This definition is disturbing when R is noncommutative. (You can skip this part if you know you are working over the reals, the complex numbers, or other commutative fields.) We would like the function cx, scaling all vectors by c, to be linear. It certainly looks linear. And indeed c(x+y) = cx+cy, so we're ok there. But premultiplication by c does not respect scaling. In other words, c(ax) is not the same as (ac)x, because a and c do not commute.

The workaround is to multiply by c on the right. Now (ax+by)c = a(xc)+b(xc). Postmultiplication by c is indeed a linear map, or a left module homomorphism if you prefer.

This assumes the vectors can be scaled on either side. If you want postmultiplication to be a linear map, our vector space is more than a left R module, it is a bimodule, with R on either side. Of course the vector space could be a right R module, whence premultiplication by a constant implements a linear map. Ok, that's enough of noncommutative rings.

A simple transformation squashes 3 space down onto the xy plane. If a line rises up from the plane at an angle, it is pushed back down into the plane, and distance is compressed by a ratio that is a function of the angle of inclination. The entire z axis is squashed into the origin, and distance is scaled by 0.

Return to our example, f(x,y,z) = x,y,0. Recall that the z axis is the kernel of this squashing transformation. Let [0,0,1] be a basis for this kernel. Extend this basis by [1,1,1] and [-1,1,1] to cover all of 3 space. The image of these two basis elements is [1,1] and [-1,1], which is a basis for the xy plane. A 3 dimensional space maps onto a 2 dimensional space with a 1 dimensional kernel.

Assume two subspaces are combined to span a larger space. Think of two planes that combine to span 3 space, and intersect in a line. Let x be a basis for the intersection. Extend this by the vectors in y to build a basis for one of the subspaces, and by the vectors in z to build a basis for the other subspace. Now x∪y∪z spans the entire space. Suppose a linear combination of vectors from this union produces 0. Since x∪y is an independent set, and x∪z is an independent set, our linear combination must include vectors from both y and z. combine the z vectors together to get a new nonzero vector that is spanned by x∪y. However, if this vector, spanned by z, is contained in the space spanned by x and y, it is in the intersection, and should be spanned by x alone. Therefore x∪y∪z is an independent set, and is a basis for the entire space. The dimension of the space is the sum of the dimensions of the two subspaces, minus the dimension of their intersection.