Linear Transformations, Function Composition












1












$begingroup$


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?






linear-algebra linear-transformations







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    $f(a,a)neq(0,0)$
    $endgroup$
    – SmileyCraft
    Jan 4 at 21:27










  • $begingroup$
    Also, $g(0,b)neq (0,0)$.
    $endgroup$
    – user3482749
    Jan 4 at 21:30


















1












$begingroup$


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?






linear-algebra linear-transformations







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    $f(a,a)neq(0,0)$
    $endgroup$
    – SmileyCraft
    Jan 4 at 21:27










  • $begingroup$
    Also, $g(0,b)neq (0,0)$.
    $endgroup$
    – user3482749
    Jan 4 at 21:30
















1












1








1





$begingroup$


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?






linear-algebra linear-transformations







share|cite|improve this question









$endgroup$




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?






linear-algebra linear-transformations




linear-algebra linear-transformations






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 4 at 21:25









JakcjonesJakcjones

828




828








  • 1




    $begingroup$
    $f(a,a)neq(0,0)$
    $endgroup$
    – SmileyCraft
    Jan 4 at 21:27










  • $begingroup$
    Also, $g(0,b)neq (0,0)$.
    $endgroup$
    – user3482749
    Jan 4 at 21:30
















  • 1




    $begingroup$
    $f(a,a)neq(0,0)$
    $endgroup$
    – SmileyCraft
    Jan 4 at 21:27










  • $begingroup$
    Also, $g(0,b)neq (0,0)$.
    $endgroup$
    – user3482749
    Jan 4 at 21:30










1




1




$begingroup$
$f(a,a)neq(0,0)$
$endgroup$
– SmileyCraft
Jan 4 at 21:27




$begingroup$
$f(a,a)neq(0,0)$
$endgroup$
– SmileyCraft
Jan 4 at 21:27












$begingroup$
Also, $g(0,b)neq (0,0)$.
$endgroup$
– user3482749
Jan 4 at 21:30






$begingroup$
Also, $g(0,b)neq (0,0)$.
$endgroup$
– user3482749
Jan 4 at 21:30












1 Answer
1






active

oldest

votes


















1












$begingroup$

$gf(a,b)=g(f(a,b))=g((0,b))=(0,0)$.
$fg(a,b)=f(g(a,b))=f((a,a))=(0,a)$.

So we have $gf=0$ but $gfne fg$. Perhaps the problem intended to say "Justify $0=gfne fg.$"






share|cite|improve this answer









$endgroup$













  • $begingroup$
    It says "Justify $0=f circ g neq g circ f$" and it's on the composite function section.
    $endgroup$
    – Jakcjones
    Jan 4 at 22:51












  • $begingroup$
    Well then that's simply incorrect. For example, take $(1,0)$, then $g(1,0)=(1,1)$ and $f(1,1)=(0,1)ne(0,0)$ so $fgne0$.
    $endgroup$
    – palmpo
    Jan 5 at 0:07










  • $begingroup$
    If you have $f(a,a)=(0,a)=(0,0)$, since $ a=0?$
    $endgroup$
    – Jakcjones
    Jan 5 at 14:01













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1 Answer
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active

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oldest

votes









1












$begingroup$

$gf(a,b)=g(f(a,b))=g((0,b))=(0,0)$.
$fg(a,b)=f(g(a,b))=f((a,a))=(0,a)$.

So we have $gf=0$ but $gfne fg$. Perhaps the problem intended to say "Justify $0=gfne fg.$"






share|cite|improve this answer









$endgroup$













  • $begingroup$
    It says "Justify $0=f circ g neq g circ f$" and it's on the composite function section.
    $endgroup$
    – Jakcjones
    Jan 4 at 22:51












  • $begingroup$
    Well then that's simply incorrect. For example, take $(1,0)$, then $g(1,0)=(1,1)$ and $f(1,1)=(0,1)ne(0,0)$ so $fgne0$.
    $endgroup$
    – palmpo
    Jan 5 at 0:07










  • $begingroup$
    If you have $f(a,a)=(0,a)=(0,0)$, since $ a=0?$
    $endgroup$
    – Jakcjones
    Jan 5 at 14:01


















1












$begingroup$

$gf(a,b)=g(f(a,b))=g((0,b))=(0,0)$.
$fg(a,b)=f(g(a,b))=f((a,a))=(0,a)$.

So we have $gf=0$ but $gfne fg$. Perhaps the problem intended to say "Justify $0=gfne fg.$"






share|cite|improve this answer









$endgroup$













  • $begingroup$
    It says "Justify $0=f circ g neq g circ f$" and it's on the composite function section.
    $endgroup$
    – Jakcjones
    Jan 4 at 22:51












  • $begingroup$
    Well then that's simply incorrect. For example, take $(1,0)$, then $g(1,0)=(1,1)$ and $f(1,1)=(0,1)ne(0,0)$ so $fgne0$.
    $endgroup$
    – palmpo
    Jan 5 at 0:07










  • $begingroup$
    If you have $f(a,a)=(0,a)=(0,0)$, since $ a=0?$
    $endgroup$
    – Jakcjones
    Jan 5 at 14:01
















1












1








1





$begingroup$

$gf(a,b)=g(f(a,b))=g((0,b))=(0,0)$.
$fg(a,b)=f(g(a,b))=f((a,a))=(0,a)$.

So we have $gf=0$ but $gfne fg$. Perhaps the problem intended to say "Justify $0=gfne fg.$"






share|cite|improve this answer









$endgroup$



$gf(a,b)=g(f(a,b))=g((0,b))=(0,0)$.
$fg(a,b)=f(g(a,b))=f((a,a))=(0,a)$.

So we have $gf=0$ but $gfne fg$. Perhaps the problem intended to say "Justify $0=gfne fg.$"







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 4 at 21:43









palmpopalmpo

4061213




4061213












  • $begingroup$
    It says "Justify $0=f circ g neq g circ f$" and it's on the composite function section.
    $endgroup$
    – Jakcjones
    Jan 4 at 22:51












  • $begingroup$
    Well then that's simply incorrect. For example, take $(1,0)$, then $g(1,0)=(1,1)$ and $f(1,1)=(0,1)ne(0,0)$ so $fgne0$.
    $endgroup$
    – palmpo
    Jan 5 at 0:07










  • $begingroup$
    If you have $f(a,a)=(0,a)=(0,0)$, since $ a=0?$
    $endgroup$
    – Jakcjones
    Jan 5 at 14:01




















  • $begingroup$
    It says "Justify $0=f circ g neq g circ f$" and it's on the composite function section.
    $endgroup$
    – Jakcjones
    Jan 4 at 22:51












  • $begingroup$
    Well then that's simply incorrect. For example, take $(1,0)$, then $g(1,0)=(1,1)$ and $f(1,1)=(0,1)ne(0,0)$ so $fgne0$.
    $endgroup$
    – palmpo
    Jan 5 at 0:07










  • $begingroup$
    If you have $f(a,a)=(0,a)=(0,0)$, since $ a=0?$
    $endgroup$
    – Jakcjones
    Jan 5 at 14:01


















$begingroup$
It says "Justify $0=f circ g neq g circ f$" and it's on the composite function section.
$endgroup$
– Jakcjones
Jan 4 at 22:51






$begingroup$
It says "Justify $0=f circ g neq g circ f$" and it's on the composite function section.
$endgroup$
– Jakcjones
Jan 4 at 22:51














$begingroup$
Well then that's simply incorrect. For example, take $(1,0)$, then $g(1,0)=(1,1)$ and $f(1,1)=(0,1)ne(0,0)$ so $fgne0$.
$endgroup$
– palmpo
Jan 5 at 0:07




$begingroup$
Well then that's simply incorrect. For example, take $(1,0)$, then $g(1,0)=(1,1)$ and $f(1,1)=(0,1)ne(0,0)$ so $fgne0$.
$endgroup$
– palmpo
Jan 5 at 0:07












$begingroup$
If you have $f(a,a)=(0,a)=(0,0)$, since $ a=0?$
$endgroup$
– Jakcjones
Jan 5 at 14:01






$begingroup$
If you have $f(a,a)=(0,a)=(0,0)$, since $ a=0?$
$endgroup$
– Jakcjones
Jan 5 at 14:01




















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