# Proof that this function is an isomorphism

If $$K = (k_1, k_2, k_3, k_4)$$ contains the basis of the linear function $$f: R^4 \rightarrow R^4$$ with $$f(k_1) = k_4 , f(k_2) = k_1 + 2k_2 , f(k_3) = 2k_1 + k_2 + k_3 , f(k_4) = 2k_2 - k_3$$

show that f is an Isomorphism.

So it got to be bijective to be an Isomorphism, i would start with $$Ker(f)= 0$$ to prove that it is injective and $$dim(Im(f))$$ would be then equal to $$dim(Im(R^4))$$ so that it would be also surjective. But im out of ideas on how to show $$Ker(f)=0$$

• Linear combinations ... Jun 25, 2020 at 15:42
• Hint: Any linear mapping on finite-dimensions (i.e. finite number of variables) can be represented by a matrix. Jun 25, 2020 at 15:52

We will solve $$f(c_1 k_1+c_2 k_2+c_3k_3+ c_4k_4)=c_1 f(k_1)+c_2 f(k_2)+c_3 f(k_3)+c_4 f(k_4)=c_1 k_4+c_2 (k_1+2k_2)+c_3 (2k_1+k_2+k_3)+c_4 (2k_2-k_3)=0$$.
Since $$(k_1,k_2,k_3,k_4)$$ forms a basis, they are linearly independent. Thus $$c_1=0, c_2+2c_3=0,2c_2+c_3+2c_4=0,c_3-c_4=0$$. Solving this system gives $$c_1=c_2=c_3=c_4=0$$, so $$\ker f=0.$$
• how did you get to this? $c_1=0,c_2+2c_3=0,2c_2+c_3+2c_4=0,c_3−c_4=0$ Jun 25, 2020 at 18:12