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Is (5, 1, 4) in the span of (1, 0, 1), (2, 1, 1)

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

Step by step

The question

Question: is (5, 1, 4) in the span of (1, 0, 1) and (2, 1, 1)?

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

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

b = [5; 1; 4]

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; 0; 1] + c2[2; 1; 1] = [5; 1; 4]

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, 2, 5; 0, 1, 1; 1, 1, 4]

We call the rows R 1, R 2 and R 3.

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, 2, 5; 0, 1, 1; 0, −1, −1]
→ [1, 0, 3; 0, 1, 1; 0, 0, 0]

4 Read the answer

Each row is an equation again.

Row 1 says c₁ = 3.

c1 = 3

Row 2 says c₂ = 1.

c2 = 1

So yes: b = 3v₁ + v₂.

b is in the span.

b = 3v1 + v2

5 Check

Let's check: we work out 3v₁ + v₂.

= 3 × [1; 0; 1] + [2; 1; 1]
= [3 × 1; 3 × 0; 3 × 1] + [2; 1; 1]
= [3; 0; 3] + [2; 1; 1]
= [3 + 2; 0 + 1; 3 + 1]
= [5; 1; 4]

We get (5, 1, 4): 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 = 3v1 + v2

Your turn

Try this one yourself, then check:

Is [9; 2; 3] in the span of [1; 0; 1], [4; 1; 1] ?
Show the answer
b = v1 + 2v2

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