Linear Transformations, Function Composition

Question

Considering the transformations

$\ f:\Re^{2} \rightarrow \Re^{2}$ and $ g:\Re^{2} \rightarrow \Re^{2}$ defined by

$f(a,b)= (0,b)$, for any $(a,b) \in \Re^{2}$

and

$g(a,b)=(a,a)$, for any $(a,b) \in \Re^{2}$

Justify that $0$ = $f \circ g$ $\neq$ $g \circ f$

$0$ is the null transformation of $\Re^{2}$

My Resolution:

First I worked out $(f \circ g)$:

$(f \circ g)(a,b)$ = $f(g(a,b))$ = $f(a,a) = (0,0)$

Then I worked out $(g \circ f)(a,b)=g(f(a,b))=g(0,b)=(0,0)???$

What am I dong wrong?

Answer 1

$gf(a,b)=g(f(a,b))=g((0,b))=(0,0)$.
$fg(a,b)=f(g(a,b))=f((a,a))=(0,a)$.
So we have $gf=0$ but $gf\ne fg$. Perhaps the problem intended to say "Justify $0=gf\ne fg.$"