sure it does. suppose x + y + z + 0w = 0.

well surely you can see that "w" can be anything. also, we can pick two of {x,y,z} to be anything we please, but then the third choice is forced upon us: if we specify y and z, then x = -y - z.

so the kernel of f is the matrices (in the standard basis) of the form:

.

let's pick three arbitrary numbers for y,z and w: how about 7,32, and -5? so we're asserting that the matrix:

is in the kernel of f.

let's write M in the standard basis for R^{2x2}:

.

so M = -39E_{1}+ 7E_{2}+ 32E_{3}- 5E_{4}.

in the standard basis, this is the (column) vector (-39,7,32,-5)^{T}. so the standard basis representation of f(M) is:

(and 1x1 matrices are just scalars).

it should be obvious that the kernel of f has dimension 3 (because its image has dimension 1), here is one basis:

.

(the "trick" here is that 2x2 real MATRICES can be thought of as "4-vectors" (elements of R^{4}), just line up the columns "end-to-end" to make one long column).