Composition of linear transformation and domain and codomain












1












$begingroup$


I am trying to do a problem regarding composition of linear transformation from my book. I want to know if my reasoning is correct.



$$T_1(x,y,z)=(2x+3y-z,x-y+z,2x+9z,3y-10z)$$



$$T_2(x_1,x_2,x_3,x_4)=(9x_1 - 3x_4,2x_2 + 4x_3 - 2x_4,2x_3 + 2x_4)$$



Q1. State domain and codomain of $T_2circ T_1$ and $T_1 circ T_2$.



Since for $T_1$ we are going from $mathbb R^3to mathbb R^4$ and for $T_2$ from $mathbb R^4to mathbb R^3$
,domain of $T_2 circ T_1$ will be $mathbb R^3$ and Codomain will also be $mathbb R^3$



Domain of $T_1 circ T_2$ will be $mathbb R^4$ and codomain will also be $mathbb R^4$.



Q2 Why are $T_2 circ T_1$ and $T_1 circ T_2$ well defined?



$T_2 circ T_1$ is well defined because the range of $T_1$ is $mathbb R^4$ and $mathbb R^4$ lies in the domain of $T_2$.



Similarly for $T_1 circ T_2$, the range of $T_2$ is $mathbb R^3$ and lies in the domain of $T_1$ which is $mathbb R^3$.



Is my reasoning correct? Any help will be appreciated.






linear-algebra







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    You could improve the readability of your question by using MathJax to type the equations.
    $endgroup$
    – Matti P.
    Jan 30 at 10:36
















1












$begingroup$


I am trying to do a problem regarding composition of linear transformation from my book. I want to know if my reasoning is correct.



$$T_1(x,y,z)=(2x+3y-z,x-y+z,2x+9z,3y-10z)$$



$$T_2(x_1,x_2,x_3,x_4)=(9x_1 - 3x_4,2x_2 + 4x_3 - 2x_4,2x_3 + 2x_4)$$



Q1. State domain and codomain of $T_2circ T_1$ and $T_1 circ T_2$.



Since for $T_1$ we are going from $mathbb R^3to mathbb R^4$ and for $T_2$ from $mathbb R^4to mathbb R^3$
,domain of $T_2 circ T_1$ will be $mathbb R^3$ and Codomain will also be $mathbb R^3$



Domain of $T_1 circ T_2$ will be $mathbb R^4$ and codomain will also be $mathbb R^4$.



Q2 Why are $T_2 circ T_1$ and $T_1 circ T_2$ well defined?



$T_2 circ T_1$ is well defined because the range of $T_1$ is $mathbb R^4$ and $mathbb R^4$ lies in the domain of $T_2$.



Similarly for $T_1 circ T_2$, the range of $T_2$ is $mathbb R^3$ and lies in the domain of $T_1$ which is $mathbb R^3$.



Is my reasoning correct? Any help will be appreciated.






linear-algebra







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    You could improve the readability of your question by using MathJax to type the equations.
    $endgroup$
    – Matti P.
    Jan 30 at 10:36














1












1








1





$begingroup$


I am trying to do a problem regarding composition of linear transformation from my book. I want to know if my reasoning is correct.



$$T_1(x,y,z)=(2x+3y-z,x-y+z,2x+9z,3y-10z)$$



$$T_2(x_1,x_2,x_3,x_4)=(9x_1 - 3x_4,2x_2 + 4x_3 - 2x_4,2x_3 + 2x_4)$$



Q1. State domain and codomain of $T_2circ T_1$ and $T_1 circ T_2$.



Since for $T_1$ we are going from $mathbb R^3to mathbb R^4$ and for $T_2$ from $mathbb R^4to mathbb R^3$
,domain of $T_2 circ T_1$ will be $mathbb R^3$ and Codomain will also be $mathbb R^3$



Domain of $T_1 circ T_2$ will be $mathbb R^4$ and codomain will also be $mathbb R^4$.



Q2 Why are $T_2 circ T_1$ and $T_1 circ T_2$ well defined?



$T_2 circ T_1$ is well defined because the range of $T_1$ is $mathbb R^4$ and $mathbb R^4$ lies in the domain of $T_2$.



Similarly for $T_1 circ T_2$, the range of $T_2$ is $mathbb R^3$ and lies in the domain of $T_1$ which is $mathbb R^3$.



Is my reasoning correct? Any help will be appreciated.






linear-algebra







share|cite|improve this question











$endgroup$




I am trying to do a problem regarding composition of linear transformation from my book. I want to know if my reasoning is correct.



$$T_1(x,y,z)=(2x+3y-z,x-y+z,2x+9z,3y-10z)$$



$$T_2(x_1,x_2,x_3,x_4)=(9x_1 - 3x_4,2x_2 + 4x_3 - 2x_4,2x_3 + 2x_4)$$



Q1. State domain and codomain of $T_2circ T_1$ and $T_1 circ T_2$.



Since for $T_1$ we are going from $mathbb R^3to mathbb R^4$ and for $T_2$ from $mathbb R^4to mathbb R^3$
,domain of $T_2 circ T_1$ will be $mathbb R^3$ and Codomain will also be $mathbb R^3$



Domain of $T_1 circ T_2$ will be $mathbb R^4$ and codomain will also be $mathbb R^4$.



Q2 Why are $T_2 circ T_1$ and $T_1 circ T_2$ well defined?



$T_2 circ T_1$ is well defined because the range of $T_1$ is $mathbb R^4$ and $mathbb R^4$ lies in the domain of $T_2$.



Similarly for $T_1 circ T_2$, the range of $T_2$ is $mathbb R^3$ and lies in the domain of $T_1$ which is $mathbb R^3$.



Is my reasoning correct? Any help will be appreciated.






linear-algebra




linear-algebra






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 30 at 10:45









kelvin hong 方

83919




83919










asked Jan 30 at 10:34









SchniederSchnieder

62




62








  • 1




    $begingroup$
    You could improve the readability of your question by using MathJax to type the equations.
    $endgroup$
    – Matti P.
    Jan 30 at 10:36














  • 1




    $begingroup$
    You could improve the readability of your question by using MathJax to type the equations.
    $endgroup$
    – Matti P.
    Jan 30 at 10:36








1




1




$begingroup$
You could improve the readability of your question by using MathJax to type the equations.
$endgroup$
– Matti P.
Jan 30 at 10:36




$begingroup$
You could improve the readability of your question by using MathJax to type the equations.
$endgroup$
– Matti P.
Jan 30 at 10:36










1 Answer
1






active

oldest

votes


















0












$begingroup$

Your answer to Q1 is correct!



For Q2, first note that the range (image) of $T_1$ cannot be $mathbb{R} ^4$, since $dimoperatorname{Ran} T_1 leqdim operatorname{Dom} T_1 = 3$.
The range of $T_1$ consists only of those vectors that are of the form $v = T_1 w$ for some $winmathbb{R}^3$.



To show that $T_2circ T_1$ is well-defined, it suffices to see that the range (or just the codomain, since the range is a subset of the codomain) of $T_1$ is a subset of the domain of $T_2$. This means that $T_2$ can be applied to any output of $T_1$. But you have already seen that the $operatorname{Dom} T_2 = operatorname{Codom} T_1$.



The same reasoning goes through for $T_1circ T_2$ being well-defined. Again, domain of $T_1$ and codomain of $T_2$ coïncide.






share|cite|improve this answer









$endgroup$














    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%2f3093357%2fcomposition-of-linear-transformation-and-domain-and-codomain%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









    0












    $begingroup$

    Your answer to Q1 is correct!



    For Q2, first note that the range (image) of $T_1$ cannot be $mathbb{R} ^4$, since $dimoperatorname{Ran} T_1 leqdim operatorname{Dom} T_1 = 3$.
    The range of $T_1$ consists only of those vectors that are of the form $v = T_1 w$ for some $winmathbb{R}^3$.



    To show that $T_2circ T_1$ is well-defined, it suffices to see that the range (or just the codomain, since the range is a subset of the codomain) of $T_1$ is a subset of the domain of $T_2$. This means that $T_2$ can be applied to any output of $T_1$. But you have already seen that the $operatorname{Dom} T_2 = operatorname{Codom} T_1$.



    The same reasoning goes through for $T_1circ T_2$ being well-defined. Again, domain of $T_1$ and codomain of $T_2$ coïncide.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Your answer to Q1 is correct!



      For Q2, first note that the range (image) of $T_1$ cannot be $mathbb{R} ^4$, since $dimoperatorname{Ran} T_1 leqdim operatorname{Dom} T_1 = 3$.
      The range of $T_1$ consists only of those vectors that are of the form $v = T_1 w$ for some $winmathbb{R}^3$.



      To show that $T_2circ T_1$ is well-defined, it suffices to see that the range (or just the codomain, since the range is a subset of the codomain) of $T_1$ is a subset of the domain of $T_2$. This means that $T_2$ can be applied to any output of $T_1$. But you have already seen that the $operatorname{Dom} T_2 = operatorname{Codom} T_1$.



      The same reasoning goes through for $T_1circ T_2$ being well-defined. Again, domain of $T_1$ and codomain of $T_2$ coïncide.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Your answer to Q1 is correct!



        For Q2, first note that the range (image) of $T_1$ cannot be $mathbb{R} ^4$, since $dimoperatorname{Ran} T_1 leqdim operatorname{Dom} T_1 = 3$.
        The range of $T_1$ consists only of those vectors that are of the form $v = T_1 w$ for some $winmathbb{R}^3$.



        To show that $T_2circ T_1$ is well-defined, it suffices to see that the range (or just the codomain, since the range is a subset of the codomain) of $T_1$ is a subset of the domain of $T_2$. This means that $T_2$ can be applied to any output of $T_1$. But you have already seen that the $operatorname{Dom} T_2 = operatorname{Codom} T_1$.



        The same reasoning goes through for $T_1circ T_2$ being well-defined. Again, domain of $T_1$ and codomain of $T_2$ coïncide.






        share|cite|improve this answer









        $endgroup$



        Your answer to Q1 is correct!



        For Q2, first note that the range (image) of $T_1$ cannot be $mathbb{R} ^4$, since $dimoperatorname{Ran} T_1 leqdim operatorname{Dom} T_1 = 3$.
        The range of $T_1$ consists only of those vectors that are of the form $v = T_1 w$ for some $winmathbb{R}^3$.



        To show that $T_2circ T_1$ is well-defined, it suffices to see that the range (or just the codomain, since the range is a subset of the codomain) of $T_1$ is a subset of the domain of $T_2$. This means that $T_2$ can be applied to any output of $T_1$. But you have already seen that the $operatorname{Dom} T_2 = operatorname{Codom} T_1$.



        The same reasoning goes through for $T_1circ T_2$ being well-defined. Again, domain of $T_1$ and codomain of $T_2$ coïncide.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 30 at 11:38









        ffffforallffffforall

        36028




        36028






























            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%2f3093357%2fcomposition-of-linear-transformation-and-domain-and-codomain%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

            Can a sorcerer learn a 5th-level spell early by creating spell slots using the Font of Magic feature?

            Image
            28 6 $begingroup$ Per the Font of Magic feature, sorcerer can use Flexible Casting to create 5th-level spell slots at level 7, even though they are not typically available until level 9. You can transform unexpended sorcery points into one spell slot as a bonus action on your turn. The Creating Spell Slots table shows the cost of creating a spell slot of [5th level is 7] If such a sorcerer levels up while still having this 5th-level spell slot, can they choose a 5th-level spell as the spell they gain upon levelling up? The Spellcasting feature states: Additionally, when you gain a level in this class, you can choose one of the sorcerer spells you know and replace it with another spell from the sorcerer spell list, which also must be of a level for which you have spell slots.
            Read more

            Does disintegrating a polymorphed enemy still kill it after the 2018 errata?

            Image
            up vote 13 down vote favorite 1 After the 2018 errata, the disintegrate spell description now reads: A creature targeted by this spell must make a Dexterity saving throw. On a failed save, the target takes 10d6 + 40 force damage. The target is disintegrated if this damage leaves it with 0 hit points. If you were to polymorph an enemy into a rat and then disintegrate it, would the enemy be disintegrated or would it just return to its original form? I know that for Druids, it’s not an instant kill anymore, but is this the case for polymorph as well? dnd-5e spells polymorph errata share | improve this question edited 18 hours ago
            Read more

            A Topological Invariant for $pi_3(U(n))$

            Image
            3 3 $begingroup$ Recently, I saw a construction of topological invariant for $pi_3(U(n))$ with $ngeq 2$ : $$ N=frac{1}{24pi^2}int_{S^3} d^3x epsilon^{ijk} Tr[(U^{-1}partial_{x_i}U)(U^{-1}partial_{x_j}U)(U^{-1}partial_{x_k}U)] , $$ where $Uin U(n)$ depends on $boldsymbol{x}=(x_1,x_2,x_3)in S^3$ , $epsilon^{ijk}$ is the Levi-Civita symbol, $i,j,k=1,2,3$ , and the duplicated indexes are summed over. It is claimed that $N$ is an integer, but why? Update 02/02/2019 I think I got an argument for $n=2$ . In this case, $U=e^{i varphi} q$ with $qin SU(2)$ . Due to the trace and the Levi-Civita symbol in $N$ , $varphi$ does not contribute to $N$ . As $Tr[q^{dagger}partial_i q]=0$ and $(q^{dagger}partial_i q)^{dagger}=-q^{dagger}partial_i q$ , $q^{dagger}partial_i q$ in geneeral has the form $q^{dagger}partia
            Read more