# How to Find the Basis of an Image

1. Jul 29, 2015

### 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!

2. Jul 29, 2015

### 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

3. Jul 29, 2015

### jasonRF

4. Jul 30, 2015

### 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

5. Jul 30, 2015

### 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.

6. Jul 31, 2015

### HallsofIvy

Staff Emeritus
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.

7. Jul 31, 2015

### Abtinnn

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?