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Let $V=\{(x,y,z)\epsilon \mathbb R^3: x-y=z\}$ and $W=\{(x,y,z)\epsilon \mathbb R^3: x+z=2y\}$ subspaces of $\mathbb R^3$. Show that $V+W= \mathbb R^3$. Also check if the sum is direct, $V \oplus W= \mathbb R^3$.

At first i found the basis of $V$ and $W$ which are $V=<(1,1,0),(1,0,1)>$ and $W=<(2,1,0),(-1,0,1)>$.

By definition the sum is $V+W=$ where $x_1,x_2$ is the vectors of $V$ and $x_3,x_4$ the vectors of $W$.

Then, i put the vectors on a matrix and proceeded to gauss elimination and i got an equation between $x_1,x_2,x_3,x_4$ equals $0$ but i don't know how to proceed from this point.

Any help will be appreciated.

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    Please do include what you've obtained so far. – 2017-02-08

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To show that $V+W=\mathbb{R}^3$ you need to show that the span of the four basis vectors you've found is all of $\mathbb{R}^3$. One way to do this is, as you mention, to consider a matrix whose columns are these four vectors, and apply the Gauss-Jordan elimination method to this matrix. If the resulting matrix (after GJE) has three pivots, this means that your four vectors span a 3-dimensional subspace of $\mathbb{R}^3$, which must be $\mathbb{R}^3$. Note that even though you have four vectors, you can't have four pivots at the end, since the matrix has just three rows.

Next, recall that a vector space sum can only be direct if $V\cap W=\{0\}$. Once you've shown that $V+W=\mathbb{R}^3$, you can use the following dimension equation: \begin{equation} \dim(V+W) = \dim(V) + \dim(W) - \dim(V\cap W) \end{equation} to determine whether or not $V\cap W=\{0\}$.

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    yes i did the GJE to the matrix but i don't know how to proceed to find the basis of the sum. Will the basis be extract from the equation that i get from $x_1,x_2,x_3,x_4$ ? – 2017-02-08
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    Once you've got the matrix in reduced row-echelon form, the pivots should correspond to the vectors that go into your basis. That is, if you have a pivot in columns 1,2, and 4, then $x_1,x_2$, and $x_4$ give a basis. But notice that you don't necessarily need a basis for $V+W$. As long as you know that it's three-dimensional, it must be all of $\mathbb{R}^3$. – 2017-02-08