Do you have definitions handy for image and kernel?
If so, we can see how we can apply them.
Let's start with the image.
In my book the image is the set of all vectors that can be "reached" by f.
Can for instance (1,0) be "reached"?
Or put otherwise, can you find an (x,y,z,w) in $\mathbb R^4$ that has (1,0) as its image?
And how about (0,1)?
If both can be reached, the image is $\mathbb R^2$, since (1,0) and (0,1) "span" $\mathbb R^2$.
The kernel is the set of vectors that the linear tranformation maps to the 0 vector. (Note that if the linear transformation is from vector space U to vector space V, then the kernel is a subspace of U and the image is a subspace of V)
Since f maps (x, y, z, w) to (2x+y+z+w,x+z-w) so the kernel must have 2x+ y+ z+ w= 0, x+ z- w= 0. If we add the two equations the "w"s cancel and 3x+ y+ 2z= 0 so that y= -3x-2z. Clearly w= x+ z. So (x, y, z, w)= (x, -3x- 2z, z, x+ z)- (x, -3x, 0, x)+ (0, -2z, z, z)= x(1, -3, 0, 1)+ z(0, -2, 1, 1) which tells you immediately what the dimension of the kernel and a basis for the kernel is.
Notice that this satifies the "rank nullity" theorem: if a linear transformation is from a vector space of dimension n to a vector space of dimension m, the rank (dimension of the image) and nullity (dimension of the kernel) add to n.
Here, that is 2+ 2= 4.
Given an arbitrary linear mapping between two vector spaces of finite dimension, my first impulse is to choose bases and create a matrix for the linear map. My choice of bases is nothing special, I pick:
It should be clear immediately that this matrix has rank 2 (the two rows are linearly independent), so the image of f is of dimension 2, and the ONLY 2-dimensional subspace of \(\displaystyle \Bbb R^2\) is, of course, \(\displaystyle \Bbb R^2\) itself.
This settles the image question (f is surjective), but only tells us (via the rank-nullity theorem) that the kernel has dimension 2. So we actually need to FIND a basis for the kernel (or: equivalently, the null space of our matrix above). How do we do that?
We need to solve the following HOMOGENEOUS system of equations:
2x + y + z + w = 0
x + z - w = 0
Putting our 2x4 matrix into rref form, we get the matrix: