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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 $Ker(f)=0$

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  • $\begingroup$ Linear combinations ... $\endgroup$
    – Divide1918
    Jun 25, 2020 at 15:42
  • $\begingroup$ Hint: Any linear mapping on finite-dimensions (i.e. finite number of variables) can be represented by a matrix. $\endgroup$
    – Chrystomath
    Jun 25, 2020 at 15:52

1 Answer 1

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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 $\ker f=0.$

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  • $\begingroup$ how did you get to this? $c_1=0,c_2+2c_3=0,2c_2+c_3+2c_4=0,c_3−c_4=0$ $\endgroup$
    – Payvand
    Jun 25, 2020 at 18:12
  • $\begingroup$ By definition of linearly indep $\endgroup$
    – Divide1918
    Jun 25, 2020 at 18:14

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