If A is a nxn singular matrix, then it has a singular value = 0












0












$begingroup$


This is a question on a testexam.



But am I correct in assuming that a singular matrix has det = 0, which gives it an eigenvalue of 0 and that gives it a singular value of 0?






linear-algebra matrices singularvalues







share|cite|improve this question









$endgroup$












  • $begingroup$
    How does a zero eigenvalue imply a zero singular value? If you explain this, then you are okay.
    $endgroup$
    – астон вілла олоф мэллбэрг
    Jan 21 at 8:07












  • $begingroup$
    I thought that singular values were the squareroot of the A(transpose)A. And I thought that At and A have the same eigenvalues?
    $endgroup$
    – Tourna
    Jan 21 at 8:11










  • $begingroup$
    Yes, that is correct, and your argument is fine.
    $endgroup$
    – астон вілла олоф мэллбэрг
    Jan 21 at 8:19
















0












$begingroup$


This is a question on a testexam.



But am I correct in assuming that a singular matrix has det = 0, which gives it an eigenvalue of 0 and that gives it a singular value of 0?






linear-algebra matrices singularvalues







share|cite|improve this question









$endgroup$












  • $begingroup$
    How does a zero eigenvalue imply a zero singular value? If you explain this, then you are okay.
    $endgroup$
    – астон вілла олоф мэллбэрг
    Jan 21 at 8:07












  • $begingroup$
    I thought that singular values were the squareroot of the A(transpose)A. And I thought that At and A have the same eigenvalues?
    $endgroup$
    – Tourna
    Jan 21 at 8:11










  • $begingroup$
    Yes, that is correct, and your argument is fine.
    $endgroup$
    – астон вілла олоф мэллбэрг
    Jan 21 at 8:19














0












0








0





$begingroup$


This is a question on a testexam.



But am I correct in assuming that a singular matrix has det = 0, which gives it an eigenvalue of 0 and that gives it a singular value of 0?






linear-algebra matrices singularvalues







share|cite|improve this question









$endgroup$




This is a question on a testexam.



But am I correct in assuming that a singular matrix has det = 0, which gives it an eigenvalue of 0 and that gives it a singular value of 0?






linear-algebra matrices singularvalues




linear-algebra matrices singularvalues






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 21 at 7:59









TournaTourna

126




126












  • $begingroup$
    How does a zero eigenvalue imply a zero singular value? If you explain this, then you are okay.
    $endgroup$
    – астон вілла олоф мэллбэрг
    Jan 21 at 8:07












  • $begingroup$
    I thought that singular values were the squareroot of the A(transpose)A. And I thought that At and A have the same eigenvalues?
    $endgroup$
    – Tourna
    Jan 21 at 8:11










  • $begingroup$
    Yes, that is correct, and your argument is fine.
    $endgroup$
    – астон вілла олоф мэллбэрг
    Jan 21 at 8:19


















  • $begingroup$
    How does a zero eigenvalue imply a zero singular value? If you explain this, then you are okay.
    $endgroup$
    – астон вілла олоф мэллбэрг
    Jan 21 at 8:07












  • $begingroup$
    I thought that singular values were the squareroot of the A(transpose)A. And I thought that At and A have the same eigenvalues?
    $endgroup$
    – Tourna
    Jan 21 at 8:11










  • $begingroup$
    Yes, that is correct, and your argument is fine.
    $endgroup$
    – астон вілла олоф мэллбэрг
    Jan 21 at 8:19
















$begingroup$
How does a zero eigenvalue imply a zero singular value? If you explain this, then you are okay.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 21 at 8:07






$begingroup$
How does a zero eigenvalue imply a zero singular value? If you explain this, then you are okay.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 21 at 8:07














$begingroup$
I thought that singular values were the squareroot of the A(transpose)A. And I thought that At and A have the same eigenvalues?
$endgroup$
– Tourna
Jan 21 at 8:11




$begingroup$
I thought that singular values were the squareroot of the A(transpose)A. And I thought that At and A have the same eigenvalues?
$endgroup$
– Tourna
Jan 21 at 8:11












$begingroup$
Yes, that is correct, and your argument is fine.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 21 at 8:19




$begingroup$
Yes, that is correct, and your argument is fine.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 21 at 8:19










2 Answers
2






active

oldest

votes


















2












$begingroup$

A $ntimes n$ matrix $A$ is singular if and only if there is a non-zero vector $v$ such that $A.v=0$. But then $v$ is an eigenvector with eigenvalue $0$.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    If $A$ is a square matrix, then $ lambda $ is a singular value of $A$ , then there is an eigenvalue $mu$ of $A^*A$ such that $lambda= mu^{1/2}$.



    If $A$ is singular, then $A^*A$ is singular.... Conclusion ?






    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%2f3081622%2fif-a-is-a-nxn-singular-matrix-then-it-has-a-singular-value-0%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









      2












      $begingroup$

      A $ntimes n$ matrix $A$ is singular if and only if there is a non-zero vector $v$ such that $A.v=0$. But then $v$ is an eigenvector with eigenvalue $0$.






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        A $ntimes n$ matrix $A$ is singular if and only if there is a non-zero vector $v$ such that $A.v=0$. But then $v$ is an eigenvector with eigenvalue $0$.






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          A $ntimes n$ matrix $A$ is singular if and only if there is a non-zero vector $v$ such that $A.v=0$. But then $v$ is an eigenvector with eigenvalue $0$.






          share|cite|improve this answer









          $endgroup$



          A $ntimes n$ matrix $A$ is singular if and only if there is a non-zero vector $v$ such that $A.v=0$. But then $v$ is an eigenvector with eigenvalue $0$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 21 at 8:22









          José Carlos SantosJosé Carlos Santos

          163k22131234




          163k22131234























              0












              $begingroup$

              If $A$ is a square matrix, then $ lambda $ is a singular value of $A$ , then there is an eigenvalue $mu$ of $A^*A$ such that $lambda= mu^{1/2}$.



              If $A$ is singular, then $A^*A$ is singular.... Conclusion ?






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                If $A$ is a square matrix, then $ lambda $ is a singular value of $A$ , then there is an eigenvalue $mu$ of $A^*A$ such that $lambda= mu^{1/2}$.



                If $A$ is singular, then $A^*A$ is singular.... Conclusion ?






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  If $A$ is a square matrix, then $ lambda $ is a singular value of $A$ , then there is an eigenvalue $mu$ of $A^*A$ such that $lambda= mu^{1/2}$.



                  If $A$ is singular, then $A^*A$ is singular.... Conclusion ?






                  share|cite|improve this answer









                  $endgroup$



                  If $A$ is a square matrix, then $ lambda $ is a singular value of $A$ , then there is an eigenvalue $mu$ of $A^*A$ such that $lambda= mu^{1/2}$.



                  If $A$ is singular, then $A^*A$ is singular.... Conclusion ?







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 21 at 8:13









                  FredFred

                  46.9k1848




                  46.9k1848






























                      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%2f3081622%2fif-a-is-a-nxn-singular-matrix-then-it-has-a-singular-value-0%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