Inverse of symmetric matrix $AA^top$ where $rank(A)=m leq n$?












0












$begingroup$


Suppose I have a matrix $Ainmathbb{R^{mtimes n}}$ where $mleq n$ and $rank(A)=m$. Is the matrix $AA^top$ singular?



My hunch is that the matrix is only non-singular when $n=m$.






linear-algebra matrices linear-transformations matrix-rank







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    Suppose I have a matrix $Ainmathbb{R^{mtimes n}}$ where $mleq n$ and $rank(A)=m$. Is the matrix $AA^top$ singular?



    My hunch is that the matrix is only non-singular when $n=m$.






    linear-algebra matrices linear-transformations matrix-rank







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      Suppose I have a matrix $Ainmathbb{R^{mtimes n}}$ where $mleq n$ and $rank(A)=m$. Is the matrix $AA^top$ singular?



      My hunch is that the matrix is only non-singular when $n=m$.






      linear-algebra matrices linear-transformations matrix-rank







      share|cite|improve this question











      $endgroup$




      Suppose I have a matrix $Ainmathbb{R^{mtimes n}}$ where $mleq n$ and $rank(A)=m$. Is the matrix $AA^top$ singular?



      My hunch is that the matrix is only non-singular when $n=m$.






      linear-algebra matrices linear-transformations matrix-rank




      linear-algebra matrices linear-transformations matrix-rank






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 4 '18 at 1:38







      UnchartedWaters

















      asked Dec 4 '18 at 1:31









      UnchartedWatersUnchartedWaters

      103




      103






















          3 Answers
          3






          active

          oldest

          votes


















          1












          $begingroup$

          Here's the proof that the $mtimes m$ matrix $AA^top$ is nonsingular. Suppose $AA^top x = 0$ for some vector $xinBbb R^m$. Then
          $$0=(AA^top x)cdot x = A^top xcdot A^top x$$
          by the fundamental property of transpose. This means that $A^top x = 0$. Since $A^top$ is an $ntimes m$ matrix with rank $m$, the only solution of this equation is $x=0$. (For example, the only linear combination of the columns of $A^top$ that gives the zero vector is the zero linear combination.) This means that $AA^top$ is in fact nonsingular.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            yeah, I was thinking of mentioning positive definite, with a comment that row rank equals column rank. At that point, I was taking a shower, and had not taken the keyboard with me.
            $endgroup$
            – Will Jagy
            Dec 4 '18 at 2:13






          • 1




            $begingroup$
            I'm sure your keyboard is grateful, @WillJagy!
            $endgroup$
            – Ted Shifrin
            Dec 4 '18 at 3:14



















          0












          $begingroup$

          Actually, with your symbols, $A A^T$ is non-singular, it is $m$ by $m.$



          The square $A^T A$ is a square matrix of size $n$ by $n$ but rank $m,$ therfore singular when $m < n$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks for your answer. But how do I prove that $AA^top$ is non-singular, in general? I'm from a CS background, not a maths background, so I might be missing some fundamental knowledge here.
            $endgroup$
            – UnchartedWaters
            Dec 4 '18 at 1:44





















          0












          $begingroup$

          Let $$A= begin{bmatrix}1 & 0 end{bmatrix}$$



          then we have $AA^T=1$.



          In general,
          $$rank(AA^T)=rank(A)=m,$$



          hence, it is nonsingular.






          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%2f3025002%2finverse-of-symmetric-matrix-aa-top-where-ranka-m-leq-n%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









            1












            $begingroup$

            Here's the proof that the $mtimes m$ matrix $AA^top$ is nonsingular. Suppose $AA^top x = 0$ for some vector $xinBbb R^m$. Then
            $$0=(AA^top x)cdot x = A^top xcdot A^top x$$
            by the fundamental property of transpose. This means that $A^top x = 0$. Since $A^top$ is an $ntimes m$ matrix with rank $m$, the only solution of this equation is $x=0$. (For example, the only linear combination of the columns of $A^top$ that gives the zero vector is the zero linear combination.) This means that $AA^top$ is in fact nonsingular.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              yeah, I was thinking of mentioning positive definite, with a comment that row rank equals column rank. At that point, I was taking a shower, and had not taken the keyboard with me.
              $endgroup$
              – Will Jagy
              Dec 4 '18 at 2:13






            • 1




              $begingroup$
              I'm sure your keyboard is grateful, @WillJagy!
              $endgroup$
              – Ted Shifrin
              Dec 4 '18 at 3:14
















            1












            $begingroup$

            Here's the proof that the $mtimes m$ matrix $AA^top$ is nonsingular. Suppose $AA^top x = 0$ for some vector $xinBbb R^m$. Then
            $$0=(AA^top x)cdot x = A^top xcdot A^top x$$
            by the fundamental property of transpose. This means that $A^top x = 0$. Since $A^top$ is an $ntimes m$ matrix with rank $m$, the only solution of this equation is $x=0$. (For example, the only linear combination of the columns of $A^top$ that gives the zero vector is the zero linear combination.) This means that $AA^top$ is in fact nonsingular.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              yeah, I was thinking of mentioning positive definite, with a comment that row rank equals column rank. At that point, I was taking a shower, and had not taken the keyboard with me.
              $endgroup$
              – Will Jagy
              Dec 4 '18 at 2:13






            • 1




              $begingroup$
              I'm sure your keyboard is grateful, @WillJagy!
              $endgroup$
              – Ted Shifrin
              Dec 4 '18 at 3:14














            1












            1








            1





            $begingroup$

            Here's the proof that the $mtimes m$ matrix $AA^top$ is nonsingular. Suppose $AA^top x = 0$ for some vector $xinBbb R^m$. Then
            $$0=(AA^top x)cdot x = A^top xcdot A^top x$$
            by the fundamental property of transpose. This means that $A^top x = 0$. Since $A^top$ is an $ntimes m$ matrix with rank $m$, the only solution of this equation is $x=0$. (For example, the only linear combination of the columns of $A^top$ that gives the zero vector is the zero linear combination.) This means that $AA^top$ is in fact nonsingular.






            share|cite|improve this answer









            $endgroup$



            Here's the proof that the $mtimes m$ matrix $AA^top$ is nonsingular. Suppose $AA^top x = 0$ for some vector $xinBbb R^m$. Then
            $$0=(AA^top x)cdot x = A^top xcdot A^top x$$
            by the fundamental property of transpose. This means that $A^top x = 0$. Since $A^top$ is an $ntimes m$ matrix with rank $m$, the only solution of this equation is $x=0$. (For example, the only linear combination of the columns of $A^top$ that gives the zero vector is the zero linear combination.) This means that $AA^top$ is in fact nonsingular.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 4 '18 at 2:04









            Ted ShifrinTed Shifrin

            63.2k44489




            63.2k44489












            • $begingroup$
              yeah, I was thinking of mentioning positive definite, with a comment that row rank equals column rank. At that point, I was taking a shower, and had not taken the keyboard with me.
              $endgroup$
              – Will Jagy
              Dec 4 '18 at 2:13






            • 1




              $begingroup$
              I'm sure your keyboard is grateful, @WillJagy!
              $endgroup$
              – Ted Shifrin
              Dec 4 '18 at 3:14


















            • $begingroup$
              yeah, I was thinking of mentioning positive definite, with a comment that row rank equals column rank. At that point, I was taking a shower, and had not taken the keyboard with me.
              $endgroup$
              – Will Jagy
              Dec 4 '18 at 2:13






            • 1




              $begingroup$
              I'm sure your keyboard is grateful, @WillJagy!
              $endgroup$
              – Ted Shifrin
              Dec 4 '18 at 3:14
















            $begingroup$
            yeah, I was thinking of mentioning positive definite, with a comment that row rank equals column rank. At that point, I was taking a shower, and had not taken the keyboard with me.
            $endgroup$
            – Will Jagy
            Dec 4 '18 at 2:13




            $begingroup$
            yeah, I was thinking of mentioning positive definite, with a comment that row rank equals column rank. At that point, I was taking a shower, and had not taken the keyboard with me.
            $endgroup$
            – Will Jagy
            Dec 4 '18 at 2:13




            1




            1




            $begingroup$
            I'm sure your keyboard is grateful, @WillJagy!
            $endgroup$
            – Ted Shifrin
            Dec 4 '18 at 3:14




            $begingroup$
            I'm sure your keyboard is grateful, @WillJagy!
            $endgroup$
            – Ted Shifrin
            Dec 4 '18 at 3:14











            0












            $begingroup$

            Actually, with your symbols, $A A^T$ is non-singular, it is $m$ by $m.$



            The square $A^T A$ is a square matrix of size $n$ by $n$ but rank $m,$ therfore singular when $m < n$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thanks for your answer. But how do I prove that $AA^top$ is non-singular, in general? I'm from a CS background, not a maths background, so I might be missing some fundamental knowledge here.
              $endgroup$
              – UnchartedWaters
              Dec 4 '18 at 1:44


















            0












            $begingroup$

            Actually, with your symbols, $A A^T$ is non-singular, it is $m$ by $m.$



            The square $A^T A$ is a square matrix of size $n$ by $n$ but rank $m,$ therfore singular when $m < n$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thanks for your answer. But how do I prove that $AA^top$ is non-singular, in general? I'm from a CS background, not a maths background, so I might be missing some fundamental knowledge here.
              $endgroup$
              – UnchartedWaters
              Dec 4 '18 at 1:44
















            0












            0








            0





            $begingroup$

            Actually, with your symbols, $A A^T$ is non-singular, it is $m$ by $m.$



            The square $A^T A$ is a square matrix of size $n$ by $n$ but rank $m,$ therfore singular when $m < n$






            share|cite|improve this answer









            $endgroup$



            Actually, with your symbols, $A A^T$ is non-singular, it is $m$ by $m.$



            The square $A^T A$ is a square matrix of size $n$ by $n$ but rank $m,$ therfore singular when $m < n$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 4 '18 at 1:42









            Will JagyWill Jagy

            102k5101199




            102k5101199












            • $begingroup$
              Thanks for your answer. But how do I prove that $AA^top$ is non-singular, in general? I'm from a CS background, not a maths background, so I might be missing some fundamental knowledge here.
              $endgroup$
              – UnchartedWaters
              Dec 4 '18 at 1:44




















            • $begingroup$
              Thanks for your answer. But how do I prove that $AA^top$ is non-singular, in general? I'm from a CS background, not a maths background, so I might be missing some fundamental knowledge here.
              $endgroup$
              – UnchartedWaters
              Dec 4 '18 at 1:44


















            $begingroup$
            Thanks for your answer. But how do I prove that $AA^top$ is non-singular, in general? I'm from a CS background, not a maths background, so I might be missing some fundamental knowledge here.
            $endgroup$
            – UnchartedWaters
            Dec 4 '18 at 1:44






            $begingroup$
            Thanks for your answer. But how do I prove that $AA^top$ is non-singular, in general? I'm from a CS background, not a maths background, so I might be missing some fundamental knowledge here.
            $endgroup$
            – UnchartedWaters
            Dec 4 '18 at 1:44













            0












            $begingroup$

            Let $$A= begin{bmatrix}1 & 0 end{bmatrix}$$



            then we have $AA^T=1$.



            In general,
            $$rank(AA^T)=rank(A)=m,$$



            hence, it is nonsingular.






            share|cite|improve this answer











            $endgroup$


















              0












              $begingroup$

              Let $$A= begin{bmatrix}1 & 0 end{bmatrix}$$



              then we have $AA^T=1$.



              In general,
              $$rank(AA^T)=rank(A)=m,$$



              hence, it is nonsingular.






              share|cite|improve this answer











              $endgroup$
















                0












                0








                0





                $begingroup$

                Let $$A= begin{bmatrix}1 & 0 end{bmatrix}$$



                then we have $AA^T=1$.



                In general,
                $$rank(AA^T)=rank(A)=m,$$



                hence, it is nonsingular.






                share|cite|improve this answer











                $endgroup$



                Let $$A= begin{bmatrix}1 & 0 end{bmatrix}$$



                then we have $AA^T=1$.



                In general,
                $$rank(AA^T)=rank(A)=m,$$



                hence, it is nonsingular.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 4 '18 at 1:55

























                answered Dec 4 '18 at 1:41









                Siong Thye GohSiong Thye Goh

                100k1466117




                100k1466117






























                    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%2f3025002%2finverse-of-symmetric-matrix-aa-top-where-ranka-m-leq-n%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

                    Quarter-circle Tiles

                    Image
                    up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2
                    Read more

                    build a pushdown automaton that recognizes the reverse language of a given pushdown automaton?

                    Image
                    0 I think it's similiar to NFAs. I replace the finite states of the given automaton for start-states for my new automaton. I do it with $epsilon$ -transition from the start state to the actual finite states of the given automaton. The start state of the given automaton will be my accepting(finite) state of my new one. All transitions will be reversed of course to recognize the reversed word. But how do I have to change the stack of the pushdown automaton? That's tricky. formal-languages automata context-free-grammar share | cite | improve this question edited Nov 29 '18 at 14:20 joee
                    Read more

                    Mont Emei

                    Image
                    Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog
                    Read more