# The dimensions theorem and different spaces

I'm trying to wrap my head around the meaning of different dimensions in lin alg. Consider the matrix:

$$\left[ \begin{array}{ccc} 1 & -1 \\ 2 & -2\\ 3 & -3 \end{array} \right]$$

I've found that $C(A) =$ span of $\left[ \begin{array}{ccc} 1 \\ 2\\ 3 \end{array} \right]$ and $N(A) =$ span of $\left[ \begin{array}{ccc} 1 \\ -1 \end{array} \right]$

and similarly I can find the $C(A^T)$ and $N(A^T)$. However, what if I'm asked for the $R(A^T)$ and $R(A)^\perp$ i.e. the orthogonal complement of the row space of A.

I believe the column space of $A$ is the row space of $A^T$, so that's fine. The row space of $A$ is the column space of $A^T$, but to find its orthogonal complement, do I find the null space of the column space of $A$ i.e. the null space of $A$?

My question boils down to:

Is $N(A)$ = orthogonal complement of the row space of $A$?

How do they all tie together? I know the number of vectors given by the dimension's theorem, i.e. $r$, $n-r$, but I'm not sure how they fit in intuitively.

The nullspace of $A$ is all the $\bf v$ such that $A{\bf v}={\bf0}$. If you think about how matrix multiplication works, you'll see this is the same thing as saying $\bf v$ is orthogonal to each row of $A$, hence, orthogonal to the row spacce of $A$. So, yes, the nullspace of $A$ is the orthogonal complement of the row space of $A$.