# Homework Help: Kernel and image of a matrix A

1. Dec 25, 2007

### Niles

[SOLVED] Kernel and image of a matrix A

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

If I have a matrix A, then the kernel of A is the solution to Ax=0?

The image of A is just the vectors that span the column space?

I have looked through my book and searched the WWW, but I can't find the answer to these questions anywhere. I hope you guys can help.

2. Dec 25, 2007

### Defennder

Loosely interpreted, I think you got it right, but you should include the fact that A is the standard matrix associated with the linear transformation. This is a question on linear transformations, right?

Your first statement should read "The kernel of the linear transformation T, which is associated with standard matrix A, is the nullspace of A, which the the solution set of Ax=0".

Your second statement should read "The set of images of the linear transformation T, which is associated with the standard matrix A, also known as the range of T, is given by the column space of A"

I know it seems pedantic but sometimes confusion is caused when we use imprecise language in mathematics.

3. Dec 25, 2007

### mathboy

Suppose A is a mxn matrix. View A as a linear transformation that left-multiplies a column vector in R^n to a column vector in R^m. Since A:R^n->R^m is now a linear transformation (prove it!), then by definition ker(A) is the solution space to Ax=0.

Now im(A) is a subspace of R^m (prove it!). What set generates this subspace? Well take the n standard basis vectors of R^n. Then the image of these n vectors by the linear transformation A are simply the columns of the matrix A, right? So im(A) is the span of the columns of A (since the n standard basis vectors form a basis of R^n), i.e. the column space of A.

Furthermore, suppose rankA (the dimension of im(A)) = r. Then by the dimension theorem, we have the number of linearly independent solutions to Ax=0 to be n-r, which you've probably learned in high school but now see the proof of. Also, rankA, which is the dimension of im(A)=column space of A, is now the number of linearly independent columns of A (and also the number of linearly independent rows of A).

Last edited: Dec 25, 2007
4. Dec 25, 2007

### Niles

Thanks to you both - it's so great that you guys can help.

So it doesn't make sense talking about the image and kernel of a matrix alone (and not associating it with a linear transformation)?

5. Dec 25, 2007

### mathboy

That's right.

ker: Hom(V,W) -> V
im: Hom(V,W) -> W

where Hom(V,W) is the set of all linear transformations from a vector space V to a vector space W. So ker, and im do not act on matrices.