Linear Transformations, Function Composition












1














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


















  • 1




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










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


















1














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


















  • 1




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










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
















1












1








1







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













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

588




588








  • 1




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










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
















  • 1




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










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










1




1




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




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












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






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












1 Answer
1






active

oldest

votes


















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 $gfne fg$. Perhaps the problem intended to say "Justify $0=gfne fg.$"






share|cite|improve this answer





















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












  • 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 $fgne0$.
    – palmpo
    Jan 5 at 0:07










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













Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3062130%2flinear-transformations-function-composition%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









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 $gfne fg$. Perhaps the problem intended to say "Justify $0=gfne fg.$"






share|cite|improve this answer





















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












  • 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 $fgne0$.
    – palmpo
    Jan 5 at 0:07










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


















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 $gfne fg$. Perhaps the problem intended to say "Justify $0=gfne fg.$"






share|cite|improve this answer





















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












  • 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 $fgne0$.
    – palmpo
    Jan 5 at 0:07










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
















1












1








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 $gfne fg$. Perhaps the problem intended to say "Justify $0=gfne fg.$"






share|cite|improve this answer












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

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

3861113




3861113












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












  • 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 $fgne0$.
    – palmpo
    Jan 5 at 0:07










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




















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












  • 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 $fgne0$.
    – palmpo
    Jan 5 at 0:07










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


















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






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














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 $fgne0$.
– palmpo
Jan 5 at 0:07




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 $fgne0$.
– palmpo
Jan 5 at 0:07












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






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




















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3062130%2flinear-transformations-function-composition%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

1300-talet

Image
Den här artikeln handlar om hundraårsperioden 1300–1399. För decenniet 1300–1309, se 1300-talet (decennium). .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output ...
Read more

1300-talet

Image
Den här artikeln handlar om hundraårsperioden 1300–1399. För decenniet 1300–1309, se 1300-talet (decennium). .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output ...
Read more

Display a custom attribute below product name in the front-end Magento 1.9.3.8

Image
0 I need to display a custom attribute bellow the product name and I don't know how can I call this attribute. I am editing the file /app/design/frontend/templatemela/MAG100219_1/template/catalog/product/view.phtml But how can I call this attribute in php? I tried calling a string echo with the name I put in the attribute, but it didn't work. Code is as below: <div class="page-title product-title"> <h1><?php echo $_helper->productAttribute($_product, $_product->getName(), 'name') ?></h1> </div> <div class="codigoinmetro"> <h2><?php echo $_helper->productAttribute($_product, $_product->getName(), 'registroinmetro') ?></h2> </div> php attrib...
Read more