column/row space of matrix powers I have to challenge you with two questions that I gave up trying:
1. Is it true that the column/row space of $A^2$ is a subspace of the column/row space of $A$? If so, how?
2. Suppose I have a square matrix A of non-negative entries. Let B be another matrix obtained from A by replacing one or more (but not all) of the columns with the corresponding negatives as shown in the eg. When do the columns of $A^k$ and $B^k$ span the same vector spaces? [for k=1, clear!]
eg. only one column change, i.e. the last column of B is the negative of the last column of A
$A =\left(\begin{array}{cc}
1 & 2 & 3 \\
4 & 5 & 6 \\
7 & 8 & 9 \\
\end{array}\right)$ , 
$B =\left(\begin{array}{cc}
1 & 2 & -3 \\
4 & 5 & -6 \\
7 & 8 & -9 \\
\end{array}\right)$
Thanks for your help!
 A: The column and row spaces of $A^k$ will always be a subspace of the column and row spaces of $A$. This is easy to see since the columnspace of $A$ is the range of the linear mapping $T$ induced by multiplication by $A$
$$T(\mathbf{v}) = A\mathbf{v}$$
Clearly then
$$\mathrm{range}(T^2) \subseteq \mathrm{range}(T)$$
The result for the rowspace can be shown by considering $A^\mathrm{T}$. 
For your second question, if we assume that $A$ is invertible, then the answer is that the columnspaces will be the same for all $k$. The columns of $A^k$ will be linear combinations of the columns of $A$ and it is rather clear that the columnspace of $B^k$ is also a subspace of the columnspace of $A$. Since $B^k$ is invertible, the columns are all linearly independent and hence
$$\mathrm{col}(A) = \mathrm{col}(B) = \mathrm{col}(B^2) = \mathrm{col}(B^3) = \cdots $$
As for non-invertible matrices, I'm doubtful that there are any trivial relationships. Even for $2\times 2$ matrices, there doesn't seem to be an obvious pattern. For example
$$A=\begin{pmatrix} 1 & 2 \\ 2 & 4 \end{pmatrix}$$
will have the columnspace spanned by the first column vector for all powers $k$, while
$$B = \begin{pmatrix} 1 & -2 \\ 2 & -4 \end{pmatrix}$$
will give you the zero matrix for all powers greater than or equal to $2$. On the other hand, the matrices
$$A = \begin{pmatrix} 1 & 3 \\ 2 & 6 \end{pmatrix},\ \ \ B = \begin{pmatrix} 1 & -3 \\ 2 & -6 \end{pmatrix}$$ will have the same columnspace for all powers.
