We want to find the nullspace of \(A\)

\[\begin{equation} A = \begin{bmatrix} 1 & 1 & 1 & 1 \\ 1 & 2 & 3 & 4 \\ 4 & 3 & 2 & 1 \end{bmatrix} \end{equation}\]

\(A\) and \(U\) have the same nullspace. Elimination brings us to \(U\)

\[\begin{equation*} A \implies \begin{bmatrix} 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 3 \\ 0 & -1 & -2 & -3 \end{bmatrix} \implies \begin{bmatrix} 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 0 & 0 \end{bmatrix} =U \end{equation*}\]

We identify our pivot columns and our free colums. The pivot columns, \(x_1\), and \(x_2\) contain our pivots. The other columns, \(x_3\) and \(x_4\), are free. Now we have two choices: back subsitution, and solving by row reduced echelon form.

Back Subsitution

For each free column, \(x_n\):

  1. Set that col to 1
  2. Set all the other free columns to 0
  3. Solve all equations for 0 with back subsitution

The upper triangular matrix above, \(U\), becomes:

\[\begin{align} x_1 &+& x_2 &+& x_3 &+& x_4 &= 0 \\ && x_2 &+& 2x_3 &+& 3x_4 &= 0 \\ && && && 0 &= 0 \end{align}\]

Do back subsitution with \(x_3 = 1\) and \(x_4 = 0\):

\[\begin{align} 0 &= x_2 + 2 &\implies& x_2 = -2 \\ 0 &= x_1 -2 + 1 &\implies& x_1 = 1 \end{align}\] \[\begin{equation*} A \begin{bmatrix} 1 \\ -2 \\ 1 \\ 0\end{bmatrix} = 0 \end{equation*}\]

Do back subsitution with \(x_3 = 0\) and \(x_4 = 1\):

\[\begin{align} 0 &= x_2 + 3 &\implies& x_2 = -3 \\ 0 &= x_1 -3 + 1 &\implies& x_1 = 2 \end{align}\] \[\begin{equation*} A \begin{bmatrix} 2 \\ -3 \\ 0 \\ 1 \end{bmatrix} = 0 \end{equation*}\]

One way to write the complete solution:

\[\begin{equation*} x_3 \begin{bmatrix} 1 \\ -2 \\ 1 \\ 0 \end{bmatrix} + x_4 \begin{bmatrix} 2 \\ -3 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} x_3 + 2x_4 \\ -2x_3 - 3x_4 \\ x_3 \\ x_4 \end{bmatrix} \end{equation*}\]

Row Reduced Echelon Form

However, it is quicker to "complete" elimination, and find the solution from rref. \(A\), \(U\), and \(R\) have the same nullspace. Our example matrix, \(A\), is very close to rref already (the pivots are already 1).

\[\begin{equation*} \begin{bmatrix} 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 0 & 0 \end{bmatrix} \implies \begin{bmatrix} 1 & 0 & -1 & -2 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 0 & 0 \end{bmatrix} \end{equation*}\]

Notice how if we set \(x_3 = 1\) and \(x_4 = 0\) we can read the values \(x_1\) and \(x_2\) from the third column (if we reverse the sign)

\(A \begin{bmatrix} 1 \\ -2 \\ 1 \\ 0\end{bmatrix} = 0\)

You cannot do back subsitution on a matrix in rref. This yields incorrect results.