Rank of [1, 1, 0, 0; 0, 1, 1, 0; 0, 0, 1, 1] Step by Step

Step by step

The question

Question: find the rank of A.

A = [1, 1, 0, 0; 0, 1, 1, 0; 0, 0, 1, 1]

The rank of A is how many leading 1s it has, after row reducing.

It tells how many columns are really new.

rank: how many leading 1s after row reducing

1 Row reduce

We row reduce A.

= [1, 1, 0, 0; 0, 1, 1, 0; 0, 0, 1, 1]

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

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, 0, −1, 0; 0, 1, 1, 0; 0, 0, 1, 1]

→ [1, 0, 0, 1; 0, 1, 0, −1; 0, 0, 1, 1]

2 Ring the leading 1s

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

We call it a leading 1, and ring it.

The leading 1s are in columns 1, 2 and 3.

Column 4 has none.

leading 1s in columns 1, 2 and 3

3 Count them

3 leading 1s, so the rank of A is 3.

rank A = 3

Every column either has a leading 1 or is free: 3 + 1 = 4.

The rank is also the dimension of the column space: 3.

rank + free columns = columns: 3 + 1 = 4

Careful!

Count the leading 1s, not the rows.

A has 3 rows, and here every row has one.

Careful: count the leading 1s, not the rows

4 Check

Let's check that column 4 adds nothing new.

In the reduced matrix it holds 1, −1 and 1.

So column 4 should be 1 times column 1, −1 times column 2 and 1 times column 3.

= [1; 0; 0] − [1; 1; 0] + [0; 1; 1]

= [1 − 1 + 0; 0 − 1 + 1; 0 − 0 + 1]

= [0; 0; 1]

We get (0, 0, 1): that is column 4 of A.

It is made from the leading columns, so it adds nothing new.

The answer is right!

Let's look back.

First: row reduce.

Then: ring the leading 1s.

Then: count them.

Last: check.

Answer