Sum of two subspaces: representing it with equations I found the following excercise:

Let $W_1 = \{(x_1, ..., x_6) : x_1 + x_2 + x_3 = 0, x_4 + x_5 + x_6 = 0 \}$. Let $W_2$ be the span of $S := \{(1, -1, 1, -1, 1, -1), (1, 0, 2, 1, 0, 0), (1, 0, -1, -1, 0, 1),  (2, 1, 0, 0, 0, 0)\}$.
Give a base, a dimension and an equation representation of $W_1 + W_2$

I'm new to the concept of sum of subspaces. But as I understand it, the first step would be to note that any $\textbf{x} = (x_1, ..., x_6) \in W_1$ satisfies
\begin{equation*}
    \begin{cases}
        x_1 = -x_2 - x_3 \\
        x_4 = -x_5 - x_6
    \end{cases}
\end{equation*}
so that its general form is
\begin{equation*} \textbf{x} =
    \begin{pmatrix}
        x_1 \\
        x_2 \\
        x_3 \\
        x_4 \\
        x_5 \\
        x_6
    \end{pmatrix} = \begin{pmatrix}
         -x_2 - x_3 \\
        x_2 \\
        x_3 \\
        -x_5 - x_6 \\
        x_5 \\
        x_6
    \end{pmatrix}
\end{equation*}
We also know any $\textbf{y} \in W_1$ is of the general form
\begin{align*} \textbf{y} =
    \begin{pmatrix}
         y_1 \\
        y_2 \\
        y_3 \\
        y_4 \\
        y_5 \\
        y_6
    \end{pmatrix} = \begin{pmatrix}x + y +z + 2w \\
    -x + w \\ 
    x + 2y - z \\
    -x + y - z \\
    x\\
    -x + z\end{pmatrix}
\end{align*}
Then for generals $\mathbf{x}, \mathbf{y}$ we have
\begin{align*}
    \textbf{x} + \textbf{y} &= \begin{pmatrix}
        x + y +z + 2w + (-x_2 - x_3)\\
    -x + w + x_2\\ 
    x + 2y - z + x_3\\
    -x + y - z + (-x_5 - x_6)\\
    x + x_5\\
    -x + z + x_6
    \end{pmatrix} 
\end{align*}
One can then conclude
$$W_1 + W_2 = \Big\{\big(x + y + z + 2w - x_2 - x_3\big), \big(-x + w + x_2 \big), \big(x +2y - z + x_3 \big), \big(-x + y - z - x_5 - x_6 \big), \big(x + x_5 \big), \big(-x + z + x_6 \big) \mid x, y, z, x_2, x_3 \in \mathbb{R}  \Big\}$$
But what would be an representation via equations of this system? I'm very new to linear algebra so go easy on me!
 A: Note that $(1,0,−1,−1,0,1)\in W_1$. Therefore, if$$W_3=\operatorname{span}\bigl\{(1,−1,1,−1,1,−1),(1,0,2,1,0,0),(2,1,0,0,0,0)\bigr\},$$then $W_1+W_2=W_1+W_3$.
Now, let us see what $W_1\cap W_3$ is. Asserting that$$\overbrace{\alpha(1,−1,1,−1,1,−1)+\beta(1,0,2,1,0,0)+\gamma(2,1,0,0,0,0)}^{\text{arbitrary element of }W_3}\in W_1$$means that $\alpha+3\beta+3\gamma=-\alpha+\beta=0$. This occurs if and only if $\beta=\alpha$ and $\gamma=-\frac{4\alpha}3$. Therefore,$$W_1\cap W_3=\left\{\left(-\frac{2 \alpha }{3},-\frac{7 \alpha }{3},3 \alpha,0,\alpha ,-\alpha \right)\,\middle|\,\alpha\in\Bbb R\right\};$$in particular, $\dim(W_1\cap W_3)=1$. So,\begin{align}\dim(W_1+W_3)&=\dim(W_1)+\dim(W_3)-\dim(W_1\cap W_3)\\&=4+3-1\\&=6.\end{align}Therefore, $W_1+W_3$ is the whole space $\Bbb R^6$.
A: In contrast to Jose's and Anne's answers using dimension, let me show you the rote method you could use. From your step here:

\begin{equation*} \mathbf{x} =
    \begin{pmatrix}
        x_1 \\
        x_2 \\
        x_3 \\
        x_4 \\
        x_5 \\
        x_6
    \end{pmatrix} = \begin{pmatrix}
         -x_2 - x_3 \\
        x_2 \\
        x_3 \\
        -x_5 - x_6 \\
        x_5 \\
        x_6
    \end{pmatrix},
\end{equation*}

you could then find a spanning set for $W_1$ like so:
\begin{align*}
\mathbf{x} &= \pmatrix{-x_2\\x_2\\0\\0\\0\\0} + \pmatrix{-x_3\\0\\x_3\\0\\0\\0} + \pmatrix{0\\0\\0\\-x_5\\x_5\\0} + \pmatrix{0\\0\\0\\-x_6\\0\\x_6} \\
&= x_2\pmatrix{-1\\1\\0\\0\\0\\0} + x_3\pmatrix{-1\\0\\1\\0\\0\\0} + x_5\pmatrix{0\\0\\0\\-1\\1\\0} + x_6\pmatrix{0\\0\\0\\-1\\0\\1}.
\end{align*}
The vector $\mathbf{x}$ is arbitrary in $W_1$, so we've shown that
$$W_1 \subseteq \operatorname{span}\left\{\pmatrix{-1\\1\\0\\0\\0\\0},\pmatrix{-1\\0\\1\\0\\0\\0},\pmatrix{0\\0\\0\\-1\\1\\0},\pmatrix{0\\0\\0\\-1\\0\\1}\right\}.$$
Equality holds, because every vector in the spanning set also lies in $W_1$. It's also easy to see the above is linearly independent, so we have a basis, but this is not necessary to observe!
Next, once you have spanning sets for $W_1$ and $W_2$, you can form a spanning set for $W_1 + W_2$ by unioning the sets; every vector in $W_1 + W_2$ is the sum of a vector in $W_1$ (a linear combination of the first spanning set) and a vector in $W_2$ (a linear combination of the second spanning set), so the union will span the sum. That is,
$$W_1 + W_2 = \operatorname{span}\left\{\pmatrix{-1\\1\\0\\0\\0\\0},\pmatrix{-1\\0\\1\\0\\0\\0},\pmatrix{0\\0\\0\\-1\\1\\0},\pmatrix{0\\0\\0\\-1\\0\\1}, \pmatrix{1\\-1\\1\\-1\\1\\-1},\pmatrix{1\\0\\2\\1\\0\\0},\pmatrix{1\\0\\-1\\-1\\0\\1},\pmatrix{2\\1\\0\\0\\0\\0}\right\}.$$
This is already a technically correct description of $W_1 + W_2$, but it's usually best to reduce the spanning set down to a basis. There are two standard ways to do this, both involving row-reduction:

*

*Place them as rows in a matrix, row-reduce down to row-echelon form (reduced, if you prefer) and keep only the non-zero rows. These rows will be linearly independent, but retain the same span as the original set, thus producing a basis, or

*Place them as columns in a matrix, and row-reduce down to row-echelon form. Note the columns where the pivots (leading $1$s) appear, and retain only the vectors from the original set that you placed in those columns.

Either way, we get an element in the basis of $W_1 + W_2$ for every pivot in the row-reduced matrix. And, no matter which method you use, you will get $6$ pivots, which tells you that the dimension of $W_1 + W_2$ is $6$, and must be all of $\Bbb{R}^6$. Indeed, if you apply the first method, reducing until reduced row-echelon form, you should find that you get back the standard basis!
A: You already found that a vector $(a,b,c,d,e,f)$ belongs to $W_1+W_2$ if and only if there exist real numbers $x_2,x_3,x_5,x_6,x,y,z,w$ such that
$$\begin{cases}a&=x+y+z+2w-x_2-x_3\\b&=-x+w-x_2\\c&=x + 2y - z + x_3\\
    d&=-x + y - z-x_5 - x_6\\
    e&=x + x_5\\
    f&=-x + z + x_6.
\end{cases}$$
Using for instance the 2nd, 3rd, 5th, and 6th equations to first eliminate the $x_i$'s, this is equivalent to the existence of real numbers $x,y,z,w$ such that
$$\begin{cases}a&=3x+3y+w+b-c\\
    d&=-x + y-e -f
\end{cases}$$
For any $(a,b,c,d,e,f)\in\Bbb R^6,$ such numbers $x,y,z,w$ always exist (e.g. choose $x,z$ arbitrarily and take $y=d+e+f+x,$ and $w=a-b+c-3x-3(d+e+f+x)$) hence
$$W_1+W_2=\Bbb R^6.$$
You could see it more rapidly since you already knew that $\dim(W_1\cap W_2)=2$: $\dim(W_1)$ is $6-2=4$ (since its 2 equations are clearly independent), $\dim(W_2)=4$ (since its 4 generating vectors are - less clearly - independent) hence by Grassmann's formula,
$$\dim(W_1+W_2)=4+4-2=6.$$
