Test if a vector belongs to a subspace

Question

I have to test if vector (U5) belongs to subspace R. U5=(1,6,-7,8) and R=(a+b,-2a,a-b,2b). Therefore R is made up by two vectors(R1+R2). Vector R1=a(1,-2,1,0) and vector R2=b(1,0,-1,2).

So far i've gaussed them and found that the system is inconsistend and therefore they are lineary independent and u5 does not belong to R. Is this right?

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Okay so i just split the R into two vectors. And from there i Gauss it down so i get -3 and 4. I think i understand that part :) The end of your answer confuses me though. Do i need to transpose the unknown x? or what do you mean by the last sentence x=(a,b)^T?

Answer 1

I don't understand what your image means. But, all you have to show is that there exists $a,b \in \mathbb{R}$ (I suppose that you are talking about a vector subspace of $\mathbb{R}^5$) such that:

$$\begin{cases} a+b=1\\ -2a=6\\ a-b=-7\\ 2b=8 \end{cases} $$ and the solution is simple: $a=-3$ and $b=4$.

Answer 2

Your $R$ is the set $$ R = \{ (a+b, -2a, a-b, 2b) \mid a, b \in \mathbb{R} \} $$ the task is to determine, if there are real $a$ and $b$ which generate $u_5 = (1,6,-7,8)$.

This leads to the system $$ A x = y \iff \\ [A|y] \iff \\ \left[ \begin{array}{rr|r} 1 & 1 & 1 \\ -2 & 0 & 6 \\ 1 & -1 & -7 \\ 0 & 2 & 8 \end{array} \right] \to \left[ \begin{array}{rr|r} 1 & 1 & 1 \\ 1 & 0 & -3 \\ 1 & -1 & -7 \\ 0 & 2 & 8 \end{array} \right] \to \left[ \begin{array}{rr|r} 0 & 1 & 4 \\ 1 & 0 & -3 \\ 0 & -1 & -4 \\ 0 & 1 & 4 \end{array} \right] \to \left[ \begin{array}{rr|r} 1 & 0 & -3 \\ 0 & 1 & 4 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right] $$ in the unknown $x = (a, b)^\top$, with $A$ derived from the definition of $R$ and $y = u_5^\top$.

This system has the solution $x = (a, b)^\top = (-3, 4)^\top$. So $u_5 \in R$, because there is at least one solution $x$.