# Free variables, nullspace for a matrix with the sum of certain columns = zero vector [Strang P142 3.2.20]

Suppose column 1 + column 3 + column 5 $\mathbf{ = 0} \quad (\bigstar)$ in a 4 by 5 matrix with four pivots. Which column is sure to have no pivot (and which variable is free)? What is the special solution? What is the nullspace?

Solution : (I think there's a typo; column 2 in green should be column 5) $\Large{{1.}}$ I accept that because the rank of a matrix $\le \min\{\text{# of rows, # of columns}\}$,
thus rank = # of pivots $\le 4.$ I don't perceive how it must be column $5$ which lacks a pivot?

Say we're in the event where we do have $4$ pivots, the maximum number.
Then by reason of the starred equation, only one of those columns $1, 3, 5$ lacks a pivot.
Each of columns $2,4$ must contain one pivot. Why? If they don't, each of columns $1, 3, 5$ will contain a pivot, which contradicts the starred equation. What did I misconstrue?

$\Large{{2.}}$ Much as the solution checks that $(1, 0, 1, 0, 1)$ by dint of column-row matrix multiplication, how would you compute/deduce it? The solution didn't expound on the intermediate steps?

An analogous question is on P4 of http: // 140.113.13.233/la/exam12s-2-sol.pdf but it uncloaks less.

1. As far as x5 = -x1 - x3, so the x5 don't have pivot because it depends on x1 and x3.

2. Remember how N(A) matrix looks like. It has two components: -F (in the upper part of the matrix) and I (in the bottom part of the matrix). Also lets recall reduced echelon form R(A). It has "1" in pivot column and "0" below and above it.

1, 0, 0, 0, ?

0, 1, 0, 0, ?

0, 0, 1, 0, ?

0, 0, 0, 1, ?

So we need to compute values for x5. To do this just substitute known values for x1 and x3 in every row to the equation above.

For the 1st row: 1, 0, 0, 0, ? x1 = 1; x3 = 0; x5=-1 - 0 = -1

And so on.

As a result you'll get (this is F):

-1

0

-1

0

But for special solution you need -F in the upper part and I in the bottom: -F is (1, 0, 1, 0) I is (1)

So there is our special solution: 1, 0, 1, 0, 1.