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 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.
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.
3 Take those columns of A
We take column 1 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, 2).
Every column either has a leading 1 or is free: 1 + 1 = 2.
The rank is also the dimension of the column space: 1.
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.
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: take those columns of A.
Last: check.