Find the basis and the dimension of the column space for the following matrix \(A = \begin{bmatrix} 1 & 2 & 3 \ \ 4 & 5 & 6 \ \ 7 & 8 & 9 \end{bmatrix}\)

Step 1 ：Step 1: We first perform row operations to bring matrix A into its row echelon form.

Step 2 ：Performing row operations, we get, \(R2 = R2 - 4 * R1\) and \(R3 = R3 - 7 * R1\), the resulting matrix is \(\begin{bmatrix} 1 & 2 & 3 \ \ 0 & -3 & -6 \ \ 0 & -6 & -12 \end{bmatrix}\)

Step 3 ：Continuing, we perform \(R3 = R3 - 2 * R2\) to get \(\begin{bmatrix} 1 & 2 & 3 \ \ 0 & -3 & -6 \ \ 0 & 0 & 0 \end{bmatrix}\)

Step 4 ：Step 2: Now that we have our row echelon form, we find the basis for the column space of A by taking the columns in A that correspond to the leading 1's in the row echelon form. This gives us the first and second columns of A.

Step 5 ：So, the basis for the column space of A is given by the vectors \(\begin{bmatrix} 1 \ \ 4 \ \ 7 \end{bmatrix}\) and \(\begin{bmatrix} 2 \ \ 5 \ \ 8 \end{bmatrix}\)

Step 6 ：Step 3: The dimension of the column space of A is just the number of vectors in our basis. So, the dimension is 2.