Rank of linear mapping

[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 :(

 

[micromass]

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

 

[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.

 

[micromass]

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.

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?

 

[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?

 

[micromass]

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?

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

 

[aaaa202]

Yes exactly. We have 3 pivots

 

[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 ?

 

[micromass]

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.

 

[aaaa202]

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

 

[micromass]

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.

 

[aaaa202]

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

 

[micromass]

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.

 

[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?

 

[HallsofIvy]

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.