s-t \\ r =t-2s \end{cases}


{p=tmqp=str+qp=s\begin{cases} p = t-m\\ q-p= s-t \\ r+q-p =-s \end{cases}


{p=tmr+2q2p=tr+qp=s\begin{cases} p = t-m\\ r+2q-2p= -t \\ r+q-p =-s \end{cases}


{m=tpt=2p2qrs=pqr\begin{cases} m = t-p\\ t= 2p-2q-r \\ s=p-q-r \end{cases}


{m=p2qrt=2p2qrs=pqr\begin{cases} m = p-2q-r\\ t= 2p-2q-r \\ s=p-q-r \end{cases}


Taking into account that f:R3Rf:\mathbb R^3\to \mathbb R is a linear function, we conclude that


f(a)=f(ta1+sa2+ma3)=tf(a1)+sf(a2)+mf(a3)=(2p2qr)1+(pqr)(1)+(p2qr)3=4p7q3r.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)=4p7q3r.f(a)=f(p,q,r)=4p-7q-3r.




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