# Linear subspace

• Apr 22nd 2010, 07:20 AM
lehder
Linear subspace
Hi everybody,

I don't know how to show that the following sets aren't linear subspaces:

$F_1$={ $(x;y;z)/x+y+z=1$}
$F_2$={ $(x;y;z)/xy=0$}
$F_3$={ $(x;y;z)/x^2=z^2$}
...

• Apr 22nd 2010, 07:55 AM
Swlabr
Quote:

Originally Posted by lehder
Hi everybody,

I don't know how to show that the following sets aren't linear subspaces:

$F_1$={ $(x;y;z)/x+y+z=1$}
$F_2$={ $(x;y;z)/xy=0$}
$F_3$={ $(x;y;z)/x^2=z^2$}
...

To show that $U$ is a subspace of a vector space $V$ where $U \subset V$ you have two things to show,

that it is closed under scalar multiplication (for $u \in U$ you need to show that $\alpha u \in U$),

that it is closed under addition (for $u, v \in U$ you need to show that $u+v \in U$).

You need to show these two things as a subspace is a vector space contained in a bigger space, and this bigger space lends the smaller space the other properties. So all you need to prove is these two things.

So, for $F_1$ you need to show that $(x_1+x_2)+(y_1+y_2)+(z_1+z_2) = 1$ and that $\alpha x_1 + \alpha y_1 + \alpha z_1=1$ for $x_1+y_1+z_1=1$ and $x_2+y_2+z_2=1$.

Do you understand why this is what you need to show?

Now, does this hold?

The other two questions can be attacked similarly.