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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!

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(1) should be fairly obvious. Column space = image. – wj32 Oct 18 '12 at 23:47

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.

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Thank you EuYu. The first one have been answered. The second one is where I am. I was just wondering if there could be non-trivial restrictions that would make it. Suppose $$A=\begin{pmatrix} a & b \\ c & d \end{pmatrix}$$ and $$B=\begin{pmatrix} a & -b \\ c & -d \end{pmatrix}$$ where $a,b,c,d\ge 0$. Seeing the relationship that $A^2=C+D$ and $B^2=C-D$, I feel that there might be some non-trivial special cases where $col(A^2)=col(B^2)$ – basta Oct 19 '12 at 5:19
If $A$ is diagonalizable, then $range(A)=range(A^k)$ for all $k \in \mathbb{N}$. – chaohuang Oct 19 '12 at 5:42

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