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

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          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











          $endgroup$













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            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

            63.5k44591




            63.5k44591












            • $begingroup$
              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

            103k5101200




            103k5101200












            • $begingroup$
              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

                101k1466117




                101k1466117






























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