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

Step by step

The question

Question: find the null space of A.

A = [2, 1; 1, 1]

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 2 columns, so v has 2 entries: x and y.

We want every v that gives 0s.

[2, 1; 1, 1] × [x; y] = [0; 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.

= [2, 1; 1, 1]

We call the rows R 1 and R 2.

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; 2, 1]

→ [1, 1; 0, −1]

→ [1, 1; 0, 1]

→ [1, 0; 0, 1]

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.

Every letter has a leading 1.

No letter is free.

leading 1: x, y

Each row says one letter is 0.

So only v = 0: the null space is just the zero vector.

Its dimension is 0.

v = [0; 0]

4 Check

Let's check another way.

A is square, so only 0 goes to 0 exactly when its determinant is not 0.

D = |2, 1; 1, 1|

So D = 2 × 1 − 1 × 1.

D = 2 × 1 − 1 × 1

2 times 1 is 2.

1 times 1 is 1.

D = 2 − 1

2 minus 1 is 1.

So D = 1.

D = 1

D is 1, not 0.

So the answer is right!

Let's look back.

First: write A v = 0.

Then: row reduce.

Then: find the free letters.

Last: check.

Answer