Rational scalar multiplication in Linear transformation using addition property [duplicate] Linear transformation: Map T:V→W is said to be Linear transformation when it satisfy property: 1) L(u+v)=L(u)+L(v)∀u,v∈V 2) L(cv)=cL(v)∀c∈F I know when c∈N then property 2 can be obtained from property 1, since L(nv)=L(v)+L(v)…..+L(v) (n times where n∈N) Now I have read in somewhere that if our scalar c is rational then property 2 also follows from property 1, but I am unable to prove this. How can I show that if c∈Q, when (2) follows from (1)? I know that if our scalar is irrational, then we cannot use property 1 to prove property 2.

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Rational scalar multiplication in Linear transformation using addition property [duplicate]  Linear transformation: Map T:V→W is said to be Linear transformation when it satisfy property:  1) L(u+v)=L(u)+L(v)∀u,v∈V  2) L(cv)=cL(v)∀c∈F  I know when c∈N then property 2 can be obtained from property 1, since  L(nv)=L(v)+L(v)…..+L(v) (n times where n∈N)  Now I have read in somewhere that if our scalar c is rational then property 2 also follows from property 1, but I am unable to prove this. How can I show that if c∈Q, when (2) follows from (1)?   I know that if our scalar is irrational, then we cannot use property 1 to prove property 2.

Ijezid8t · Accepted Answer

Suppose our scalar c is rational; that is c=pq for some integers p and q. Then, if L satisfies property (1),  L(cv)=L(pqv)=pL(1qv)  by your previous result for integer scalars. On the other hand,  L(v)=L(q⋅1qv)=qL(1qv)  so, multiplying both sides by 1q,  L(1qv)=1qL(v)  returning to our first equation, this gives us that  L(pqv)=pqL(v)  so property (2) follows from property (1).

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