Answer to Question #266198 in Linear Algebra for qissy

Question

determine whether the set s={(a,b,c) ; ab=1} is a subspace of R3

Answer 1

If $\begin{pmatrix} 0 \\ 0\\ 0 \end{pmatrix}$ $\in\>V$ then given $ab=1,$ then $(0)(0)\neq1$

$\therefore \begin{pmatrix} 0 \\ 0\\0 \end{pmatrix}$ is not V $\implies$ V is empty

2). Let $x,y\in$ V then

$x=\begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}$ and $y=\begin{pmatrix} y_1 \\ y_2 \\ y_3 \end{pmatrix}$ $x_iy_i\in\R$

$x_1x_2=1$ and $y_1y_2=1$

$x_2=\frac{1}{x_{1}}$ $y_2=\frac{1}{y_1}$

$x+y= \begin{pmatrix} x_1+y_1 \\ x_2+y_2 \\ x_3+y_3 \end{pmatrix}$

$(x_2+y_2)(x_1+y_1)=1\implies(x_2+y_2)=\frac{1}{x_1+y_1}$

$x_2+y_2=\frac{1}{x_1}+\frac{1}{y_1}\neq\frac{1}{x_1+y_1}$

S is not closed under addition

3) Take $x\in$ V and $\alpha\in\R$ then

$x=\begin{pmatrix} x_1 \\ x_2\\ x_3 \end{pmatrix}$ for $x_i \in\R$

$(x_1)(x_2)=1\implies\alpha\begin{Bmatrix} (x_1)(x_2) \end{Bmatrix}=\alpha(1)$

$\alpha\>x=\begin{pmatrix} \alpha\>x_1 \\ \alpha\>x_2\\ \alpha\>x_3 \end{pmatrix}$

$\therefore(\alpha\>x_2)(\alpha\>x_1)=\alpha^2(x_1)(x_2)\ne\alpha(x_1)(x_2)$

S is not closed under scalar multiplication

S is not a subspace of $\R^3$