• 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.