mistake in calculations for ImT and KerT












3














I was asked to calculate ImT and KerT for the following linear transformation. I got to a result for both, but as can be seen from my conclusions, I know there must be a mistake somewhere along the way. Yet I cannot find it… Can anyone help out?



The linear transformation $T:M_{2x2}(R) to R_3[x]$ is defined by:
$$Tbegin{pmatrix}a&b\c&d end{pmatrix}$$$= (a-d)x^2+(b+c)x+5a-5d$



for every $$begin{pmatrix}a&b\c&d end{pmatrix}$$$in M_{2x2}(R)$



I reasoned in the following way:



For KerT:



a=d



b=-c



kerT:
$$begin{pmatrix}a&b\-b&a end{pmatrix}$$=$$abegin{pmatrix}1&0\0&1 end{pmatrix}$$+$$bbegin{pmatrix}0&1\-1&0 end{pmatrix}$$



So KerT = Sp{$$begin{pmatrix}1&0\0&1 end{pmatrix}$$,$$begin{pmatrix}0&1\-1&0 end{pmatrix}$$}
and as the matrices in KerT are clearly linearly independent, this is also a basis with dimension 2.



For ImT:
$(a-d)x^2+(b+c)x+5a-5d$ is a linear combination of the standard basis of R3[x], namely $(x^2,x,1)$. So $ImT = Sp{x^2,x,1}$. This set is per definition linearly independent and therefore $Sp{x^2,x,1}$ is a basis to ImT and its dimension is 3.



The problem is that according to my logic, n = dimKerT + dimImT = 5 and this is contradictory to the fact that n = dimKerT + dimImT = 4, as the dimension of $M_{2x2}(R)=4$






linear-algebra linear-transformations







share|cite|improve this question
























  • Use ${ X}$ for ${ X}$.
    – Shaun
    yesterday
















3














I was asked to calculate ImT and KerT for the following linear transformation. I got to a result for both, but as can be seen from my conclusions, I know there must be a mistake somewhere along the way. Yet I cannot find it… Can anyone help out?



The linear transformation $T:M_{2x2}(R) to R_3[x]$ is defined by:
$$Tbegin{pmatrix}a&b\c&d end{pmatrix}$$$= (a-d)x^2+(b+c)x+5a-5d$



for every $$begin{pmatrix}a&b\c&d end{pmatrix}$$$in M_{2x2}(R)$



I reasoned in the following way:



For KerT:



a=d



b=-c



kerT:
$$begin{pmatrix}a&b\-b&a end{pmatrix}$$=$$abegin{pmatrix}1&0\0&1 end{pmatrix}$$+$$bbegin{pmatrix}0&1\-1&0 end{pmatrix}$$



So KerT = Sp{$$begin{pmatrix}1&0\0&1 end{pmatrix}$$,$$begin{pmatrix}0&1\-1&0 end{pmatrix}$$}
and as the matrices in KerT are clearly linearly independent, this is also a basis with dimension 2.



For ImT:
$(a-d)x^2+(b+c)x+5a-5d$ is a linear combination of the standard basis of R3[x], namely $(x^2,x,1)$. So $ImT = Sp{x^2,x,1}$. This set is per definition linearly independent and therefore $Sp{x^2,x,1}$ is a basis to ImT and its dimension is 3.



The problem is that according to my logic, n = dimKerT + dimImT = 5 and this is contradictory to the fact that n = dimKerT + dimImT = 4, as the dimension of $M_{2x2}(R)=4$






linear-algebra linear-transformations







share|cite|improve this question
























  • Use ${ X}$ for ${ X}$.
    – Shaun
    yesterday














3












3








3







I was asked to calculate ImT and KerT for the following linear transformation. I got to a result for both, but as can be seen from my conclusions, I know there must be a mistake somewhere along the way. Yet I cannot find it… Can anyone help out?



The linear transformation $T:M_{2x2}(R) to R_3[x]$ is defined by:
$$Tbegin{pmatrix}a&b\c&d end{pmatrix}$$$= (a-d)x^2+(b+c)x+5a-5d$



for every $$begin{pmatrix}a&b\c&d end{pmatrix}$$$in M_{2x2}(R)$



I reasoned in the following way:



For KerT:



a=d



b=-c



kerT:
$$begin{pmatrix}a&b\-b&a end{pmatrix}$$=$$abegin{pmatrix}1&0\0&1 end{pmatrix}$$+$$bbegin{pmatrix}0&1\-1&0 end{pmatrix}$$



So KerT = Sp{$$begin{pmatrix}1&0\0&1 end{pmatrix}$$,$$begin{pmatrix}0&1\-1&0 end{pmatrix}$$}
and as the matrices in KerT are clearly linearly independent, this is also a basis with dimension 2.



For ImT:
$(a-d)x^2+(b+c)x+5a-5d$ is a linear combination of the standard basis of R3[x], namely $(x^2,x,1)$. So $ImT = Sp{x^2,x,1}$. This set is per definition linearly independent and therefore $Sp{x^2,x,1}$ is a basis to ImT and its dimension is 3.



The problem is that according to my logic, n = dimKerT + dimImT = 5 and this is contradictory to the fact that n = dimKerT + dimImT = 4, as the dimension of $M_{2x2}(R)=4$






linear-algebra linear-transformations







share|cite|improve this question















I was asked to calculate ImT and KerT for the following linear transformation. I got to a result for both, but as can be seen from my conclusions, I know there must be a mistake somewhere along the way. Yet I cannot find it… Can anyone help out?



The linear transformation $T:M_{2x2}(R) to R_3[x]$ is defined by:
$$Tbegin{pmatrix}a&b\c&d end{pmatrix}$$$= (a-d)x^2+(b+c)x+5a-5d$



for every $$begin{pmatrix}a&b\c&d end{pmatrix}$$$in M_{2x2}(R)$



I reasoned in the following way:



For KerT:



a=d



b=-c



kerT:
$$begin{pmatrix}a&b\-b&a end{pmatrix}$$=$$abegin{pmatrix}1&0\0&1 end{pmatrix}$$+$$bbegin{pmatrix}0&1\-1&0 end{pmatrix}$$



So KerT = Sp{$$begin{pmatrix}1&0\0&1 end{pmatrix}$$,$$begin{pmatrix}0&1\-1&0 end{pmatrix}$$}
and as the matrices in KerT are clearly linearly independent, this is also a basis with dimension 2.



For ImT:
$(a-d)x^2+(b+c)x+5a-5d$ is a linear combination of the standard basis of R3[x], namely $(x^2,x,1)$. So $ImT = Sp{x^2,x,1}$. This set is per definition linearly independent and therefore $Sp{x^2,x,1}$ is a basis to ImT and its dimension is 3.



The problem is that according to my logic, n = dimKerT + dimImT = 5 and this is contradictory to the fact that n = dimKerT + dimImT = 4, as the dimension of $M_{2x2}(R)=4$






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








edited yesterday

























asked yesterday









dalta

577




577












  • Use ${ X}$ for ${ X}$.
    – Shaun
    yesterday


















  • Use ${ X}$ for ${ X}$.
    – Shaun
    yesterday
















Use ${ X}$ for ${ X}$.
– Shaun
yesterday




Use ${ X}$ for ${ X}$.
– Shaun
yesterday










2 Answers
2






active

oldest

votes


















4















For ImT:
$(a-d)x^2+(b+c)x+5a-5d$ is a linear combination of the standard basis of R3[x], namely $(x^2,x,1)$. So $ImT = Sp{x^2,x,1}$. This set is per definition linearly independent and therefore $Sp{x^2,x,1}$ is a basis to ImT and its dimension is 3.




It's not an arbitrary linear combination of those standard basis vectors, as the coefficients aren't independent. You could rewrite:
$$(a-d)x^2+(b+c)x+5a-5d = (a-d)(x^2+5)+(b+c)x$$
so it can also be written as a linear combination of (only) $x^2+5$ and $x$, so...




Or for a more general approach, rewrite as (split in all the coefficients):
$$(a-d)x^2+(b+c)x+5a-5d = a(color{blue}{x^2+5})+bcolor{red}{x}+ccolor{purple}{x}+d(color{green}{-x^2-5})$$
and reduce the spanning set $left{color{blue}{x^2+5},color{red}{x},color{purple}{x},color{green}{-x^2-5}right}$ to a basis by eliminating the linearly dependent elements.






share|cite|improve this answer



















  • 1




    Thank you! So my calculations for KerT were correct?
    – dalta
    yesterday






  • 1




    Yes, although you want to write a comma (instead of a $+$) between the two matrices that span the kernel (the basis has two elements, not the sum of two matrices); in the spanning set.
    – StackTD
    yesterday





















2














$$text{Im}(T)netext{Sp}{1,x,x^2}$$



This is because $TBig(begin{bmatrix}a&b\c&d end{bmatrix}Big)=(a-d)(x^2+5)+(b+c)x=k_1(x^2+5)+k_2x$, which means $$text{Im}(T)=text{Sp}{x,x^2+5}$$ which is also a basis with dimension $2$.






share|cite|improve this answer





















    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%2f3060493%2fmistake-in-calculations-for-imt-and-kert%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4















    For ImT:
    $(a-d)x^2+(b+c)x+5a-5d$ is a linear combination of the standard basis of R3[x], namely $(x^2,x,1)$. So $ImT = Sp{x^2,x,1}$. This set is per definition linearly independent and therefore $Sp{x^2,x,1}$ is a basis to ImT and its dimension is 3.




    It's not an arbitrary linear combination of those standard basis vectors, as the coefficients aren't independent. You could rewrite:
    $$(a-d)x^2+(b+c)x+5a-5d = (a-d)(x^2+5)+(b+c)x$$
    so it can also be written as a linear combination of (only) $x^2+5$ and $x$, so...




    Or for a more general approach, rewrite as (split in all the coefficients):
    $$(a-d)x^2+(b+c)x+5a-5d = a(color{blue}{x^2+5})+bcolor{red}{x}+ccolor{purple}{x}+d(color{green}{-x^2-5})$$
    and reduce the spanning set $left{color{blue}{x^2+5},color{red}{x},color{purple}{x},color{green}{-x^2-5}right}$ to a basis by eliminating the linearly dependent elements.






    share|cite|improve this answer



















    • 1




      Thank you! So my calculations for KerT were correct?
      – dalta
      yesterday






    • 1




      Yes, although you want to write a comma (instead of a $+$) between the two matrices that span the kernel (the basis has two elements, not the sum of two matrices); in the spanning set.
      – StackTD
      yesterday


















    4















    For ImT:
    $(a-d)x^2+(b+c)x+5a-5d$ is a linear combination of the standard basis of R3[x], namely $(x^2,x,1)$. So $ImT = Sp{x^2,x,1}$. This set is per definition linearly independent and therefore $Sp{x^2,x,1}$ is a basis to ImT and its dimension is 3.




    It's not an arbitrary linear combination of those standard basis vectors, as the coefficients aren't independent. You could rewrite:
    $$(a-d)x^2+(b+c)x+5a-5d = (a-d)(x^2+5)+(b+c)x$$
    so it can also be written as a linear combination of (only) $x^2+5$ and $x$, so...




    Or for a more general approach, rewrite as (split in all the coefficients):
    $$(a-d)x^2+(b+c)x+5a-5d = a(color{blue}{x^2+5})+bcolor{red}{x}+ccolor{purple}{x}+d(color{green}{-x^2-5})$$
    and reduce the spanning set $left{color{blue}{x^2+5},color{red}{x},color{purple}{x},color{green}{-x^2-5}right}$ to a basis by eliminating the linearly dependent elements.






    share|cite|improve this answer



















    • 1




      Thank you! So my calculations for KerT were correct?
      – dalta
      yesterday






    • 1




      Yes, although you want to write a comma (instead of a $+$) between the two matrices that span the kernel (the basis has two elements, not the sum of two matrices); in the spanning set.
      – StackTD
      yesterday
















    4












    4








    4







    For ImT:
    $(a-d)x^2+(b+c)x+5a-5d$ is a linear combination of the standard basis of R3[x], namely $(x^2,x,1)$. So $ImT = Sp{x^2,x,1}$. This set is per definition linearly independent and therefore $Sp{x^2,x,1}$ is a basis to ImT and its dimension is 3.




    It's not an arbitrary linear combination of those standard basis vectors, as the coefficients aren't independent. You could rewrite:
    $$(a-d)x^2+(b+c)x+5a-5d = (a-d)(x^2+5)+(b+c)x$$
    so it can also be written as a linear combination of (only) $x^2+5$ and $x$, so...




    Or for a more general approach, rewrite as (split in all the coefficients):
    $$(a-d)x^2+(b+c)x+5a-5d = a(color{blue}{x^2+5})+bcolor{red}{x}+ccolor{purple}{x}+d(color{green}{-x^2-5})$$
    and reduce the spanning set $left{color{blue}{x^2+5},color{red}{x},color{purple}{x},color{green}{-x^2-5}right}$ to a basis by eliminating the linearly dependent elements.






    share|cite|improve this answer















    For ImT:
    $(a-d)x^2+(b+c)x+5a-5d$ is a linear combination of the standard basis of R3[x], namely $(x^2,x,1)$. So $ImT = Sp{x^2,x,1}$. This set is per definition linearly independent and therefore $Sp{x^2,x,1}$ is a basis to ImT and its dimension is 3.




    It's not an arbitrary linear combination of those standard basis vectors, as the coefficients aren't independent. You could rewrite:
    $$(a-d)x^2+(b+c)x+5a-5d = (a-d)(x^2+5)+(b+c)x$$
    so it can also be written as a linear combination of (only) $x^2+5$ and $x$, so...




    Or for a more general approach, rewrite as (split in all the coefficients):
    $$(a-d)x^2+(b+c)x+5a-5d = a(color{blue}{x^2+5})+bcolor{red}{x}+ccolor{purple}{x}+d(color{green}{-x^2-5})$$
    and reduce the spanning set $left{color{blue}{x^2+5},color{red}{x},color{purple}{x},color{green}{-x^2-5}right}$ to a basis by eliminating the linearly dependent elements.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited yesterday

























    answered yesterday









    StackTD

    22.3k2049




    22.3k2049








    • 1




      Thank you! So my calculations for KerT were correct?
      – dalta
      yesterday






    • 1




      Yes, although you want to write a comma (instead of a $+$) between the two matrices that span the kernel (the basis has two elements, not the sum of two matrices); in the spanning set.
      – StackTD
      yesterday
















    • 1




      Thank you! So my calculations for KerT were correct?
      – dalta
      yesterday






    • 1




      Yes, although you want to write a comma (instead of a $+$) between the two matrices that span the kernel (the basis has two elements, not the sum of two matrices); in the spanning set.
      – StackTD
      yesterday










    1




    1




    Thank you! So my calculations for KerT were correct?
    – dalta
    yesterday




    Thank you! So my calculations for KerT were correct?
    – dalta
    yesterday




    1




    1




    Yes, although you want to write a comma (instead of a $+$) between the two matrices that span the kernel (the basis has two elements, not the sum of two matrices); in the spanning set.
    – StackTD
    yesterday






    Yes, although you want to write a comma (instead of a $+$) between the two matrices that span the kernel (the basis has two elements, not the sum of two matrices); in the spanning set.
    – StackTD
    yesterday













    2














    $$text{Im}(T)netext{Sp}{1,x,x^2}$$



    This is because $TBig(begin{bmatrix}a&b\c&d end{bmatrix}Big)=(a-d)(x^2+5)+(b+c)x=k_1(x^2+5)+k_2x$, which means $$text{Im}(T)=text{Sp}{x,x^2+5}$$ which is also a basis with dimension $2$.






    share|cite|improve this answer


























      2














      $$text{Im}(T)netext{Sp}{1,x,x^2}$$



      This is because $TBig(begin{bmatrix}a&b\c&d end{bmatrix}Big)=(a-d)(x^2+5)+(b+c)x=k_1(x^2+5)+k_2x$, which means $$text{Im}(T)=text{Sp}{x,x^2+5}$$ which is also a basis with dimension $2$.






      share|cite|improve this answer
























        2












        2








        2






        $$text{Im}(T)netext{Sp}{1,x,x^2}$$



        This is because $TBig(begin{bmatrix}a&b\c&d end{bmatrix}Big)=(a-d)(x^2+5)+(b+c)x=k_1(x^2+5)+k_2x$, which means $$text{Im}(T)=text{Sp}{x,x^2+5}$$ which is also a basis with dimension $2$.






        share|cite|improve this answer












        $$text{Im}(T)netext{Sp}{1,x,x^2}$$



        This is because $TBig(begin{bmatrix}a&b\c&d end{bmatrix}Big)=(a-d)(x^2+5)+(b+c)x=k_1(x^2+5)+k_2x$, which means $$text{Im}(T)=text{Sp}{x,x^2+5}$$ which is also a basis with dimension $2$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered yesterday









        Shubham Johri

        3,961717




        3,961717






























            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.





            Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


            Please pay close attention to the following guidance:


            • 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.


            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%2f3060493%2fmistake-in-calculations-for-imt-and-kert%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