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Are (2, 1) and (4, 2) independent

Step by step, with the reason for every step · Linear independence · Linear Algebra

Step by step

The question

Question: are (2, 1) and (4, 2) linearly independent?

v1 = [2; 1], v2 = [4; 2]

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[2; 1] + c2[4; 2] = [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.

= [2, 4, 0; 1, 2, 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; 2, 4, 0]
→ [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 × [2; 1] + [4; 2]
= −[2 × 2; 2 × 1] + [4; 2]
= −[4; 2] + [4; 2]
= [−4 + 4; −2 + 2]
= [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.

Answer
= dependent

Your turn

Try this one yourself, then check:

Are [1; 3], [2; 6] independent?
Show the answer
dependent

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