Find the basis and dimension for the row space of the matrix \( A = \begin{bmatrix} 1 & 2 & 3 & 4 \ \ 2 & 4 & 6 & 8 \ \ 3 & 6 & 9 & 12 \ \ 4 & 8 & 12 & 16 \end{bmatrix} \)

Step 1 ：First, we perform row operations to put the matrix in row echelon form. We begin by subtracting 2 times the first row from the second row, 3 times the first row from the third row, and 4 times the first row from the fourth row to get \( \begin{bmatrix} 1 & 2 & 3 & 4 \ \ 0 & 0 & 0 & 0 \ \ 0 & 0 & 0 & 0 \ \ 0 & 0 & 0 & 0 \end{bmatrix} \)

Step 2 ：Next, we can see that the second, third, and fourth rows are all zero vectors, so we eliminate them to get \( \begin{bmatrix} 1 & 2 & 3 & 4 \end{bmatrix} \)

Step 3 ：The basis for the row space of A is the set of all non-zero rows in the row echelon form of A, which in this case is \(\{\begin{bmatrix} 1 & 2 & 3 & 4 \end{bmatrix}\}\)

Step 4 ：The dimension of the row space of A is the number of vectors in its basis, which in this case is 1