Given a matrix A = \(\begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 3 & 6 & 9 \end{bmatrix}\), find the basis and dimension for the column space of the matrix.

Step 1 ：First, we perform the Gaussian elimination to row reduce the matrix to its row echelon form. The row echelon form of the matrix A is \(\begin{bmatrix} 1 & 2 & 3 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}\).

Step 2 ：The column space of the matrix A is spanned by the columns of A that correspond to the leading 1's in the row echelon form. In this case, only the first column has a leading 1, so the column space is spanned by the first column of A.

Step 3 ：Therefore, the basis for the column space of the matrix A is \(\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}\).

Step 4 ：The dimension of the column space of a matrix is the number of vectors in its basis. Thus, the dimension for the column space of the matrix A is 1.