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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?

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  • $\begingroup$ Shouldn't it be $(x,y,z)$ instead of $(3,4,5)$? $\endgroup$ – jiyanez Sep 14 '16 at 4:06
  • $\begingroup$ Oh yes! Sorry, my mistake! $\endgroup$ – John Appleseed Sep 14 '16 at 4:10
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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 $X$.

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  • $\begingroup$ would the basis not be something of the sort: $x_1=\begin{pmatrix}-2\\0\\1\end{pmatrix}$ and $x_2=\begin{pmatrix}-1\\1\\0\end{pmatrix}$ As they both span X and are linearly independent? $\endgroup$ – John Appleseed Sep 14 '16 at 6:03
  • $\begingroup$ But also the $x_1$ and $x_2$ values from the second equation $\endgroup$ – John Appleseed Sep 14 '16 at 6:05
  • $\begingroup$ Notice that your vector $x_2$ is not in $X$, so you are not spanning $X$. $\endgroup$ – jiyanez Sep 14 '16 at 18:27
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1) Note that you have a homogeneous system $Ax=0$, where $x=(x,y,z)^T$. As such, your set $X$ is a set of all solutions of this homogeneous system. Therefore, it is subspace of of $\mathbb{R}^3$ (assuming that you field is $\mathbb{R}$).

2) you can see that $(1,1,2)$ and $(2,1,1)$, which are the coordinates of $x$, are linearly independent. So, you have one $df$. Thus, $\dim X=1$ and the basis of $X$ would be some particular solution of $Ax=0$.

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