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Rank and column space

Linear Algebra · 21 problems, easy to harder

▶ Watch the first one solved

How we do it

  1. Row reduce
  2. Ring the leading 1s
  3. Count them
  4. Check

The steps for Rank of [1, 2; 2, 4]. Every solution names its steps, so you always know what comes next.

Worked examples

Rank of [1, 2; 2, 4] Easy

The question

Question: find the rank of A.

A = [1, 2; 2, 4]

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, 2; 2, 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, 0]

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 column 1.

Column 2 has none.

leading 1s in column 1

3 Count them

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

rank A = 1

Every column either has a leading 1 or is free: 1 + 1 = 2.

The rank is also the dimension of the column space: 1.

rank + free columns = columns: 1 + 1 = 2

Careful!

Count the leading 1s, not the rows.

A has 2 rows, but only 1 of them have a leading 1.

Careful: count the leading 1s, not the rows

4 Check

Let's check that column 2 adds nothing new.

In the reduced matrix it holds 2.

So column 2 should be 2 times column 1.

= 2 × [1; 2]
= [2 × 1; 2 × 2]
= [2; 4]
= [2; 4]

We get (2, 4): that is column 2 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
Careful: count the leading 1s, not the rows

▶ Watch it solved

Rank of [1, 2, 1; 2, 4, 3; 3, 6, 4] Medium

The question

Question: find the rank of A.

A = [1, 2, 1; 2, 4, 3; 3, 6, 4]

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, 2, 1; 2, 4, 3; 3, 6, 4]

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, 2, 1; 0, 0, 1; 0, 0, 1]
→ [1, 2, 0; 0, 0, 1; 0, 0, 0]

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 and 3.

Column 2 has none.

leading 1s in columns 1 and 3

3 Count them

2 leading 1s, so the rank of A is 2.

rank A = 2

Every column either has a leading 1 or is free: 2 + 1 = 3.

The rank is also the dimension of the column space: 2.

rank + free columns = columns: 2 + 1 = 3

Careful!

Count the leading 1s, not the rows.

A has 3 rows, but only 2 of them have a leading 1.

Careful: count the leading 1s, not the rows

4 Check

Let's check that column 2 adds nothing new.

In the reduced matrix it holds 2 and 0.

So column 2 should be 2 times column 1.

= 2 × [1; 2; 3]
= [2 × 1; 2 × 2; 2 × 3]
= [2; 4; 6]
= [2; 4; 6]

We get (2, 4, 6): that is column 2 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
Careful: count the leading 1s, not the rows

▶ Watch it solved

Rank of [1, 2, 3; 2, 4, 6; 3, 6, 9] Harder

The question

Question: find the rank of A.

A = [1, 2, 3; 2, 4, 6; 3, 6, 9]

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, 2, 3; 2, 4, 6; 3, 6, 9]

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, 2, 3; 0, 0, 0; 0, 0, 0]

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 column 1.

Columns 2 and 3 have none.

leading 1s in column 1

3 Count them

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

rank A = 1

Every column either has a leading 1 or is free: 1 + 2 = 3.

The rank is also the dimension of the column space: 1.

rank + free columns = columns: 1 + 2 = 3

Careful!

Count the leading 1s, not the rows.

A has 3 rows, but only 1 of them have a leading 1.

Careful: count the leading 1s, not the rows

4 Check

Let's check that column 2 adds nothing new.

In the reduced matrix it holds 2.

So column 2 should be 2 times column 1.

= 2 × [1; 2; 3]
= [2 × 1; 2 × 2; 2 × 3]
= [2; 4; 6]
= [2; 4; 6]

We get (2, 4, 6): that is column 2 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
Careful: count the leading 1s, not the rows

▶ Watch it solved

Practice

Try each one, then open the answer. Stuck? Open the problem to watch every step.

Easy

  1. rank of [1, 2; 2, 4]
    Answer
    Careful: count the leading 1s, not the rows
  2. rank of [1, 2; 3, 4]
    Answer
    Careful: count the leading 1s, not the rows
  3. rank of [1, 2, 3; 2, 4, 6]
    Answer
    Careful: count the leading 1s, not the rows
  4. rank of [1, 0, 2; 0, 1, 3]
    Answer
    Careful: count the leading 1s, not the rows
  5. basis for the column space of [1, 2; 2, 4]
    Answer
    rank + free columns = columns: 1 + 1 = 2
  6. basis for the column space of [1, 2, 1; 2, 4, 3]
    Answer
    rank + free columns = columns: 2 + 1 = 3
  7. rank of [1, -1; -2, 2]
    Answer
    Careful: count the leading 1s, not the rows

Medium

  1. rank of [1, 2, 1; 2, 4, 3; 3, 6, 4]
    Answer
    Careful: count the leading 1s, not the rows
  2. rank of [1, 1, 1; 1, 2, 3; 2, 3, 4]
    Answer
    Careful: count the leading 1s, not the rows
  3. rank of [1, 0, 1; 0, 1, 1; 1, 1, 3]
    Answer
    Careful: count the leading 1s, not the rows
  4. basis for the column space of [1, 1, 2; 1, 2, 3; 2, 3, 5]
    Answer
    rank + free columns = columns: 2 + 1 = 3
  5. basis for the column space of [1, 2, 0; 0, 0, 1; 1, 2, 1]
    Answer
    rank + free columns = columns: 2 + 1 = 3
  6. rank of [2, 4; 1, 2; 3, 6]
    Answer
    Careful: count the leading 1s, not the rows
  7. rank of [1, 3; 2, 6; -1, -3]
    Answer
    Careful: count the leading 1s, not the rows
  8. basis for the column space of [1, 2, 3; 0, 1, 1; 1, 3, 4]
    Answer
    rank + free columns = columns: 2 + 1 = 3
  9. rank of [2, 1; 1, 1]
    Answer
    Careful: count the leading 1s, not the rows

Harder

  1. rank of [1, 2, 3; 2, 4, 6; 3, 6, 9]
    Answer
    Careful: count the leading 1s, not the rows
  2. rank of [1, 2, 0, 1; 2, 4, 1, 4; 1, 2, 1, 3]
    Answer
    Careful: count the leading 1s, not the rows
  3. basis for the column space of [1, 0, 1, 2; 0, 1, 1, 1; 1, 1, 2, 3]
    Answer
    rank + free columns = columns: 2 + 2 = 4
  4. rank of [1, 1, 0, 0; 0, 1, 1, 0; 0, 0, 1, 1]
    Answer
    Careful: count the leading 1s, not the rows
  5. basis for the column space of [1, 3; 2, 6]
    Answer
    rank + free columns = columns: 1 + 1 = 2