linear algebra - The effect of row exchange in the four fundamental subspaces

I am investigating the effect of row exchanges in the four fundamental spaces of a matrix $A$. Consider the permutation matrix $P_{12}$ that permutes(exchanges) the firt two rows of a matrix. So the question I have is, "which subspaces stay the same?"

For the nullspace of $A$, i.e $N(A)$, it is easy to understand that $N(A)=N(P_{12}A)$ because:

$$Ax=0 \iff P_{12}Ax = 0$$ thus $A$ and $P_{12}$ have the same nullspace.

For the row space $C(A^T)$ I understand intuitivelly that stays the same because we end up to work with the same rows that defines it, but what is the mathematical proof similat to this of the $N(A)$.

For, the collumn space so far I have,

$$Ax=b \iff P_{12}Ax=P_{12}b$$ $P_{12}$ changes the column space that is all $b$'s to $P_{12}b$. However, taking an example of $2\times 2$ invertible matrix, both column spaces are $\mathbb{R}^2$. Is that a contradiction?

So, could you please help me to make correct relevant proofs for the $C(A)$, $C(A^T)$, and $N(A^T)$?

Thanks!!

Answer

Let $(a_1, \dotsc, a_n)$ denote the rows of $A$. Then $C(A^T)$ is spanned by $\{a_1, \dotsc, a_n\}$ (or $\{a_1^T, \dotsc, a_n^T\}$, depending on your definition). The rows of $P_{12} A$ are $(a_2, a_1, a_3, \dotsc, a_n)$, so $C((P_{12}A)^T)$ is spanned by $\{a_2, a_1, a_3, \dotsc, a_n\}$ (or $\{a_2^T, a_1^T, a_3^T, \dotsc, a_n^T\}$, depending on your definition). Both generating sets are the same, so their span is the same. Therefore $C(A^T) = C((P_{12}A)^T)$.

The column spaces are not necessarily the same, as you have already noticed. If $A$ is an invertible $n \times n$ square matrix then the columns span $K^n$. Then $P_{12} A$ is also invertible, so the colums of $P_{12} A$ also span $K^n$. So for an invertible matrix $A$ it just happens to be the case that $C(A) = C(P_{12} A)$.

Lastly $N(A^T)$ is not invariant. Take for example
$$

A =
\begin{pmatrix}
1 & 1 \\
0 & 0
\end{pmatrix}.
$$
Then $(0,1)^T \in N(A^T)$ but $(0,1) \notin N((P_{12}A)^T)$.