Null space of [1, 2, 3] Step by Step

Step by step

The question

Question: find the null space of A.

A = [1, 2, 3]

The null space of A is every vector v that A sends to 0: A times v is the zero vector.

null space of A: every v with A v = 0

1 Write A v = 0

A has 3 columns, so v has 3 entries: x, y and z.

We want every v that gives 0s.

[1, 2, 3] × [x; y; z] = [0]

2 Row reduce

As an augmented matrix the part after the bar is all 0s, and every move keeps it 0.

So we leave it out and row reduce A itself.

= [1, 2, 3]

We call the rows R 1.

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

The matrix already has its leading 1s, with 0s above and below.

No moves are needed.

3 Find the free letters

The first number that is not 0 in each row is a 1.

We call it a leading 1, and ring it.

The letter x has a leading 1.

The letters y and z are free: any number.

leading 1: x

free: y, z

4 Write the leading letters with the free ones

Row 1 as an equation: x + 2y + 3z = 0.

x + 2y + 3z = 0

We want x alone.

To remove + 2y and + 3z, we take away 2y and take away 3z on both sides.

x ~~+ 2y + 3z − 2y − 3z~~ = −2y − 3z

They cancel on the left.

So x = −2y − 3z.

x = −2y − 3z

5 Write it as vectors

We stack the letters in a column.

The free letters stay as they are.

[x; y; z] = [−2y − 3z; y; z]

We split it up: what goes with y makes one vector, (−2, 1, 0).

What goes with z makes another, (−3, 0, 1).

[x; y; z] = y[−2; 1; 0] + z[−3; 0; 1], y and z any numbers

Careful!

In the vector of y, the y entry itself is 1.

Do not leave it out.

Careful: the free letter's own entry is 1

6 Write the basis

Every vector of the null space is made from these vectors, and they are independent.

So they are a basis: (−2, 1, 0) and (−3, 0, 1).

= basis: [−2; 1; 0], [−3; 0; 1]

The dimension of the null space is how many basis vectors: 2, one for each free letter.

dimension = 2 (one for each free letter)

7 Check

Let's check: A times each basis vector must be 0.

= [1, 2, 3] × [−2; 1; 0]

= [1 × (−2) + 2 × 1 + 3 × 0]

We work out every times.

= [−2 + 2 + 0]

We get 0s: (−2, 1, 0) is in the null space.

= [0]

Now the next one.

= [1, 2, 3] × [−3; 0; 1]

= [1 × (−3) + 2 × 0 + 3 × 1]

We work out every times.

= [−3 + 0 + 3]

We get 0s: (−3, 0, 1) is in the null space.

= [0]

A sends every basis vector to 0.

So the answer is right!

Let's look back.

First: write A v = 0.

Then: row reduce.

Then: find the free letters.

Then: write the leading letters with the free ones.

Then: write it as vectors.

Then: write the basis.

Last: check.

Answer