Worked examples
The question
Question: find the null space of A.
A = [1, 2, 3]
The null space of A is every vector v that A sends to 0: A times v is the zero vector.
null space of A: every v with A v = 0
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.
[1, 2, 3] × [x; y; z] = [0]
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.
= [1, 2, 3]
We call the rows R 1.
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.
swap two rows
× a row by a number (not 0)
add a multiple of one row to another
The matrix already has its leading 1s, with 0s above and below.
No moves are needed.
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 letter x has a leading 1.
The letters y and z are free: any number.
leading 1: x
free: y, z
4 Write the leading letters with the free ones
Row 1 as an equation: x + 2y + 3z = 0.
x + 2y + 3z = 0
We want x alone.
To remove + 2y and + 3z, we take away 2y and take away 3z on both sides.
x + 2y + 3z − 2y − 3z = −2y − 3z
They cancel on the left.
So x = −2y − 3z.
x = −2y − 3z
5 Write it as vectors
We stack the letters in a column.
The free letters stay as they are.
[x; y; z] = [−2y − 3z; y; z]
We split it up: what goes with y makes one vector, (−2, 1, 0).
What goes with z makes another, (−3, 0, 1).
[x; y; z] = y[−2; 1; 0] + z[−3; 0; 1], y and z any numbers
Careful!
In the vector of y, the y entry itself is 1.
Do not leave it out.
Careful: the free letter's own entry is 1
6 Write the basis
Every vector of the null space is made from these vectors, and they are independent.
So they are a basis: (−2, 1, 0) and (−3, 0, 1).
= basis: [−2; 1; 0], [−3; 0; 1]
The dimension of the null space is how many basis vectors: 2, one for each free letter.
dimension = 2 (one for each free letter)
7 Check
Let's check: A times each basis vector must be 0.
= [1, 2, 3] × [−2; 1; 0]
= [1 × (−2) + 2 × 1 + 3 × 0]
We work out every times.
= [−2 + 2 + 0]
We get 0s: (−2, 1, 0) is in the null space.
= [0]
Now the next one.
= [1, 2, 3] × [−3; 0; 1]
= [1 × (−3) + 2 × 0 + 3 × 1]
We work out every times.
= [−3 + 0 + 3]
We get 0s: (−3, 0, 1) is in the null space.
= [0]
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.
Answer dimension = 2 (one for each free letter)
▶ Watch it solved
The question
Question: find the null space of A.
A = [1, 0, −1; 0, 1, 2]
The null space of A is every vector v that A sends to 0: A times v is the zero vector.
null space of A: every v with A v = 0
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.
[1, 0, −1; 0, 1, 2] × [x; y; z] = [0; 0]
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.
= [1, 0, −1; 0, 1, 2]
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.
swap two rows
× a row by a number (not 0)
add a multiple of one row to another
The matrix already has its leading 1s, with 0s above and below.
No moves are needed.
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.
leading 1: x, y
free: z
4 Write the leading letters with the free ones
Row 1 as an equation: x − z = 0.
x − z = 0
We want x alone.
To remove − z, we add z on both sides.
x − z + z = z
They cancel on the left.
So x = z.
x = z
Row 2 as an equation: y + 2z = 0.
y + 2z = 0
We want y alone.
To remove + 2z, we take away 2z on both sides.
y + 2z − 2z = −2z
They cancel on the left.
So y = −2z.
y = −2z
5 Write it as vectors
We stack the letters in a column.
The letter z is free, so it stays z.
[x; y; z] = [z; −2z; z]
We split it up: what goes with z makes one vector, (1, −2, 1).
[x; y; z] = z[1; −2; 1], z any number
Careful!
In the vector of z, the z entry itself is 1.
Do not leave it out.
Careful: the free letter's own entry is 1
6 Write the basis
Every vector of the null space is made from this one vector.
So it is a basis: (1, −2, 1).
= basis: [1; −2; 1]
The dimension of the null space is how many basis vectors: 1, one for each free letter.
dimension = 1 (one for each free letter)
7 Check
Let's check: A times each basis vector must be 0.
= [1, 0, −1; 0, 1, 2] × [1; −2; 1]
= [1 × 1 + 0 × (−2) + (−1) × 1; 0 × 1 + 1 × (−2) + 2 × 1]
We work out every times.
= [1 + 0 − 1; 0 − 2 + 2]
We get 0s: (1, −2, 1) is in the null space.
= [0; 0]
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.
Answer dimension = 1 (one for each free letter)
▶ Watch it solved
The question
Question: find the null space of A.
A = [1, 2, 0, 1; 0, 0, 1, 2]
The null space of A is every vector v that A sends to 0: A times v is the zero vector.
null space of A: every v with A v = 0
1 Write A v = 0
A has 4 columns, so v has 4 entries: x, y, z and w.
We want every v that gives 0s.
[1, 2, 0, 1; 0, 0, 1, 2] × [x; y; z; w] = [0; 0]
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.
= [1, 2, 0, 1; 0, 0, 1, 2]
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.
swap two rows
× a row by a number (not 0)
add a multiple of one row to another
The matrix already has its leading 1s, with 0s above and below.
No moves are needed.
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 z have a leading 1.
The letters y and w are free: any number.
leading 1: x, z
free: y, w
4 Write the leading letters with the free ones
Row 1 as an equation: x + 2y + w = 0.
x + 2y + w = 0
We want x alone.
To remove + 2y and + w, we take away 2y and take away w on both sides.
x + 2y + w − 2y − w = −2y − w
They cancel on the left.
So x = −2y − w.
x = −2y − w
Row 2 as an equation: z + 2w = 0.
z + 2w = 0
We want z alone.
To remove + 2w, we take away 2w on both sides.
z + 2w − 2w = −2w
They cancel on the left.
So z = −2w.
z = −2w
5 Write it as vectors
We stack the letters in a column.
The free letters stay as they are.
[x; y; z; w] = [−2y − w; y; −2w; w]
We split it up: what goes with y makes one vector, (−2, 1, 0, 0).
What goes with w makes another, (−1, 0, −2, 1).
[x; y; z; w] = y[−2; 1; 0; 0] + w[−1; 0; −2; 1], y and w any numbers
Careful!
In the vector of y, the y entry itself is 1.
Do not leave it out.
Careful: the free letter's own entry is 1
6 Write the basis
Every vector of the null space is made from these vectors, and they are independent.
So they are a basis: (−2, 1, 0, 0) and (−1, 0, −2, 1).
= basis: [−2; 1; 0; 0], [−1; 0; −2; 1]
The dimension of the null space is how many basis vectors: 2, one for each free letter.
dimension = 2 (one for each free letter)
7 Check
Let's check: A times each basis vector must be 0.
= [1, 2, 0, 1; 0, 0, 1, 2] × [−2; 1; 0; 0]
= [1 × (−2) + 2 × 1 + 0 × 0 + 1 × 0; 0 × (−2) + 0 × 1 + 1 × 0 + 2 × 0]
We work out every times.
= [−2 + 2 + 0 + 0; 0 + 0 + 0 + 0]
We get 0s: (−2, 1, 0, 0) is in the null space.
= [0; 0]
Now the next one.
= [1, 2, 0, 1; 0, 0, 1, 2] × [−1; 0; −2; 1]
= [1 × (−1) + 2 × 0 + 0 × (−2) + 1 × 1; 0 × (−1) + 0 × 0 + 1 × (−2) + 2 × 1]
We work out every times.
= [−1 + 0 + 0 + 1; 0 + 0 − 2 + 2]
We get 0s: (−1, 0, −2, 1) is in the null space.
= [0; 0]
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.
Answer dimension = 2 (one for each free letter)
▶ Watch it solved
