Hi, I hope I;m writing this correctly:
I need to proof that there is no existence liner system over R that this is the set of the answers:
$\displaystyle \left \{ \left ( a,a^{2},b \right )\mid a,b \in \mathbb{R}\right \}$
Hi, I hope I;m writing this correctly:
I need to proof that there is no existence liner system over R that this is the set of the answers:
$\displaystyle \left \{ \left ( a,a^{2},b \right )\mid a,b \in \mathbb{R}\right \}$
It is obvious that the "linear system" x= a, $\displaystyle y= a^2$, z= b has the solution set $\displaystyle \{a, a^2, b\}$ so what you say you are trying to prove is not true! Please check your problem again.
basically, I'm struggling to translate this question, I know that there is no linear equation system in R that this is the solution set of her, I don't know how to prove that.
sorry, this is the question I've been asked for, to prove that a system of linear equations doesn't exist over $\displaystyle \mathbb{R}$ for the solutions set $\displaystyle \left \{ \left ( a,a^2,b \right )|a,b \in \mathbb{R} \right \}$. I can't understand why you find this question wrong.
Is it possible that she/he means something completely other by linear system?
Could it be a linear operator?.
I asked for a working definition be posted, but not get it.
I think the previous responders didn't understand your question.
Given a linear system $AX=B$ where $A$ is a 3 by 3 matrix, the solution set $S$ is a subspace of $R^3$ if $B=0$ or the set of differences of members of $S$ is a subspace. So first the set $S=\{(a,a^2,b)\,:a\in R,\,b\in R\}$ is not a subspace:$(0,0,1)\text{ and }(1,1,0)\in S$, but the difference $(0,0,1)-(1,1,0)=(-1,-1,1)\not\in S$ since there is no $a\in R$ with $a^2=-1$. Next the set of differences $S_1=\{(a_1-a_2,a_1^2-a_2^2,b_1-b_2)\,:\,a_1,b_1,a_2,b_2\in R\}$. Suppose $S_1$ is a subspace. Then $(0,0,1)-(0,0,0)=(0,0,1)\in S_1$, $(1,1,0)-(-1,1,0)=(2,0,0)\in S_1$ and $(1,1,0)-(0,0,0)=(1,1,0)\in S_1$. Clearly these 3 vectors are linearly independent members of $S_1$. So $S_1=R^3$. In particular, $(0,1,0)\in S_1$ or for suitable real values $(0,1,0)=(a_1-a_2,a_1^2-a_2^2,b_1-b_2)$. But this says $a_1=a_2$ and so $a_1^2-a_2^2=1$ is a contradiction. QED.
I hope that is just a poorly composed sentence (poor choice of words). I think that both of both of us just wanted the poster to take responsibility for making the question clear. It is of great interest to me, at least, if the posters can understand your reply. I hope she/he will tell us.