So I'm doing this review sheet for an exam and I've come across a question I'm stuck on.
It gives me some $3\times 4$ matrix $A$, asks me to determine whether or not $v$ is in the column space of $A$ if it is find all vectors $x$ which satisfy $Ax=v$ (which is just Span? right?) then use the fundamental theorem of linear algebra to find the orthogonal complement of $C(A)$, which stats
$$(\text{Row} A)^{\perp} = \text{Nul} A\quad \text{or} \quad(\text{Col} A)^{\perp} = (\text{Nul} A)^T.$$
So I thought that you find Nul $A$, and Col $A$ but then how do you find $(\text{Col} A)^{\perp}$?
If you wanna see the full problem let me know.
Large edit below
Ok let me try an example(since I can't answer my own questions yet)
M = \begin{array}{cc} 1&2&3&3\\ 1&2&1&1\\ 2&2&2&3\\ \end{array} v = \begin{array}{cc} 2\\1\\-3\\ \end{array} so M = v reduces to \begin{array}{cc} 1&0&-1&0&-5\\ 0&1&2&0&11\\ 0&0&0&1&5\\ \end{array} so we get that it is in the col v is in the col space of M so I'm trying to find the orthogonal complement of v right? So to find Nul(A)$^T$ is just \begin{array}{cc} 1&0&-1&0&0\\ 0&1&2&0&0\\ 0&0&0&1&0\\ \end{array} which transposes and that's my Nul(A)$^T$ ? And is that my answer? Am I done?