Is (1, 1, 0), (0, 1, 1), (1, 2, 1) a basis for R^3 Step by Step

Step by step

The question

Question: are (1, 1, 0), (0, 1, 1) and (1, 2, 1) a basis for R³?

v₁ = [1; 1; 0], v₂ = [0; 1; 1], v₃ = [1; 2; 1]

A basis for R³ is a set of vectors that are independent, and that make every vector of R³ as a linear combination.

basis for R³: independent, and every vector

is a linear combination of them

1 Count the vectors

There are 3 vectors, and R³ needs exactly 3.

Good.

Now: are they independent?

2 Write the matrix

Each vector goes in as a column: the first entries make row 1, the second entries row 2.

We test independence, so the right side is 0s.

Every move keeps them 0, so we leave them out.

= [1, 0, 1; 1, 1, 2; 0, 1, 1]

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, 1, 1; 0, 1, 1]

→ [1, 0, 1; 0, 1, 1; 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.

Careful!

For 3 vectors in R³, independent is enough.

Then they also reach every vector.

Careful: 3 vectors in R³: independent is enough

Column 3 has no leading 1.

So the vectors are dependent.

column 3: no leading 1: dependent

So they are not a basis.

= not a basis

5 Check

Let's check with the determinant of the same matrix.

A basis: it is not 0.

Not a basis: it is 0.

D = |1, 0, 1; 1, 1, 2; 0, 1, 1|1, 0; 1, 1; 0, 1

Down: 2.

down: 1 × 1 × 1 + 0 × 2 × 0 + 1 × 1 × 1 = 2

Up: 2.

up: 0 × 1 × 1 + 1 × 2 × 1 + 1 × 1 × 0 = 2

Down minus up: D = 0.

D = 2 − 2 = 0

D is 0.

So they are not a basis: the answer is right!

Let's look back.

First: count the vectors.

Then: write the matrix.

Then: row reduce.

Then: count the leading 1s.

Last: check.

Answer