Ax=0

$A =$ $\begin{bmatrix} 1&2&2&2\\ 2&4&6&8\\ 3&6&8&10 \end{bmatrix}$

Note: You may notice not all rows and columns are independent
Let's do elimination to start solving.

$\begin{bmatrix} 1&2&2&2\\ 2&4&6&8\\ 3&6&8&10 \end{bmatrix}$ $=>$ $\begin{bmatrix} 1&2&2&2\\ 0&0&2&4\\ 0&0&2&4 \end{bmatrix}$

Uh oh. Now I see a 0 in the pivot position and I cannot do a row exchange. Really that's just telling me that the now is just a combination of earlier columns. It's depending on earlier columns. Let's just go on and use the next column as the pivot for that row.

$A=$ $\begin{bmatrix} 1&2&2&2\\ 2&4&6&8\\ 3&6&8&10 \end{bmatrix}$ $=>$ $\begin{bmatrix} 1&2&2&2\\ 0&0&2&4\\ 0&0&2&4 \end{bmatrix}$ $=>$ $\begin{bmatrix} 1&2&2&2\\ 0&0&2&4\\ 0&0&0&0 \end{bmatrix}$ $=U$

Note: We have 2 pivots (aka rank 2). Row 1, Column 1 and Row 2, Column 3.

Special Solutions.

Let's write out $U$ in matrix in equation form to see what we have.

$x_1+2x_2+2x_3+2x_4=0$

$2x_3+4x_4=0$

Let's call the columns with a pivot a pivot column, and ones without a free column. So we have 2 pivot columns and 2 free columns.

Let's assign some values to the free columns, then solve for the pivot columns using back substitution. We'll call these special solutions.

$x=$ $\begin{bmatrix} \\ 1\\ \\ 0 \end{bmatrix}$ $=>$ $\begin{bmatrix} -2\\ 1\\ 0\\ 0 \end{bmatrix}$ $=>$ $C$ $\begin{bmatrix} -2\\ 1\\ 0\\ 0 \end{bmatrix}$

Let's put 1 in the other free variable.

$x=$ $\begin{bmatrix} \\ 0\\ \\ 1 \end{bmatrix}$ $=>$ $\begin{bmatrix} 2\\ 0\\ -2\\ 1 \end{bmatrix}$ $=>$ $C$ $\begin{bmatrix} 2\\ 0\\ -2\\ 1 \end{bmatrix}$

Great. Our entire nullspace is a linear combination of these 2 special solutions. There will be 1 special solution for each free variable. In this case there were 2 (free columns, rank, free variables, etc).

$C$ $\begin{bmatrix} -2\\ 1\\ 0\\ 0 \end{bmatrix}$ $+d$ $\begin{bmatrix} 2\\ 0\\ -2\\ 1 \end{bmatrix}$

Reduced Row Echelon Form ($R$)

Reduced row echelon form has zeros above and below pivots, and the pivots are = 1. Let's do this using elimination.

$U=$ $\begin{bmatrix} 1&2&2&2\\ 0&0&2&4\\ 0&0&0&0 \end{bmatrix}$ $=>$ $\begin{bmatrix} 1&2&0&-2\\ 0&0&2&4\\ 0&0&0&0 \end{bmatrix}$ $=>$ $\begin{bmatrix} 1&2&0&-2\\ 0&0&1&2\\ 0&0&0&0 \end{bmatrix}$ $=R$

Note: This row of 0’s was created because elimination identified that row 3 is just a combination of rows 1 and 2.
Why this form?

We see in the pivot columns we see $I$ $\begin{bmatrix}1&0\\0&1\end{bmatrix}$

In the free columns we see the opposite of the nullspace definintions we found above $\begin{bmatrix}2&-2\\0&2\end{bmatrix}$

Solve the transpose

$A=$ $\begin{bmatrix} 1&2&3\\ 2&4&6\\ 2&6&8\\ 2&8&10 \end{bmatrix}$

Elimination from $A$ -> $U$

$\begin{bmatrix} 1&2&3\\ 2&4&6\\ 2&6&8\\ 2&8&10 \end{bmatrix}$ $=>$ $\begin{bmatrix} 1&2&3\\ 0&0&0\\ 0&2&2\\ 0&4&4 \end{bmatrix}$ $=>$ $\begin{bmatrix} 1&2&3\\ 0&2&2\\ 0&0&0\\ 0&0&0 \end{bmatrix}$

Note: Notice that I did a row exchange to do this elimination and ended with a rank 2, which was the same as with the transpose.

Solve for Nullspace

There's 1 free column, so you fill in a 1 in the free column. $x=$ $\begin{bmatrix}\\\\1\end{bmatrix}$ Then with back substution solve for the pivot columns. The solution is any linear combination of this, which is a line in the null space.

$$x=C\begin{bmatrix}-1\\-1\\1\end{bmatrix}$$

Row Reduction from $U$ -> $R$

Row Echelon Form

$\begin{bmatrix} 1&2&3\\ 0&2&2\\ 0&0&0\\ 0&0&0 \end{bmatrix}$ $=>$ $\begin{bmatrix} 1&0&1\\ 0&2&2\\ 0&0&0\\ 0&0&0 \end{bmatrix}$ $=>$ $\begin{bmatrix} 1&0&1\\ 0&1&1\\ 0&0&0\\ 0&0&0 \end{bmatrix}$ $=R$

We see the identity matrix in the pivot columns, and the opposite of our pivot nullspace varibales in the free column. Perfect!