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Are (1, 0, 0), (0, 0, 0), (0, 1, 0) independent

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

Step by step

The question

Question: are (1, 0, 0), (0, 0, 0) and (0, 1, 0) linearly independent?

v1 = [1; 0; 0], v2 = [0; 0; 0], v3 = [0; 1; 0]

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; 0] + c2[0; 0; 0] + c3[0; 1; 0] = [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?

A shortcut: one of them is the zero vector.

1 times it already makes 0, so they are dependent.

a zero vector: 1 × 0 = 0

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, 0, 0; 0, 0, 1, 0; 0, 0, 0, 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

The matrix already has its leading 1s, with 0s above and below.

No moves are needed.

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 = 0.

So c₁ = 0.

c1 = 0

Row 2 says c₃ = 0.

So c₃ = 0.

c3 = 0

So v₂ = 0, and not all the numbers are 0.

v2 = 0

So the vectors are linearly dependent.

= dependent

6 Check

Let's check: we work out v₂.

= [0; 0; 0]
= [0; 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 [3; 2; 1], [0; 0; 0], [0; 1; 2] independent?
Show the answer
dependent

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