# Rank of linear mapping

1. Jan 18, 2012

### aaaa202

How come the rank of a matrix is equal to the amount of pivot points in the reduced row echelon form? My book denotes this a trivial point, but unfortunately I don't see it :(

2. Jan 18, 2012

### micromass

Staff Emeritus
How did you define "rank" in the first place?

3. Jan 18, 2012

### aaaa202

I don't what it is called in english but it's the dimension of the space that the linear function maps a vector onto.

4. Jan 18, 2012

### micromass

Staff Emeritus
OK, so it's the dimension of the range. Good.

Next step: Take a matrix A. If we put it in reduced echelon form, then we obtain a matrix B. Do you see why both matrices have the same rank??

In general: if $A=EBE^{-1}$ for some invertible matrix E, do you see why A and B have the same rank?

5. Jan 18, 2012

### aaaa202

Okay yes, I should have been able to figure that out myself. But then suppose you have row reduced matrix like the one on the attached picture. As a basis for the range you choose the vectors equal to columns with pivot points -i.e. column 1,2,3. However - wouldn't it be just as good to choose 1,2 and 4? Since that'd also make a 3 pivot points.
And lastly: Would it then also work to choose any other combination of 3 vectors out of the 4?

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6. Jan 18, 2012

### micromass

Staff Emeritus
Am I correct in saying that your last row is a zero row?? In that case, that doesn't count as a pivot point.

7. Jan 18, 2012

### aaaa202

Yes exactly. We have 3 pivots

8. Jan 18, 2012

### aaaa202

hmm I still don't get it tbh. Consider the matrix on the attached picture. What would a basis for the range then be? If you use the rule that the basis vectors equals the column with pivot points you'd get that (1,0) is a basis for the range. But how is (1,0) a basis for the solutions to equation x1 + 2x2 = a ?

9. Jan 18, 2012

### micromass

Staff Emeritus
Can you give me a specific matrix?? I don't really understand your picture.

Do notice that the range is being generated by the columns of the matrix.

10. Jan 18, 2012

### aaaa202

oops the reason you didn't understand the picture was that i forgot to attach it: here

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11. Jan 18, 2012

### micromass

Staff Emeritus
In this case, the range will be all vectors of the form (a,0). So the rank will be 1.

Note that the range is generated by the column vectors: So the range = span (1,0) , (2,0) in this case.

12. Jan 18, 2012

### aaaa202

But isn't the range the solutions to the equation:
x1 + 2x2 = a
Why does x2 have to be 0?

13. Jan 18, 2012

### micromass

Staff Emeritus
No, not at all. Given a matrix A, to find the range: you put up the equation Ax=y. All the possibilities for y constitute the range. That is: y is in the range if there is an x such that Ax=y.

14. Jan 18, 2012

### aaaa202

Ahh okay, it'd seem I didn't understand the basic definition of the range. But other than that I think I get it now.
Except! My real, deeper problem is perhaps that I don't understand why that, when you have a matrix like the one on the attached picture. How can you then be sure, that the columns with no pivot points can be written as a linear combination of the ones who do have pivot points? In general if you have n columns and n-a of them have pivots, how can you then know, that the a of them with no pivot points can be expressed as linear combinations of the n-a columns with pivots?

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15. Jan 19, 2012

### HallsofIvy

Staff Emeritus
If A is a linear transformation from vector space U to vector space V, then the range of A is a subspace of V, not U. It is the set of all vectors, y, in V, such that y= Ax for some x in A.