Finding the nullspace: reduce to R, then read off the special solutions

What R looks like

R is what you get when elimination goes all the way. Three things are true of R:

• Each pivot is 1 (divide a row by its pivot if needed).
• Each pivot column is zero everywhere else, above the pivot as well as below.
• Rows of zeros, if any, sit at the bottom.

Take A = 12132438. It takes two row steps:

The second step is new. It clears the entry above the pivot in column 3. Lesson 3's Gauss-Jordan did the same thing.

Why R has the same nullspace

Each row step subtracts a multiple of one equation from another, or divides a row by a nonzero number. All the right sides are 0, and they stay 0. So no solution is gained or lost. Quick check with x = −2100:

Pivot variables and free variables

The pivots are in columns 1 and 3. So x₁ and x₃ are pivot variables, and x₂ and x₄ are free. Rx = 0 reads:

Each equation has exactly one pivot variable in it. So whatever you choose for x₂ and x₄, the pivot variables follow: x₁ = −2x₂ − x₄ and x₃ = −2x₄. Choose x₂ = 3, x₄ = 1, for example: x₁ = −6 − 1 = −7, x₃ = −2. Then A−73−21 = −7 + 6 − 2 + 3−14 + 12 − 6 + 8 = 00 ✓.

The special solutions

The simplest choices are one free variable 1, the rest 0.

Every nullspace vector is a combination x = c·s₁ + d·s₂ of the two special solutions, with c and d equal to its free variables x₂ and x₄. For example, the choice x₂ = 3, x₄ = 1 from above gives 3 times s₁ plus 1 times the second special solution. Check this once you have found it.