This question is regarding the answer provided to exercise 39, section 2.2 of Friedbergs Linear Algebra Book (ISBN: 0130084514).
The exercise reads: "Prove that there is an additive function, $T:R \rightarrow R$ that is not a linear transformation."
He defines a function $f:β \rightarrow R$ where $β$ is a basis for $R \ over \ Q$ such that for some $x,y \in β, f(x) = y \ and \ f(y) = x \ and \ f(z)=z , \ otherwise$
Then he applies the theorem which states that there exists a unique linear transformation, $T:V \rightarrow V$ such that $T(u) = f(u), for \ any \ u \in β$
Finally he states that $T$ is additive but $T(c*x) \neq c*T(x), \ for \ c=y/x$
My problem is that if $c \in Q$ and $T(c*x) \neq c*T(x)$ then $T$ is not a linear transformation but we have defined it as one by the theorem.
If $c \notin Q$ and $T(c*x) \neq c*T(x)$ then this does not prove that T is not a linear transformation since $R$ was defined over $Q$, $R$ being the domain of $T$
So what am I missing?
Thanks in advance!!
Best Answer
What you're missing is that $\Bbb R$ is a vector space in two ways:
It's a 1-dimensional vector space $V$ over the field $\Bbb R$.
It's an infinite-dimensional vector space $W$ over the field $\Bbb Q$.
Your function is linear as a mapping from the second vector space ($W$) to itself, but NOT as a mapping from the first vector space ($V$) to itself.