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- An Example
Recall that a system is homogeneous if it is of the form
Ax = 0.
The solution set here goes by the name “the null space of A,” or N(A). We can speed up
the row operations a little if we notice that when doing row operations on
[A| 0 ]
the last column never changes. We can do operation just on A, as long as we remember
when converting rows back to equations to put zeros “= 0” on the right.
Example 1. Find the null space of A, where
A =
and find a basis for this null space.
We need to solve
Ax = 0.
1
Using just A, we use row ops to find reduced row echelon form:
y
R3 − 4R1 → R
R4 − R1 → R
R5 − 2R1 → R
R6 − R1 → R
y
R4 − R3 → R
R6 − 2R3 → R
yR2^ ↔^ R
y−
1 2
R2 → R
y
R4 − R3 → R
R6 − 2R3 → R
1 2 0 0 1 − 2 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
yR1 − R2 → R
1 0 0 0
1 2 0 1 0 0 −
1 2 0 0 1 − 2 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
so we have the equations
x 1 +
1 2
x 5 = 0
x 2 +
1 2
x 5 = 0
x 3 − 2 x 4 = 0
- Null Space vs Nullity
Sometimes we only want to know how big the solution set is to
Ax = 0.
Definition 1. The nullity of a matrix A is the dimension of its null space:
nullity(A) = dim(N(A)).
It is easier to find the nullity than to find the null space. This is because
The number of free variables (in the solved equations) equals the nullity of A.
- Nullity vs Basis for Null Space
There is a general method to find a basis for the null space:
(a) Use row operations to reduced echelon form.
(b) Write out corresponding simplified equations for the null space.
(c) Set first free variable to 1, the others to 0. This solution x is a basis element.
(d) Repeat (b), so each free variable takes its trun being the only 1.
Here is our first method to find just the nullity:
(a) Use row operations just to echelon form.
(b) Count the non-pivot columns.
This distinction is important in some applications, as the nullity is therefore faster to com-
pute.
Example 2. Find a basis for N(A) where
A =
Getting srarted with row ops:
↓R1 ↔ R2 ↓R3 − R2 → R
y
R2 − 2R1 → R
R3 − 3R1 → R
y
1 3
R2 → R
1 3
R3 → R
We know the nullity (it is 3) but the question asks us to really find the null space. We press
on:
↓−R3 → R
y^1 3
R2 → R
y
R1 − 2R3 → R
R2 + R3 → R
↓R1 + R2 → R
So we have three free variables, and the equations become
x 1 + x 2 − 2 x 3 − 3 x 5 = 0
x 4 − x 5 = 0
x 6 = 0
and so the null space is given by
x 1
x 2
x 3
x 4
x 5
x 6
−r + 2s + 3t
r
s
t
t
0
Set r = 1, s = 0 and t = 0 and we get
− 1
1
0
0
0
0
Set r = 0, s = 1 and t = 0 and we get
2 0 1 0 0 0
Now we can also do column operations, and can stop as soon as we see where the pivots
will be.
y
R2 − 2R1 → R
R3 − 2R1 → R
↓R2 ↔ R
↓C1 ↔ C
↓R2 − 3R3 → R
This is enough for us to see there are three non-pivot rows, so
nullity(A) = 3.
Notice that the pivots don’t end up in the same locations as before, so we have lost track
of N(A).
There is often no advantage of being able to do column and row operations together; it is
too complicated to figure out what to do next. In specific calculations, in in the theory of
rank, it tells us a lot.
