Necessary and sufficient conditions for $x'Ax = 0$












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I came across the following problem and I am having a hard time thinking about it.




Let $A$ be a $ktimes k$ real matrix. Notice that I do not require that $A$ is symmetric, positive definite or anything else. I would like to consider any real matrix $A$ of such dimensions.



Now, I am interested in necessary and sufficient conditions for $exists x neq 0$ such that $ x' A x = 0$, where $x in mathbb{R}^{k}$.




Is this a known result? Any ideas?






linear-algebra matrices reference-request quadratic-forms







share|cite|improve this question









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    0












    $begingroup$


    I came across the following problem and I am having a hard time thinking about it.




    Let $A$ be a $ktimes k$ real matrix. Notice that I do not require that $A$ is symmetric, positive definite or anything else. I would like to consider any real matrix $A$ of such dimensions.



    Now, I am interested in necessary and sufficient conditions for $exists x neq 0$ such that $ x' A x = 0$, where $x in mathbb{R}^{k}$.




    Is this a known result? Any ideas?






    linear-algebra matrices reference-request quadratic-forms







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      I came across the following problem and I am having a hard time thinking about it.




      Let $A$ be a $ktimes k$ real matrix. Notice that I do not require that $A$ is symmetric, positive definite or anything else. I would like to consider any real matrix $A$ of such dimensions.



      Now, I am interested in necessary and sufficient conditions for $exists x neq 0$ such that $ x' A x = 0$, where $x in mathbb{R}^{k}$.




      Is this a known result? Any ideas?






      linear-algebra matrices reference-request quadratic-forms







      share|cite|improve this question









      $endgroup$




      I came across the following problem and I am having a hard time thinking about it.




      Let $A$ be a $ktimes k$ real matrix. Notice that I do not require that $A$ is symmetric, positive definite or anything else. I would like to consider any real matrix $A$ of such dimensions.



      Now, I am interested in necessary and sufficient conditions for $exists x neq 0$ such that $ x' A x = 0$, where $x in mathbb{R}^{k}$.




      Is this a known result? Any ideas?






      linear-algebra matrices reference-request quadratic-forms




      linear-algebra matrices reference-request quadratic-forms






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 8 '18 at 15:16









      Raul GuariniRaul Guarini

      460211




      460211






















          1 Answer
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          $begingroup$

          Note that$$x^TAx = sum_{i,j} A_{ij}x_i x_j = sum_{i,j} 0.5(A_{ij}+A_{ji}) x_i x_j = 0.5x^T(A+A^T)x.$$
          The second equality is due to each product $x_ix_j$ occuring twice in the summation when $ineq j$. So the question is whether the symmetric matrix $A+A^T$ is positive definite.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            If it's not positive definite, then we can apply some sort of continuity argument and find such $x$, right?
            $endgroup$
            – Raul Guarini
            Dec 8 '18 at 16:13










          • $begingroup$
            @RaulGuarini no continuity argument is needed, any eigenvector $x$ belonging to a nonpositive eigenvalue of $A+A^T$ will have $x^TAx leq 0$.
            $endgroup$
            – LinAlg
            Dec 8 '18 at 16:14












          • $begingroup$
            True! Thanks a lot!
            $endgroup$
            – Raul Guarini
            Dec 8 '18 at 16:18











          Your Answer





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          1 Answer
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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          0












          $begingroup$

          Note that$$x^TAx = sum_{i,j} A_{ij}x_i x_j = sum_{i,j} 0.5(A_{ij}+A_{ji}) x_i x_j = 0.5x^T(A+A^T)x.$$
          The second equality is due to each product $x_ix_j$ occuring twice in the summation when $ineq j$. So the question is whether the symmetric matrix $A+A^T$ is positive definite.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            If it's not positive definite, then we can apply some sort of continuity argument and find such $x$, right?
            $endgroup$
            – Raul Guarini
            Dec 8 '18 at 16:13










          • $begingroup$
            @RaulGuarini no continuity argument is needed, any eigenvector $x$ belonging to a nonpositive eigenvalue of $A+A^T$ will have $x^TAx leq 0$.
            $endgroup$
            – LinAlg
            Dec 8 '18 at 16:14












          • $begingroup$
            True! Thanks a lot!
            $endgroup$
            – Raul Guarini
            Dec 8 '18 at 16:18
















          0












          $begingroup$

          Note that$$x^TAx = sum_{i,j} A_{ij}x_i x_j = sum_{i,j} 0.5(A_{ij}+A_{ji}) x_i x_j = 0.5x^T(A+A^T)x.$$
          The second equality is due to each product $x_ix_j$ occuring twice in the summation when $ineq j$. So the question is whether the symmetric matrix $A+A^T$ is positive definite.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            If it's not positive definite, then we can apply some sort of continuity argument and find such $x$, right?
            $endgroup$
            – Raul Guarini
            Dec 8 '18 at 16:13










          • $begingroup$
            @RaulGuarini no continuity argument is needed, any eigenvector $x$ belonging to a nonpositive eigenvalue of $A+A^T$ will have $x^TAx leq 0$.
            $endgroup$
            – LinAlg
            Dec 8 '18 at 16:14












          • $begingroup$
            True! Thanks a lot!
            $endgroup$
            – Raul Guarini
            Dec 8 '18 at 16:18














          0












          0








          0





          $begingroup$

          Note that$$x^TAx = sum_{i,j} A_{ij}x_i x_j = sum_{i,j} 0.5(A_{ij}+A_{ji}) x_i x_j = 0.5x^T(A+A^T)x.$$
          The second equality is due to each product $x_ix_j$ occuring twice in the summation when $ineq j$. So the question is whether the symmetric matrix $A+A^T$ is positive definite.






          share|cite|improve this answer









          $endgroup$



          Note that$$x^TAx = sum_{i,j} A_{ij}x_i x_j = sum_{i,j} 0.5(A_{ij}+A_{ji}) x_i x_j = 0.5x^T(A+A^T)x.$$
          The second equality is due to each product $x_ix_j$ occuring twice in the summation when $ineq j$. So the question is whether the symmetric matrix $A+A^T$ is positive definite.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 8 '18 at 16:04









          LinAlgLinAlg

          9,2711521




          9,2711521












          • $begingroup$
            If it's not positive definite, then we can apply some sort of continuity argument and find such $x$, right?
            $endgroup$
            – Raul Guarini
            Dec 8 '18 at 16:13










          • $begingroup$
            @RaulGuarini no continuity argument is needed, any eigenvector $x$ belonging to a nonpositive eigenvalue of $A+A^T$ will have $x^TAx leq 0$.
            $endgroup$
            – LinAlg
            Dec 8 '18 at 16:14












          • $begingroup$
            True! Thanks a lot!
            $endgroup$
            – Raul Guarini
            Dec 8 '18 at 16:18


















          • $begingroup$
            If it's not positive definite, then we can apply some sort of continuity argument and find such $x$, right?
            $endgroup$
            – Raul Guarini
            Dec 8 '18 at 16:13










          • $begingroup$
            @RaulGuarini no continuity argument is needed, any eigenvector $x$ belonging to a nonpositive eigenvalue of $A+A^T$ will have $x^TAx leq 0$.
            $endgroup$
            – LinAlg
            Dec 8 '18 at 16:14












          • $begingroup$
            True! Thanks a lot!
            $endgroup$
            – Raul Guarini
            Dec 8 '18 at 16:18
















          $begingroup$
          If it's not positive definite, then we can apply some sort of continuity argument and find such $x$, right?
          $endgroup$
          – Raul Guarini
          Dec 8 '18 at 16:13




          $begingroup$
          If it's not positive definite, then we can apply some sort of continuity argument and find such $x$, right?
          $endgroup$
          – Raul Guarini
          Dec 8 '18 at 16:13












          $begingroup$
          @RaulGuarini no continuity argument is needed, any eigenvector $x$ belonging to a nonpositive eigenvalue of $A+A^T$ will have $x^TAx leq 0$.
          $endgroup$
          – LinAlg
          Dec 8 '18 at 16:14






          $begingroup$
          @RaulGuarini no continuity argument is needed, any eigenvector $x$ belonging to a nonpositive eigenvalue of $A+A^T$ will have $x^TAx leq 0$.
          $endgroup$
          – LinAlg
          Dec 8 '18 at 16:14














          $begingroup$
          True! Thanks a lot!
          $endgroup$
          – Raul Guarini
          Dec 8 '18 at 16:18




          $begingroup$
          True! Thanks a lot!
          $endgroup$
          – Raul Guarini
          Dec 8 '18 at 16:18


















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