To find a basis for the image of matrix $A$, we need to identify the linearly independent columns of $A$. The matrix $A$ is:

$A = \begin{bmatrix} 2 & -6 & 2 \\ 2 & -3 & 1 \\ 2 & -6 & 2 \\ -2 & 6 & -2 \end{bmatrix}$

1. Row Reduction to Row Echelon Form (REF): Performing row reduction on $A$ will help us determine the pivot columns, which correspond to the linearly independent columns. These pivot columns will form a basis for the column space (image) of $A$.

2. Steps for Row Reduction:

   • Start with the first column as the pivot column and perform operations to create zeros below the pivot in the first column.
   • Move to the next pivot and repeat until the matrix is in REF.

3. Identify Pivot Columns: After row reduction, the pivot columns of the matrix correspond to the linearly independent columns of the original matrix $A$. The vectors in these columns will form a basis for the image of $A$.

Let's proceed with the row reduction:

Performing the calculations manually or via a tool, we get that the matrix $A$ has linearly independent columns corresponding to the first and second columns. Therefore, the basis for the image of $A$ consists of these columns:

$\text{Basis for the image of } A = \left\{ \begin{bmatrix} 2 \\ 2 \\ 2 \\ -2 \end{bmatrix}, \begin{bmatrix} -6 \\ -3 \\ -6 \\ 6 \end{bmatrix} \right\}$

So, the answer is:

• Column 1: $\begin{bmatrix} 2 \\ 2 \\ 2 \\ -2 \end{bmatrix}$
• Column 2: $\begin{bmatrix} -6 \\ -3 \\ -6 \\ 6 \end{bmatrix}$

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