Step by step
The question
Question: find a basis for the column space of A.
The column space of A is the span of its columns: every linear combination of them.
We want a basis: as few columns as possible that still make all of it.
1 Row reduce
We row reduce A.
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.
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 2.
Columns 3 and 4 have none.
3 Take those columns of A
We take columns 1 and 2 of the first matrix, A itself.
Careful!
The columns come from A itself, not from the reduced matrix.
Row moves change the columns, but not which ones are new.
So a basis for the column space is (1, 0, 1) and (0, 1, 1).
Every column either has a leading 1 or is free: 2 + 2 = 4.
The rank is also the dimension of the column space: 2.
4 Check
Let's check that column 3 adds nothing new.
In the reduced matrix it holds 1 and 1.
So column 3 should be 1 times column 1 and 1 times column 2.
We get (1, 1, 2): that is column 3 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: take those columns of A.
Last: check.