Worked examples
The question
Question: are (1, 2) and (2, 4) linearly independent?
v1 = [1; 2], v2 = [2; 4]
Independent means: the only way to make the zero vector from them is to take 0 of each.
Otherwise they are dependent.
independent: the only way to make 0
is c1 = c2 = 0
1 Write the vector equation
We look for numbers c₁ and c₂ that make the zero vector.
c1[1; 2] + c2[2; 4] = [0; 0]
Careful!
Taking 0 of each always gives 0.
That proves nothing.
The question is: is it the only way?
Careful: c1 = c2 = 0 always works: is it the ONLY way?
2 Write the augmented matrix
Each vector goes in as a column: the first entries make row 1, the second entries row 2.
After the bar: 0s.
Every move keeps them 0.
= [1, 2, 0; 2, 4, 0]
We call the rows R 1 and R 2.
3 Row reduce
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
→ [1, 2, 0; 0, 0, 0]
4 Count the leading 1s
The first number that is not 0 in each row is a 1.
We call it a leading 1, and ring it.
Column 2 has no leading 1.
So c₂ is free: it can be any number, not only 0.
column 2: no leading 1, so c2 is free
5 Find a way to make 0
We pick c₂ = 1.
Each row then gives a leading c.
Row 1 says c₁ + 2c₂ = 0.
So c₁ = −2.
c1 = −2
So −2v₁ + v₂ = 0, and not all the numbers are 0.
−2v1 + v2 = 0
So the vectors are linearly dependent.
= dependent
6 Check
Let's check: we work out −2v₁ + v₂.
= −2 × [1; 2] + [2; 4]
= −[2 × 1; 2 × 2] + [2; 4]
= −[2; 4] + [2; 4]
= [−2 + 2; −4 + 4]
= [0; 0]
We get the zero vector.
So the vectors are dependent: the answer is right!
Let's look back.
First: write the vector equation.
Then: write the augmented matrix.
Then: row reduce.
Then: count the leading 1s.
Then: find a way to make 0.
Last: check.
▶ Watch it solved
The question
Question: are (1, 0, 1), (0, 1, 1) and (1, 1, 3) linearly independent?
v1 = [1; 0; 1], v2 = [0; 1; 1], v3 = [1; 1; 3]
Independent means: the only way to make the zero vector from them is to take 0 of each.
Otherwise they are dependent.
independent: the only way to make 0
is c1 = c2 = c3 = 0
1 Write the vector equation
We look for numbers c₁, c₂ and c₃ that make the zero vector.
c1[1; 0; 1] + c2[0; 1; 1] + c3[1; 1; 3] = [0; 0; 0]
Careful!
Taking 0 of each always gives 0.
That proves nothing.
The question is: is it the only way?
Careful: c1 = c2 = c3 = 0 always works: is it the ONLY way?
2 Write the augmented matrix
Each vector goes in as a column: the first entries make row 1, the second entries row 2.
After the bar: 0s.
Every move keeps them 0.
= [1, 0, 1, 0; 0, 1, 1, 0; 1, 1, 3, 0]
We call the rows R 1, R 2 and R 3.
3 Row reduce
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
→ [1, 0, 1, 0; 0, 1, 1, 0; 0, 1, 2, 0]
→ [1, 0, 1, 0; 0, 1, 1, 0; 0, 0, 1, 0]
→ [1, 0, 0, 0; 0, 1, 0, 0; 0, 0, 1, 0]
4 Count the leading 1s
The first number that is not 0 in each row is a 1.
We call it a leading 1, and ring it.
Every column has a leading 1.
So each row says one c is 0: c₁ = c₂ = c₃ = 0.
That is the only way.
a leading 1 in every column: c1 = c2 = c3 = 0
So the vectors are linearly independent.
= independent
5 Check
Let's check another way.
The vectors make a square matrix.
Its determinant must not be 0.
D = |1, 0, 1; 0, 1, 1; 1, 1, 3|1, 0; 0, 1; 1, 1
Down: 3.
down: 1 × 1 × 3 + 0 × 1 × 1 + 1 × 0 × 1 = 3
Up: 2.
up: 1 × 1 × 1 + 1 × 1 × 1 + 3 × 0 × 0 = 2
Down minus up: D = 1.
D = 3 − 2 = 1
D is 1, not 0.
So the vectors are independent: the answer is right!
Let's look back.
First: write the vector equation.
Then: write the augmented matrix.
Then: row reduce.
Then: count the leading 1s.
Last: check.
▶ Watch it solved
The question
Question: are (1, 2), (3, 4) and (5, 6) linearly independent?
v1 = [1; 2], v2 = [3; 4], v3 = [5; 6]
Independent means: the only way to make the zero vector from them is to take 0 of each.
Otherwise they are dependent.
independent: the only way to make 0
is c1 = c2 = c3 = 0
1 Write the vector equation
We look for numbers c₁, c₂ and c₃ that make the zero vector.
c1[1; 2] + c2[3; 4] + c3[5; 6] = [0; 0]
Careful!
Taking 0 of each always gives 0.
That proves nothing.
The question is: is it the only way?
Careful: c1 = c2 = c3 = 0 always works: is it the ONLY way?
A shortcut: 3 vectors with only 2 entries each are always dependent.
Let's see why.
3 vectors in R²: more vectors than entries
2 Write the augmented matrix
Each vector goes in as a column: the first entries make row 1, the second entries row 2.
After the bar: 0s.
Every move keeps them 0.
= [1, 3, 5, 0; 2, 4, 6, 0]
We call the rows R 1 and R 2.
3 Row reduce
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
→ [1, 3, 5, 0; 0, −2, −4, 0]
→ [1, 3, 5, 0; 0, 1, 2, 0]
→ [1, 0, −1, 0; 0, 1, 2, 0]
4 Count the leading 1s
The first number that is not 0 in each row is a 1.
We call it a leading 1, and ring it.
Column 3 has no leading 1.
So c₃ is free: it can be any number, not only 0.
column 3: no leading 1, so c3 is free
5 Find a way to make 0
We pick c₃ = 1.
Each row then gives a leading c.
Row 1 says c₁ − c₃ = 0.
So c₁ = 1.
c1 = 1
Row 2 says c₂ + 2c₃ = 0.
So c₂ = −2.
c2 = −2
So v₁ − 2v₂ + v₃ = 0, and not all the numbers are 0.
v1 − 2v2 + v3 = 0
So the vectors are linearly dependent.
= dependent
6 Check
Let's check: we work out v₁ − 2v₂ + v₃.
= [1; 2] − 2 × [3; 4] + [5; 6]
= [1; 2] − [2 × 3; 2 × 4] + [5; 6]
= [1; 2] − [6; 8] + [5; 6]
= [1 − 6 + 5; 2 − 8 + 6]
= [0; 0]
We get the zero vector.
So the vectors are dependent: the answer is right!
Let's look back.
First: write the vector equation.
Then: write the augmented matrix.
Then: row reduce.
Then: count the leading 1s.
Then: find a way to make 0.
Last: check.
▶ Watch it solved
