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

Step by step

The question

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

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

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.

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.

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.

Every column has a leading 1.

So the vectors are independent.

So they are a basis for R³.

5 Check

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

A basis: it is not 0.

Not a basis: it is 0.

Down: 2.

Up: 0.

Down minus up: D = 2.

D is 2, not 0.

So they are 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.