How to Find the Basis of an Image

[Abtinnn]

I've been reading a book on linear algebra. It talks about finding the the basis of kernel and image of a linear transformation. I understand how to find the basis of the kernel, but I don't understand how to find the basis of the image. Could someone please explain a method of doing it? Thank you!

 

[jasonRF]

Think about the definition of the image of a transformation $A$: it is the set of all vectors that can be written $A x$. If you have a matrix representation for $A$, then any vector $Ax$ must be a linear combination of the columns of $A$. Does that help?

jason

 

[jasonRF]

I am thinking my answer wasn't very good. This is a standard thing to do for matrices (finding basis of column space): check out pages 59-62

http://www.math.brown.edu/~treil/papers/LADW/book.pdf

 

[Abtinnn]

Oh!
I get it now!
You're explanation actually cleared things up. And the book helped as well.

Thanks a lot for your help! :D

 

[mathwonk]

just reduce the columns to an independent spanning set. One way to do that is do gauss elimination via row operations until you hve echelon form. Then look where the "pivot" columns are located. Go back to the original matrix and choose the columns in those same positions and you will have a basis.

 

[HallsofIvy]

To take an easy example, suppose we have a linear transformation on R² that maps (x, y) to (4x+ 2y, 2x+ y). The null space consists of all vectors that are mapped to the 0 vector: 4x+ 2y= 0 and 2x+ y= 0. In fact, 4x+ 2y= 2(2x+ y) so those are the same equation which is equivalent to y= -2x. Rather than getting the single solution, (0, 0), any vector of the form (x, y)= (x, -2x) is in the null space. The set {(1, -2)} is a basis for the one dimensional null space.

Since this operator is maps R² to itself, and the null space has dimension 1, the image has dimension 2- 1= 1 also. But where the null space is a subspace of the domain, the image is a subspace of the range. Here, both domain and range are R² but it is useful to make this distinction. While we can write the null space in terms of "x" and "y", to look at the image we need to write u= 4x+ 2y, v= 2x+ y and write the basis of the image in terms of u and v. Here, u= 4x+ 2y= 2(2x+ y)= 2v. A vector in the image is of the form (u, v)= (2v, v)= v(2, 1). {(2, 1)} is a basis for the image.

A slightly harder example: A linear transformation maps (x, y, z) to (x+ y+ z, y+ z, x). Now, if (x, y, z) is in the null space, we have x+ y+ z= 0, y+ z= 0, x= 0. With x= 0, both the first two equations reduce to y+ z= 0 or z= -y. Any vector in the null space is of the form (x, y, z)= (0, y, -y)= y(0, 1, -1) so {(0, 1, -1)} is a basis for the one dimensional null space.

To find a basis for the image, write u= x+ y+ z, v= y+ z, w= x. From the last equation, we an write u= w+ y+ z or u- w= y+ z= v. We have the single equation u- w= v or u= v+ w that must be satisfied for all (u, v, w) in the image. That is, any vector in the image can be written in the form (u, v, w)= (v+ w, v, w)= (v, v, 0)+ (w, 0, w)= v(1, 1, 0)+ w(1, 0, 1). {(1, 1, 0), (1, 0, 1)} is a basis for the image which is two dimensional.

 

[Abtinnn]

To take an easy example, suppose we have a linear transformation on R² that maps (x, y) to (4x+ 2y, 2x+ y). The null space consists of all vectors that are mapped to the 0 vector: 4x+ 2y= 0 and 2x+ y= 0. In fact, 4x+ 2y= 2(2x+ y) so those are the same equation which is equivalent to y= -2x. Rather than getting the single solution, (0, 0), any vector of the form (x, y)= (x, -2x) is in the null space. The set {(1, -2)} is a basis for the one dimensional null space.

Since this operator is maps R² to itself, and the null space has dimension 1, the image has dimension 2- 1= 1 also. But where the null space is a subspace of the domain, the image is a subspace of the range. Here, both domain and range are R² but it is useful to make this distinction. While we can write the null space in terms of "x" and "y", to look at the image we need to write u= 4x+ 2y, v= 2x+ y and write the basis of the image in terms of u and v. Here, u= 4x+ 2y= 2(2x+ y)= 2v. A vector in the image is of the form (u, v)= (2v, v)= v(2, 1). {(2, 1)} is a basis for the image.

A slightly harder example: A linear transformation maps (x, y, z) to (x+ y+ z, y+ z, x). Now, if (x, y, z) is in the null space, we have x+ y+ z= 0, y+ z= 0, x= 0. With x= 0, both the first two equations reduce to y+ z= 0 or z= -y. Any vector in the null space is of the form (x, y, z)= (0, y, -y)= y(0, 1, -1) so {(0, 1, -1)} is a basis for the one dimensional null space.

To find a basis for the image, write u= x+ y+ z, v= y+ z, w= x. From the last equation, we an write u= w+ y+ z or u- w= y+ z= v. We have the single equation u- w= v or u= v+ w that must be satisfied for all (u, v, w) in the image. That is, any vector in the image can be written in the form (u, v, w)= (v+ w, v, w)= (v, v, 0)+ (w, 0, w)= v(1, 1, 0)+ w(1, 0, 1). {(1, 1, 0), (1, 0, 1)} is a basis for the image which is two dimensional.

Thanks a lot! Is this method the same as the methods mentioned by the other members? I fail to see a similarity. If this is a different method, how many ways of finding an the basis of an image are there in total?