In each of the following cases prove that the given function

In each of the following cases, prove that the given function is a homomorphism and describe its image

and kernel.

(b) The function f R x R R defined by f ((x,y)) x-3y.

Solution

We know that a homomorphism is a structure preserving map between two algebraic structures of the same type. Here, f: R x RR is defined by f ((x,y))= x-3y. Also, f ((x₁,y₁)+(x₂,y₂)) =f(x₁+x₂,y₁+y₂) = x₁+x₂ -3(y₁+y₂) = x₁-3y₁ +x₂-3y₂ = f(x₁,y₁)+f(x₂,y₂). Thus, the mapping f: R x RR preserves the structure of addition between R x R and R. Hence f is a homomorphism.

The kernel of f is the set of solutions to the equation f(x,y) = 0 bi.e. x -3y = 0 or, x = 3y. Then (x,y) = (3y,y) = y(3,1). Hence Ker(f) = span{(3,1)}.

Im (f) = { f(x,y) : x,y R} = {x-3y: x,y R}. Now, since every element of R can be expressed as x-3y for some, x,y R, hence Im (f) = R.