Proving a matrix is surjective











up vote
1
down vote

favorite












Is a $n times n$ matrix always surjective? If it is, how can I prove it using the rank-nullity theorem?






linear-algebra matrices







share|cite|improve this question









New contributor




Elliot Silver is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
















  • 4




    Small language point for the future. Matrices are not surjective The linear functions they determine may be.
    – Ethan Bolker
    3 hours ago















up vote
1
down vote

favorite












Is a $n times n$ matrix always surjective? If it is, how can I prove it using the rank-nullity theorem?






linear-algebra matrices







share|cite|improve this question









New contributor




Elliot Silver is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
















  • 4




    Small language point for the future. Matrices are not surjective The linear functions they determine may be.
    – Ethan Bolker
    3 hours ago













up vote
1
down vote

favorite









up vote
1
down vote

favorite











Is a $n times n$ matrix always surjective? If it is, how can I prove it using the rank-nullity theorem?






linear-algebra matrices







share|cite|improve this question









New contributor




Elliot Silver is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











Is a $n times n$ matrix always surjective? If it is, how can I prove it using the rank-nullity theorem?






linear-algebra matrices




linear-algebra matrices






share|cite|improve this question









New contributor




Elliot Silver is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




Elliot Silver is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 9 hours ago









gimusi

86.9k74393




86.9k74393






New contributor




Elliot Silver is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 9 hours ago









Elliot Silver

112




112




New contributor




Elliot Silver is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Elliot Silver is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Elliot Silver is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 4




    Small language point for the future. Matrices are not surjective The linear functions they determine may be.
    – Ethan Bolker
    3 hours ago














  • 4




    Small language point for the future. Matrices are not surjective The linear functions they determine may be.
    – Ethan Bolker
    3 hours ago








4




4




Small language point for the future. Matrices are not surjective The linear functions they determine may be.
– Ethan Bolker
3 hours ago




Small language point for the future. Matrices are not surjective The linear functions they determine may be.
– Ethan Bolker
3 hours ago










3 Answers
3






active

oldest

votes

















up vote
4
down vote













No it's not. To see this consider the zero matrix as a map from $mathbb{R}^nrightarrow mathbb{R}^n$ which is clearly not surjective.






share|cite|improve this answer




























    up vote
    3
    down vote













    Recall that surjective means that for any $bin mathbb{R^n}$ the system



    $$Ax=b$$



    has solution, that is true if and only if $dim(Im(A))=n$ that if and only if $A$ is full rank.






    share|cite|improve this answer






























      up vote
      1
      down vote













      Maybe your getting confused because there is a fact that says that a $ntimes n$ matrix which has null kernel is always surjective. This is a classical applications of rank-nullity theorem because $$n=dim Ker (A) + dim Im (A)$$ and its clear that $dim Ker(A)=0 implies dim Im(A)=n$ which in this case means that $A$ is surjective.






      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',
        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
        });


        }
        });






        Elliot Silver is a new contributor. Be nice, and check out our Code of Conduct.










         

        draft saved


        draft discarded


















        StackExchange.ready(
        function () {
        StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3011809%2fproving-a-matrix-is-surjective%23new-answer', 'question_page');
        }
        );

        Post as a guest















        Required, but never shown

























        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes








        up vote
        4
        down vote













        No it's not. To see this consider the zero matrix as a map from $mathbb{R}^nrightarrow mathbb{R}^n$ which is clearly not surjective.






        share|cite|improve this answer

























          up vote
          4
          down vote













          No it's not. To see this consider the zero matrix as a map from $mathbb{R}^nrightarrow mathbb{R}^n$ which is clearly not surjective.






          share|cite|improve this answer























            up vote
            4
            down vote










            up vote
            4
            down vote









            No it's not. To see this consider the zero matrix as a map from $mathbb{R}^nrightarrow mathbb{R}^n$ which is clearly not surjective.






            share|cite|improve this answer












            No it's not. To see this consider the zero matrix as a map from $mathbb{R}^nrightarrow mathbb{R}^n$ which is clearly not surjective.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 9 hours ago









            CyclotomicField

            2,0691312




            2,0691312






















                up vote
                3
                down vote













                Recall that surjective means that for any $bin mathbb{R^n}$ the system



                $$Ax=b$$



                has solution, that is true if and only if $dim(Im(A))=n$ that if and only if $A$ is full rank.






                share|cite|improve this answer



























                  up vote
                  3
                  down vote













                  Recall that surjective means that for any $bin mathbb{R^n}$ the system



                  $$Ax=b$$



                  has solution, that is true if and only if $dim(Im(A))=n$ that if and only if $A$ is full rank.






                  share|cite|improve this answer

























                    up vote
                    3
                    down vote










                    up vote
                    3
                    down vote









                    Recall that surjective means that for any $bin mathbb{R^n}$ the system



                    $$Ax=b$$



                    has solution, that is true if and only if $dim(Im(A))=n$ that if and only if $A$ is full rank.






                    share|cite|improve this answer














                    Recall that surjective means that for any $bin mathbb{R^n}$ the system



                    $$Ax=b$$



                    has solution, that is true if and only if $dim(Im(A))=n$ that if and only if $A$ is full rank.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited 9 hours ago

























                    answered 9 hours ago









                    gimusi

                    86.9k74393




                    86.9k74393






















                        up vote
                        1
                        down vote













                        Maybe your getting confused because there is a fact that says that a $ntimes n$ matrix which has null kernel is always surjective. This is a classical applications of rank-nullity theorem because $$n=dim Ker (A) + dim Im (A)$$ and its clear that $dim Ker(A)=0 implies dim Im(A)=n$ which in this case means that $A$ is surjective.






                        share|cite|improve this answer



























                          up vote
                          1
                          down vote













                          Maybe your getting confused because there is a fact that says that a $ntimes n$ matrix which has null kernel is always surjective. This is a classical applications of rank-nullity theorem because $$n=dim Ker (A) + dim Im (A)$$ and its clear that $dim Ker(A)=0 implies dim Im(A)=n$ which in this case means that $A$ is surjective.






                          share|cite|improve this answer

























                            up vote
                            1
                            down vote










                            up vote
                            1
                            down vote









                            Maybe your getting confused because there is a fact that says that a $ntimes n$ matrix which has null kernel is always surjective. This is a classical applications of rank-nullity theorem because $$n=dim Ker (A) + dim Im (A)$$ and its clear that $dim Ker(A)=0 implies dim Im(A)=n$ which in this case means that $A$ is surjective.






                            share|cite|improve this answer














                            Maybe your getting confused because there is a fact that says that a $ntimes n$ matrix which has null kernel is always surjective. This is a classical applications of rank-nullity theorem because $$n=dim Ker (A) + dim Im (A)$$ and its clear that $dim Ker(A)=0 implies dim Im(A)=n$ which in this case means that $A$ is surjective.







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited 3 hours ago

























                            answered 3 hours ago









                            Robson

                            688221




                            688221






















                                Elliot Silver is a new contributor. Be nice, and check out our Code of Conduct.










                                 

                                draft saved


                                draft discarded


















                                Elliot Silver is a new contributor. Be nice, and check out our Code of Conduct.













                                Elliot Silver is a new contributor. Be nice, and check out our Code of Conduct.












                                Elliot Silver is a new contributor. Be nice, and check out our Code of Conduct.















                                 


                                draft saved


                                draft discarded














                                StackExchange.ready(
                                function () {
                                StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3011809%2fproving-a-matrix-is-surjective%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

                                Contact image not getting when fetch all contact list from iPhone by CNContact

                                Image
                                up vote 0 down vote favorite I know this question already asked but not getting solution. From this code I will get all the information from the contact but image not found when open vcf files on mac os, also not getting when share this file. I use this stackoverflow link here but It's not help full. var contacts = [CNContact]() let keys = [CNContactVCardSerialization.descriptorForRequiredKeys() ] as [Any] let request = CNContactFetchRequest(keysToFetch: keys as! [CNKeyDescriptor]) do { try self.contactStore.enumerateContacts(with: request) { (contact, stop) in // Array containing all unified contacts from everywhere contacts.append(contact) } } catch { print("unable to fetch contacts") } do { let data = try
                                Read more

                                count number of partitions of a set with n elements into k subsets

                                Image
                                up vote 6 down vote favorite 1 This program is for count number of partitions of a set with n elements into k subsets I am confusing here return k*countP(n-1, k) + countP(n-1, k-1); can some one explain what is happening here? why we are multiplying with k? NOTE->I know this is not the best way to calculate number of partitions that would be DP // A C++ program to count number of partitions // of a set with n elements into k subsets #include<iostream> using namespace std; // Returns count of different partitions of n // elements in k subsets int countP(int n, int k) { // Base cases if (n == 0 || k == 0 || k > n) return 0; if (k == 1 || k == n) return 1; // S(n+1, k) = k*S(n, k) + S(n, k-1) return k*countP(n-1, k) + countP(n-1, k-1); } // Driver program int ma
                                Read more

                                A CLEAN and SIMPLE way to add appendices to Table of Contents and bookmarks

                                Image
                                0 I found a lot of questions abount appendices and ToC. Many users want appendices to be grouped in an Appendix part, however some problems arise with ToC, hyperref, PDF viewer bookmarks, and so on. There are different solutions which require extra packages, command patching and other extra code, however none of them satisfies me. I almost found an easy way to accomplish a good result, where appendices are added to bookmarks in the right way and hyperref links point to the right page. However, the number of the "Appendix" part page is wrong (it's the number of appendix A). Is there any EASY way to fix that? This is a MWE: documentclass{book} usepackage[nottoc,notlot,notlof]{tocbibind} usepackage{hyperref} begin{document} frontmatter tableofcontents mainmatter part{First} chapter{
                                Read more