What is the dimension for this subspace? For V (x,y,z)
$S$ is a subspace of $V$ satisfying following condition.


*

*$x+y+z=0$

*$x+y+z=0$ and $x-y-z=0$
I dont know about 1 
but For number 2, isn't $x$,$y$ and $z$ $0$? then dimension of number 2 should be 0, I think 
 A: Hint: can u see {$(1,0,-1),(0,1,-1)$} and {$(0,-1,1)$} forms a basis for $1$ and $2$ respectively?
A: One way to do this is to construct the matrix equation corresponding to the subspace. This approach allows you to answer seemingly more complicated, but mathematically identical, problems of this nature. Each problem is equivalent to computing the subspace of solutions to $Ax = 0$ for some matrix $A$, and by the rank-nullity theorem we know that the solution space will have dimension $3 - \mathrm{rank}(A)$. For problem 1, this is equivalent to solving the homogeneous system
\begin{align*}
\begin{bmatrix} 1 & 1 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}
\end{align*}
and for problem 2 this is equivalent to solving
\begin{align*}
\begin{bmatrix} 1 & 1 & 1 \\ 1 & -1 & -1\end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}
\end{align*}
Now compute the ranks of the relevant matrices via Gaussian elimination.
A: For (1) you can assign values of $x$ and $y$ freely, and a solution value of $z$ is determined $z=-y-x$.  Thus solutions correspond exactly to points in the $xy$-plane, and so the solution, like the $xy$-plane, is $2$-dimensional.
For (2) you can only assign one variable ($y$ or $z$; $x$ must equal $0$) freely.
