Find the basis and the dimension of the column space for the following matrix $A=\left[\begin{array}{ccccccc}1& 2& 34& 5& 67& 8& 9\end{array}\right]$

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, $R2=R2-4\ast R1$ and $R3=R3-7\ast R1$, the resulting matrix is $\left[\begin{array}{ccccccc}1& 2& 30& -3& -60& -6& -12\end{array}\right]$

Step 3 ：Continuing, we perform $R3=R3-2\ast R2$ to get $\left[\begin{array}{ccccccc}1& 2& 30& -3& -60& 0& 0\end{array}\right]$

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 $\left[\begin{array}{c}147\end{array}\right]$ and $\left[\begin{array}{c}258\end{array}\right]$

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.