Null space of [1, 2; 3, 4] Step by Step

Step by step

The question

Question: find the null space of A.

A = [1, 2; 3, 4]

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.

[1, 2; 3, 4] × [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.

= [1, 2; 3, 4]

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, 2; 0, −2]

→ [1, 2; 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 = |1, 2; 3, 4|

So D = 1 × 4 − 3 × 2.

D = 1 × 4 − 3 × 2

1 times 4 is 4.

3 times 2 is 6.

D = 4 − 6

4 minus 6 is −2.

So D = −2.

D = −2

D is −2, 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