How to show that a map is linear in $C^n$?












0














Could someone tell me if I am on right way solving Problem b)?



Problem:




Let $U,Vsubsetmathbb C^n$ be two subspaces, such that $mathbb C^n = U+V$ and further assume $Ucap V = {0}$.



a) Show that every $xinmathbb C^n$ can be written as $x=x_U+x_V$ with $x_Uin U$ and $x_Vin V$ and that this decomposition is unique.



b) Define f : $C^n$ → $C^n$, $f$($x$) := $x_U$. Show that $f$ is a linear map.



$x_u$ is a Projection of $x$ onto $U$, where $U$ is a subspace of $C^n$




My thoughts:




$f$($x$) = < $lambda $ , $x_U$> = $lambda $ . $f$($x$)



$f$($x$) = ($x$+ $y$) = $x_u$ + $y_u$= $f$($x$) + $f$($y$)







linear-algebra map-projections







share|cite|improve this question
























  • What does $x_U$ mean?
    – José Carlos Santos
    2 days ago










  • Projection of x onto u where U is a subspace
    – Kai
    2 days ago










  • And how do you define that?
    – José Carlos Santos
    2 days ago










  • Your function doesn't make sense.
    – K. Y
    2 days ago










  • @K.Y That function makes sense. The decomposition as sum of an element of $U$ and an element of $V$ is unique. This defines two projections over $U$ and $V$.
    – Crostul
    2 days ago
















0














Could someone tell me if I am on right way solving Problem b)?



Problem:




Let $U,Vsubsetmathbb C^n$ be two subspaces, such that $mathbb C^n = U+V$ and further assume $Ucap V = {0}$.



a) Show that every $xinmathbb C^n$ can be written as $x=x_U+x_V$ with $x_Uin U$ and $x_Vin V$ and that this decomposition is unique.



b) Define f : $C^n$ → $C^n$, $f$($x$) := $x_U$. Show that $f$ is a linear map.



$x_u$ is a Projection of $x$ onto $U$, where $U$ is a subspace of $C^n$




My thoughts:




$f$($x$) = < $lambda $ , $x_U$> = $lambda $ . $f$($x$)



$f$($x$) = ($x$+ $y$) = $x_u$ + $y_u$= $f$($x$) + $f$($y$)







linear-algebra map-projections







share|cite|improve this question
























  • What does $x_U$ mean?
    – José Carlos Santos
    2 days ago










  • Projection of x onto u where U is a subspace
    – Kai
    2 days ago










  • And how do you define that?
    – José Carlos Santos
    2 days ago










  • Your function doesn't make sense.
    – K. Y
    2 days ago










  • @K.Y That function makes sense. The decomposition as sum of an element of $U$ and an element of $V$ is unique. This defines two projections over $U$ and $V$.
    – Crostul
    2 days ago














0












0








0







Could someone tell me if I am on right way solving Problem b)?



Problem:




Let $U,Vsubsetmathbb C^n$ be two subspaces, such that $mathbb C^n = U+V$ and further assume $Ucap V = {0}$.



a) Show that every $xinmathbb C^n$ can be written as $x=x_U+x_V$ with $x_Uin U$ and $x_Vin V$ and that this decomposition is unique.



b) Define f : $C^n$ → $C^n$, $f$($x$) := $x_U$. Show that $f$ is a linear map.



$x_u$ is a Projection of $x$ onto $U$, where $U$ is a subspace of $C^n$




My thoughts:




$f$($x$) = < $lambda $ , $x_U$> = $lambda $ . $f$($x$)



$f$($x$) = ($x$+ $y$) = $x_u$ + $y_u$= $f$($x$) + $f$($y$)







linear-algebra map-projections







share|cite|improve this question















Could someone tell me if I am on right way solving Problem b)?



Problem:




Let $U,Vsubsetmathbb C^n$ be two subspaces, such that $mathbb C^n = U+V$ and further assume $Ucap V = {0}$.



a) Show that every $xinmathbb C^n$ can be written as $x=x_U+x_V$ with $x_Uin U$ and $x_Vin V$ and that this decomposition is unique.



b) Define f : $C^n$ → $C^n$, $f$($x$) := $x_U$. Show that $f$ is a linear map.



$x_u$ is a Projection of $x$ onto $U$, where $U$ is a subspace of $C^n$




My thoughts:




$f$($x$) = < $lambda $ , $x_U$> = $lambda $ . $f$($x$)



$f$($x$) = ($x$+ $y$) = $x_u$ + $y_u$= $f$($x$) + $f$($y$)







linear-algebra map-projections




linear-algebra map-projections






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 days ago









Omnomnomnom

126k788176




126k788176










asked 2 days ago









KaiKai

256




256












  • What does $x_U$ mean?
    – José Carlos Santos
    2 days ago










  • Projection of x onto u where U is a subspace
    – Kai
    2 days ago










  • And how do you define that?
    – José Carlos Santos
    2 days ago










  • Your function doesn't make sense.
    – K. Y
    2 days ago










  • @K.Y That function makes sense. The decomposition as sum of an element of $U$ and an element of $V$ is unique. This defines two projections over $U$ and $V$.
    – Crostul
    2 days ago


















  • What does $x_U$ mean?
    – José Carlos Santos
    2 days ago










  • Projection of x onto u where U is a subspace
    – Kai
    2 days ago










  • And how do you define that?
    – José Carlos Santos
    2 days ago










  • Your function doesn't make sense.
    – K. Y
    2 days ago










  • @K.Y That function makes sense. The decomposition as sum of an element of $U$ and an element of $V$ is unique. This defines two projections over $U$ and $V$.
    – Crostul
    2 days ago
















What does $x_U$ mean?
– José Carlos Santos
2 days ago




What does $x_U$ mean?
– José Carlos Santos
2 days ago












Projection of x onto u where U is a subspace
– Kai
2 days ago




Projection of x onto u where U is a subspace
– Kai
2 days ago












And how do you define that?
– José Carlos Santos
2 days ago




And how do you define that?
– José Carlos Santos
2 days ago












Your function doesn't make sense.
– K. Y
2 days ago




Your function doesn't make sense.
– K. Y
2 days ago












@K.Y That function makes sense. The decomposition as sum of an element of $U$ and an element of $V$ is unique. This defines two projections over $U$ and $V$.
– Crostul
2 days ago




@K.Y That function makes sense. The decomposition as sum of an element of $U$ and an element of $V$ is unique. This defines two projections over $U$ and $V$.
– Crostul
2 days ago










1 Answer
1






active

oldest

votes


















3














You did not prove a).



Since $mathbb{C}^n = U + V = { x_U + x_V mid x_U in U, x_V in V}$, we know that each $xinmathbb C^n$ can be written as $x=x_U + x_V$ with $x_Uin U$ and $x_V in V$. Assume we have another such decomposition $x = x'_U + x'_V$. Then $x_U - x'_U = x'_V - x_V$. The left hand side is an element of $U$, the right hand side an element of $V$. Since $U cap V = {0}$, we see that $x_U - x'_U = x'_V - x_V = 0$. This proves uniqueness.



Your proof of b) is not correct (although you probably had the right idea).



Let $x,y in mathbb{C}^n$ and let $x = x_U + x_V, y = y_U +y_V$ be their unique decompositions. Then $lambda x + mu y = (lambda x_u + mu y_u) + (lambda x_V + mu y_V)$ is a decomposition of $lambda x + mu y$ and by uniqueness we see that



$$f(lambda x + mu y) = (lambda x + mu y)_U = lambda x_u + mu y_u = lambda f(x) + mu f(y) .$$






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%2f3062913%2fhow-to-show-that-a-map-is-linear-in-cn%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









    3














    You did not prove a).



    Since $mathbb{C}^n = U + V = { x_U + x_V mid x_U in U, x_V in V}$, we know that each $xinmathbb C^n$ can be written as $x=x_U + x_V$ with $x_Uin U$ and $x_V in V$. Assume we have another such decomposition $x = x'_U + x'_V$. Then $x_U - x'_U = x'_V - x_V$. The left hand side is an element of $U$, the right hand side an element of $V$. Since $U cap V = {0}$, we see that $x_U - x'_U = x'_V - x_V = 0$. This proves uniqueness.



    Your proof of b) is not correct (although you probably had the right idea).



    Let $x,y in mathbb{C}^n$ and let $x = x_U + x_V, y = y_U +y_V$ be their unique decompositions. Then $lambda x + mu y = (lambda x_u + mu y_u) + (lambda x_V + mu y_V)$ is a decomposition of $lambda x + mu y$ and by uniqueness we see that



    $$f(lambda x + mu y) = (lambda x + mu y)_U = lambda x_u + mu y_u = lambda f(x) + mu f(y) .$$






    share|cite|improve this answer




























      3














      You did not prove a).



      Since $mathbb{C}^n = U + V = { x_U + x_V mid x_U in U, x_V in V}$, we know that each $xinmathbb C^n$ can be written as $x=x_U + x_V$ with $x_Uin U$ and $x_V in V$. Assume we have another such decomposition $x = x'_U + x'_V$. Then $x_U - x'_U = x'_V - x_V$. The left hand side is an element of $U$, the right hand side an element of $V$. Since $U cap V = {0}$, we see that $x_U - x'_U = x'_V - x_V = 0$. This proves uniqueness.



      Your proof of b) is not correct (although you probably had the right idea).



      Let $x,y in mathbb{C}^n$ and let $x = x_U + x_V, y = y_U +y_V$ be their unique decompositions. Then $lambda x + mu y = (lambda x_u + mu y_u) + (lambda x_V + mu y_V)$ is a decomposition of $lambda x + mu y$ and by uniqueness we see that



      $$f(lambda x + mu y) = (lambda x + mu y)_U = lambda x_u + mu y_u = lambda f(x) + mu f(y) .$$






      share|cite|improve this answer


























        3












        3








        3






        You did not prove a).



        Since $mathbb{C}^n = U + V = { x_U + x_V mid x_U in U, x_V in V}$, we know that each $xinmathbb C^n$ can be written as $x=x_U + x_V$ with $x_Uin U$ and $x_V in V$. Assume we have another such decomposition $x = x'_U + x'_V$. Then $x_U - x'_U = x'_V - x_V$. The left hand side is an element of $U$, the right hand side an element of $V$. Since $U cap V = {0}$, we see that $x_U - x'_U = x'_V - x_V = 0$. This proves uniqueness.



        Your proof of b) is not correct (although you probably had the right idea).



        Let $x,y in mathbb{C}^n$ and let $x = x_U + x_V, y = y_U +y_V$ be their unique decompositions. Then $lambda x + mu y = (lambda x_u + mu y_u) + (lambda x_V + mu y_V)$ is a decomposition of $lambda x + mu y$ and by uniqueness we see that



        $$f(lambda x + mu y) = (lambda x + mu y)_U = lambda x_u + mu y_u = lambda f(x) + mu f(y) .$$






        share|cite|improve this answer














        You did not prove a).



        Since $mathbb{C}^n = U + V = { x_U + x_V mid x_U in U, x_V in V}$, we know that each $xinmathbb C^n$ can be written as $x=x_U + x_V$ with $x_Uin U$ and $x_V in V$. Assume we have another such decomposition $x = x'_U + x'_V$. Then $x_U - x'_U = x'_V - x_V$. The left hand side is an element of $U$, the right hand side an element of $V$. Since $U cap V = {0}$, we see that $x_U - x'_U = x'_V - x_V = 0$. This proves uniqueness.



        Your proof of b) is not correct (although you probably had the right idea).



        Let $x,y in mathbb{C}^n$ and let $x = x_U + x_V, y = y_U +y_V$ be their unique decompositions. Then $lambda x + mu y = (lambda x_u + mu y_u) + (lambda x_V + mu y_V)$ is a decomposition of $lambda x + mu y$ and by uniqueness we see that



        $$f(lambda x + mu y) = (lambda x + mu y)_U = lambda x_u + mu y_u = lambda f(x) + mu f(y) .$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 2 days ago

























        answered 2 days ago









        Paul FrostPaul Frost

        9,5502632




        9,5502632






























            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%2f3062913%2fhow-to-show-that-a-map-is-linear-in-cn%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

            Mario Kart Wii

            Image
            Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro...
            Read more

            The Binding of Isaac: Rebirth/Afterbirth

            Image
            Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! The Binding of Isaac: Rebirth/Afterbirth From The Cutting Room Floor < The Binding of Isaac: Rebirth Jump to: navigation, search This is a sub-page of The Binding of Isaac: Rebirth . Afterbirth is the first expansion pack for Rebirth and features a variety of improvements. Contents 1 Unused Graphics 1.1 Old Man Keeper 1.2 Skinless Hush 1.3 Miscellaneous Unused Hush Sprites 1.4 Generosity achievement 1.5 Afterbirth+ Room Editor 1.6 Bearded Ultra Greed 1.7 Early Ultra Greed 1.8 Unused Brownie Expression 1.9 Frail's Spinning Head 1.10 Mom's Dead Hand Alt 2 Unused Audio 2.1 Unused Dead Womb Layer 2.2 Unused Hush Battle Outro 2.3 The Stain Grunt Unused Graphics Old Man Keeper Image file: gfxuimain me...
            Read more

            What does “Dominus providebit” mean?

            Image
            5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1 ...
            Read more