Find a basis for the subspace $S$ of $\mathbb{R}^4$, $S :=\lbrace (w,x,y,z) \in \mathbb{R}^4\;|\; 2x-y+5z=0\rbrace$. What is the dimension of $S$
Is there a systematic way to do this? There has got to be a better way to do this than trial and error. Would the dimension be 3 since $w$ has no effect on the condition of $S$
Best Answer
You should compute the solutions of $2x-y+5z=0$. $$x=\frac{y}{2}-\frac{5z}{2}.$$ Notice that both $y$ and $z$ are free variables (also $w$). Thus any solution will look like \begin{align*} \begin{pmatrix} w\\x\\y\\z \end{pmatrix} = \begin{pmatrix} w\\\frac{y}{2}-\frac{5z}{2}\\y\\z \end{pmatrix} = w \begin{pmatrix} 1\\0\\0\\0 \end{pmatrix} + y \begin{pmatrix} 0\\\frac{1}{2}\\1\\0 \end{pmatrix} + z \begin{pmatrix} 0\\\frac{-5}{2}\\0\\1 \end{pmatrix} \end{align*} Thus a basis of $S$ is given by these three vectors (hence dimension of $S$ is $3$).