Question

I don't understand this format. Don't copy the answer from the book. Is there other method to proof this, such as let two vectors based on the form? what does that mean x1+x2=0? does that imply 0 in there?

2. For each set that follows determine whether it is a subspace of R2. Prove your answers. 8. 3. Let

Answer 1

$\mathbb{R}^2=\{(x_1,x_2),~x_1,x_2\in \mathbb{R}\}=all~points~in~two~dimensional~space\\ (x_1,x_2)~is~called~vector~i..e.~point~in~\mathbb{R}^2~is~called~vector.~We~generally~denote\\ vector~as~\begin{pmatrix} x_1\\ x_2 \end{pmatrix}~i.e.~a~matrix~of~order~2\times~1.\\ S_1~contains~those~points~whose~sum~of~two~cordinates~is~zero~i.e.\\ S_{1}=\{(x,-x),~x\in~\mathbb{R}\}.~Now~I~use~standard~notation~of~vectors~and~answer~the~given~\\ questions:$

$(a)~Take~any~two~vectors~{\bf x}=\begin{pmatrix} x_1\\ x_2 \end{pmatrix},~{\bf y}=\begin{pmatrix} y_1\\ y_2 \end{pmatrix}\in~S_1\\ \Rightarrow x_1+x_2=0,~y_1+y_2=0........(1)\\ Now,~{\bf x}+\lambda {\bf y}=\begin{pmatrix} x_1\\ x_2 \end{pmatrix}+\lambda \begin{pmatrix} y_1\\ y_2 \end{pmatrix}=\begin{pmatrix} x_1+\lambda y_1\\ x_2+\lambda y_2 \end{pmatrix}~where,~\lambda\in \mathbb{R}\\ Now,~x_1+\lambda y_1+x_2+\lambda y_2=(x_1+x_2)+\lambda (y_1+y_2)=0~(from~(1))\\ Hence~{\bf x}+\lambda {\bf y}\in S_1.\\ Hence~S_1~is~a~subspace~of~\mathbb{R}^2.\\$

$(b)~Take~any~two~vectors~{\bf x}=\begin{pmatrix} x_1\\ x_2 \end{pmatrix},~{\bf y}=\begin{pmatrix} y_1\\ y_2 \end{pmatrix}\in~S_2\\ ~with~x_1\neq 0,~y_2\neq 0\\ \Rightarrow x_2=0,~y_1=0........(1)\\ Now,~{\bf x}+\lambda {\bf y}=\begin{pmatrix} x_1\\ x_2 \end{pmatrix}+\lambda \begin{pmatrix} y_1\\ y_2 \end{pmatrix}=\begin{pmatrix} x_1+\lambda y_1\\ x_2+\lambda y_2 \end{pmatrix}~where,~\lambda\in \mathbb{R}\\ Now,~(x_1+\lambda y_1)(x_2+\lambda y_2)=x_1x_2+\lambda y_1x_2+\lambda x_1y_2+\lambda^2y_1y_2=\lambda x_1y_2~(from~(1))\\ \neq 0~provided~\lambda\neq 0\\ Hence~{\bf x}+\lambda {\bf y}~does~not~belong~to~ S_2.\\ Hence~S_2~is~not~a~subspace~of~\mathbb{R}^2.\\$