• Use elimination to solve a \(3 \times 4\) rectangular matrix (\(A\)):
\[\begin{bmatrix} \begin{array}{rrrr} 1 & 2 & 2 & 2 \\ 2 & 4 & 6 & 8 \\ 3 & 6 & 8 & 10 \\ \end{array} \end{bmatrix}\]
  • During elimination:
    • Nullspace does not change
    • Solutions do not change
    • Columnspace IS changing
    • Rowspace does not change (not explicitly mentioned, but assume so)
  • Result of elimination (\(U\)):
\[\begin{bmatrix} \begin{array}{rrrr} 1 & 2 & 2 & 2 \\ 0 & 0 & 2 & 4 \\ 0 & 0 & 0 & 0 \\ \end{array} \end{bmatrix}\]

  • No pivot in second column means it’s free (a combo of other columns)

  • 2 total pivots; they are 1 at (1,1) and 2 at (2,3)

  • The result of elimination is in upper echelon form (\(U\))

  • Rank of matrix == pivots == 2 in this example.

  • You can solve \(Ux=0\), same solutions as \(Ax=0\)

  • 2 pivot columns, 2 free columns

  • You can assign any number to the free columns and then solve the equations for the pivots columns.

  • Assign \(x_2 = 1\) and \(x_4 = 0\) to the free variables and solve for \(x_1\) and \(x_3\). Note \(x_1=-2\) and \(x_3=0\) after back subbing. This is a vector in the nullspace and any multiple of it is in the nullspace.

  • You have two free variables, so you need another vector in the nullspace. Now Assign \(x_2 = 0\) and \(x_4 = 1\) to the free variables and solve for \(x_1\) and \(x_3\). Note \(x_1=2\) and \(x_3=-2\) after back subbing. This is anonter vector in the nullspace and any multiple of it is in the nullspace.

  • Any linear combination of those two vectors are in the nullspace. You’ll get one vector in the nullspace for each free column.

\[nullspace(A) = c_1 * \begin{bmatrix} \begin{array}{r} -2 \\ 1 \\ 0 \\ 0 \\ \end{array} \end{bmatrix} + c_2 * \begin{bmatrix} \begin{array}{r} 2 \\ 0 \\ -2 \\ 1 \\ \end{array} \end{bmatrix}\]

  • Rank is equal to the pivot variable count. Free variables is \(n-R\) for an \(m \times n\) matrix.

  • Finding the nullspace: Do elimination. Set each free variable to one (and others to zero) and solve for a vector in the nullspace.

  • Matrix \(R\) is the reduced row echelon form (rref). Use the pivots to clean up the rows above them and make pivots equal to 1. \(R\) for our above example is:

\[\begin{bmatrix} \begin{array}{rrrr} 1 & 2 & 0 & -2 \\ 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 0 \\ \end{array} \end{bmatrix}\]

  • Note the identity matrix \(I\) is mixed into the rref matrix \(R\).

  • Typical rref looks like (I is identity matrix, F is free variables, the columns from I and F may be intermixed, \(r\) pivot rows AND columns, \(m-r\) free rows, \(n-r\) free columns):

\[\begin{bmatrix} \begin{array}{rr} I & F \\ 0 & 0 \\ \end{array} \end{bmatrix}\]
  • Nullspace matrix (\(N\)) - columns are the special solutions. \(RN=0\).
\[N = \begin{bmatrix} \begin{array}{rr} -F \\ I \\ \end{array} \end{bmatrix}\]

  • I find this part of the lecture confusing. He is composing matrices (\(F\) and \(I\)) that don’t quite correspond to the example.

  • Rank of \(A^T\) is the same as \(A\). \(N(A^T)\) is dimension 1 in our example.