# Column Space

1. Apr 4, 2009

### roam

1. The problem statement, all variables and given/known data

We have a matrix A which row-reduces to:

$$A = \left[\begin{array}{ccccc} 1&2&0&0\\ 0&0&1&0\\0&0&0&1 \end{array}\right]$$

I'm asked to find the column space of A.

2. Relevant equations

3. The attempt at a solution

I'm not sure what to write down for this question... Here's what I think:

col(A) = {x1(1,0,0)+x2(2,0,0)+x3((0,1,0)+x4(0,0,1), $$x \in R^4$$}

Since the 2nd column is simply a multiple of the first maybe it makes it a subspace of R3, I'm not sure...

2. Apr 5, 2009

### rochfor1

Almost. You look at the pivot rows in the reduced form of the matrix, but take the span of the corresponding rows in the original matrix.

3. Apr 5, 2009

### roam

Hi!

Thanks, now if I write that using the corresponding rows in the original matrix, it'll look like:

col(A) = {x1(1,2,1)+x2(2,4,2)+x3((1,1,1)+x4(1,0,2), $$x \in R^4$$}

But I know that the 2nd column is a multiple of the first column, how do I need to express it? I'm not sure what to do about this.

4. Apr 5, 2009

### rochfor1

That's the beauty of it---when you row reduce the matrix, you see that the second column is in the span of the pivot columns, so it doesn't count towards the column space. Just forget about it. In other words,

col(A)=span{(1,2,1),(1,1,1),(1,0,2)}

5. Apr 7, 2009

### HallsofIvy

Staff Emeritus
Yes, since the second column is a multiple of the first, you don't need it!
The column space is just the space spanned by {(1, 0, 0), (0, 1, 0), (0, 0, 1)} and is precisely R3, not a subspace of it.