My teacher showed a certain exercise in a video conference but he didnt explain at all how he found out the span, can you shed some light about that process? I don't seem to understand how did we find the marked answer in the picture below
by the way it is known that $dimW = 2$
To begin, note that you are trying to solve the simultaneous system: $\begin{cases} \begin{bmatrix}1&-2&3&4\end{bmatrix}\bullet \begin{bmatrix}a&b&c&d\end{bmatrix}=0\\\\ \begin{bmatrix}3&-5&7&9\end{bmatrix}\bullet \begin{bmatrix}a&b&c&d\end{bmatrix}=0\end{cases}$
This can be rewritten as a matrix equation $\begin{bmatrix}1&-2&3&4\\3&-5&7&8\end{bmatrix}\begin{bmatrix}a\\b\\c\\d\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}$
Via row reductions of the augmented matrix, you arrived at $\left[\begin{array}{cccc|c}1&0&-1&-4&0\\0&1&-2&-4&0\end{array}\right]$
Interpreting this as a system of equations, this is:
$\begin{cases} 1a+0b-1c-4d=0\\ 0a+1b-2c-4d=0\end{cases}$ which rewritten is $\begin{cases} a = c+4d\\b=2c+4d\end{cases}$
Including the often forgotten lines that $c=c$ and $d=d$, this implies:
$\begin{cases} a = 1c+4d\\b=2c+4d\\c=1c+0d\\d=0c+1d\end{cases}$ or written another way $\begin{bmatrix}a\\b\\c\\d\end{bmatrix} = \begin{bmatrix}c+4d\\2c+4d\\1c+0d\\0c+1d\end{bmatrix}$
Rewritten one last time, that is $\begin{bmatrix}a\\b\\c\\d\end{bmatrix} = c\begin{bmatrix}1\\2\\1\\0\end{bmatrix}+d\begin{bmatrix}4\\4\\0\\1\end{bmatrix}$
That is to say, the solution set to the original question of finding which vectors are perpendicular to $u_1$ and $u_2$ is the set of vectors spanned by the basis $\begin{bmatrix}1\\2\\1\\0\end{bmatrix}$ and $\begin{bmatrix}4\\4\\0\\1\end{bmatrix}$, the vectors appearing in the image.
