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

Step by step

The question

Question: are (1, 1) and (1, −1) a basis for R²?

v₁ = [1; 1], v₂ = [1; −1]

A basis for R² is a set of vectors that are independent, and that make every vector of R² as a linear combination.

basis for R²: independent, and every vector

is a linear combination of them

1 Count the vectors

There are 2 vectors, and R² needs exactly 2.

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, 1; 1, −1]

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, 1; 0, −2]

→ [1, 1; 0, 1]

→ [1, 0; 0, 1]

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 2 vectors in R², independent is enough.

Then they also reach every vector.

Careful: 2 vectors in R²: independent is enough

Every column has a leading 1.

So the vectors are independent.

a leading 1 in all 2 columns: independent

So they are a basis for R².

= a basis for R²

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, 1; 1, −1|

So D = 1 × (−1) − 1 × 1.

D = 1 × (−1) − 1 × 1

1 times (−1) is −1.

1 times 1 is 1.

D = −1 − 1

−1 minus 1 is −2.

So D = −2.

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.

Answer