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!

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!
