Let $X=\left\{ \left.\begin{pmatrix}x\\y\\x\end{pmatrix}\right\vert x+y+2z=0=2x+y+z\right\}$
a) Show that X is a vector space
b) Find a basis for X.
c) Find dim X.
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So far I have for a):
1) X is non-empty since $0\in X$
2) Let $v=\begin{pmatrix}x_1\\y_1\\z_1\end{pmatrix}$ and $w=\begin{pmatrix}x_2\\y_2\\z_2\end{pmatrix}$
$v+w=\begin{pmatrix}x_1+x_2\\y_1+y_2\\z_1+z_2\end{pmatrix}$ -> For $x_1+y_1=-2z_1$
$\therefore(x_1+y_1)+(x_2+y_2)=-2z_1+-2z_2=-2(z_1+z_2)$
$\therefore$ True under vector addition…
Am I doing this correctly? The same will be done for the other side of the equation.
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How do I then find the basis and dim of X?
Best Answer
You have to verify more things to check that is a vector space. You need to check the second condition of the definition of $X$, and also what happens when you multiply by a scalar. Maybe it is easier a different approach.
Your set $X$ are the solutions of the system $$\left. \begin{array}{lll} x+y+2z & = &0\\ 2x+y+z & = & 0\end{array}\right\}$$ Using your favorite way to solve this system, you find that the solutions are $y=-3z$, $x=z$, and $z$ a free variable. Then $$X = \left\{ \left(\begin{array}{r} z\\-3z\\z \end{array}\right):z\in k\right\},$$ where $k$ is the field you are working on (I suppose it is $\mathbb{R}$). Then you can notice that all the vectors in $X$ are of the form $\lambda\cdot(1,-3,1)$, so the set $\{(1,-3,1)\}$ generates all $X$, and obviously it can't be smaller. Hence $\{(1,-3,1)\}$ is a base of $X$.