I am given the following :
Let $A$ be a $4 \times 4$ matrix with RREF given by:
$$
U = \begin{bmatrix}
1 & 0 & 2 & 1 \\
0 & 1 & 1 & 4 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
\end{bmatrix}.
$$
If
$$
a_1 = \begin{bmatrix} -3 \\ 5 \\ 2 \\ 1 \end{bmatrix} \text{ and } a_2 = \begin{bmatrix} 4 \\ -3 \\ 7 \\ -1 \end{bmatrix},
$$
find $a_3$ and $a_4$. [Note: $a_1,a_2,a_3,a_4$ are the columns of $A$.]
This is in the section with column, row, and null space. I don't know how those apply no this problem though… I have absolutely no idea how to do this. Can anyone help point me in the right direction?
Best Answer
The (homogeneous) linear relations which hold between the columns of the RREF also hold among the columns of the original matrix.
So your final two columns are precisely $2a_1+a_2$ and $a_1+4a_2$.
$$A = \begin{bmatrix} a_1 & a_2 & a_3 & a_4 \end{bmatrix} = \begin{bmatrix} a_1 & a_2 & 2a_1+a_2 & a_1+4a_2 \end{bmatrix} = \begin{bmatrix} -3 & 4 & -2 & 13 \\ 5 & -3 & 7 & 7 \\ 2 & 7 & 11 & 30 \\ 1 & -1 & 3 & -3 \end{bmatrix}$$
As you preform row operation, the columns are never mixed. So if column 3 is 2 times column 1 plus column 2 in the RREF, it must have been that way in the original matrix.
This means that given the RREF and a list of pivot columns (the columns corresponding to leading non-zero row entries) one can easily reconstruct the original matrix.
This also explains why the pivot columns form a basis for the column space.