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Is (4, 7) in the span of (1, 1), (1, 2)

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

Step by step

The question

Question: is (4, 7) in the span of (1, 1) and (1, 2)?

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

We call them v₁ and v₂, and the vector we test b.

b = [4; 7]

The span of some vectors is every linear combination of them: any numbers times them, added.

span: every c1v1 + c2v2

1 Write the vector equation

b is in the span if some numbers c₁ and c₂ make this true.

We look for them.

c1[1; 1] + c2[1; 2] = [4; 7]

2 Write the augmented matrix

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

b goes after the bar.

= [1, 1, 4; 1, 2, 7]

We call the rows R 1 and R 2.

Careful!

Each vector is a column of the matrix, not a row.

Row 1 holds the first entries of v₁ and v₂, and of b.

Careful: the vectors are the COLUMNS

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, 4; 0, 1, 3]
→ [1, 0, 1; 0, 1, 3]

4 Read the answer

Each row is an equation again.

Row 1 says c₁ = 1.

c1 = 1

Row 2 says c₂ = 3.

c2 = 3

So yes: b = v₁ + 3v₂.

b is in the span.

b = v1 + 3v2

5 Check

Let's check: we work out v₁ + 3v₂.

= [1; 1] + 3 × [1; 2]
= [1; 1] + [3 × 1; 3 × 2]
= [1; 1] + [3; 6]
= [1 + 3; 1 + 6]
= [4; 7]

We get (4, 7): that is b.

So the answer is right!

Let's look back.

First: write the vector equation.

Then: write the augmented matrix.

Then: row reduce.

Then: read the answer.

Last: check.

Answer
b = v1 + 3v2

Your turn

Try this one yourself, then check:

Is [6; 15] in the span of [1; 2], [1; 3] ?
Show the answer
b = 3v1 + 3v2

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