Step by step
The question
Question: find the null space of A.
The null space of A is every vector v that A sends to 0: A times v is the zero vector.
1 Write A v = 0
A has 3 columns, so v has 3 entries: x, y and z.
We want every v that gives 0s.
2 Row reduce
As an augmented matrix the part after the bar is all 0s, and every move keeps it 0.
So we leave it out and row reduce A itself.
We call the rows R 1 and R 2.
We use the three row moves: swap two rows, times a row by a number that is not 0, add a multiple of one row to another.
They never change the answers.
Our goal: a leading 1 in each row, with 0s above and below it.
3 Find the free letters
The first number that is not 0 in each row is a 1.
We call it a leading 1, and ring it.
The letters x and y have a leading 1.
The letter z is free: any number.
4 Write the leading letters with the free ones
Row 1 as an equation: x − z = 0.
We want x alone.
To remove − z, we add z on both sides.
They cancel on the left.
So x = z.
Row 2 as an equation: y + z = 0.
We want y alone.
To remove + z, we take away z on both sides.
They cancel on the left.
So y = −z.
5 Write it as vectors
We stack the letters in a column.
The letter z is free, so it stays z.
We split it up: what goes with z makes one vector, (1, −1, 1).
Careful!
In the vector of z, the z entry itself is 1.
Do not leave it out.
6 Write the basis
Every vector of the null space is made from this one vector.
So it is a basis: (1, −1, 1).
The dimension of the null space is how many basis vectors: 1, one for each free letter.
7 Check
Let's check: A times each basis vector must be 0.
We work out every times.
We get 0s: (1, −1, 1) is in the null space.
A sends every basis vector to 0.
So the answer is right!
Let's look back.
First: write A v = 0.
Then: row reduce.
Then: find the free letters.
Then: write the leading letters with the free ones.
Then: write it as vectors.
Then: write the basis.
Last: check.