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.
▶ Watch it solved
The question
Question: are (1, 1, 0), (0, 1, 1) and (1, 0, 1) a basis for R³?
v1 = [1; 1; 0], v2 = [0; 1; 1], v3 = [1; 0; 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, 0; 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, 2]
→ [1, 0, 1; 0, 1, −1; 0, 0, 1]
→ [1, 0, 0; 0, 1, 0; 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 3 vectors in R³, independent is enough.
Then they also reach every vector.
Careful: 3 vectors in R³: independent is enough
Every column has a leading 1.
So the vectors are independent.
a leading 1 in all 3 columns: independent
So they are a basis for R³.
= 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.
D = |1, 0, 1; 1, 1, 0; 0, 1, 1|1, 0; 1, 1; 0, 1
Down: 2.
down: 1 × 1 × 1 + 0 × 0 × 0 + 1 × 1 × 1 = 2
Up: 0.
up: 0 × 1 × 1 + 1 × 0 × 1 + 1 × 1 × 0 = 0
Down minus up: D = 2.
D = 2 − 0 = 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.
▶ Watch it solved
The question
Question: are (1, 2, 3) and (0, 1, 1) a basis for R³?
v1 = [1; 2; 3], v2 = [0; 1; 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 2 vectors, but every basis for R³ has exactly 3.
2 vectors cannot reach every vector of R³.
2 vectors, but R³ needs 3
So they are not a basis.
= not a basis
2 Check
Let's check with one vector: (1, 0, 0).
We try to make it from them.
The vectors as columns, then this one after the bar.
= [1, 0, 1; 2, 1, 0; 3, 1, 0]
Row reduced, as before.
Row 3 says 0 = −1: false.
→ [1, 0, 1; 0, 1, −2; 0, 0, −1]
So (1, 0, 0) cannot be made from them.
They do not reach every vector: the answer is right!
Let's look back.
First: count the vectors.
Last: check.
▶ Watch it solved
