If a matrix $Q$ is symmetric and positie definite, is it possible to show that the matrix...












6














If I have a symmetric and positive definite $ntimes n$ matrix $Q$ and a full row-rank totally unimodular $mtimes n$, where $m<n$, matrix $A$, is it posible to show that the matrix
$$Q-A^T(AQ^{-1}A^T)^{-1}A$$
is also positive definite?



I have show that it is true when the matrix $Q$ has dimention $2times 2$, by doing all the posible situations. Also, If $m=n$, I have that the matrix $Q-A^T(AQ^{-1}A^T)^{-1}A$ can be zero.






positive-definite positive-semidefinite







share|cite|improve this question





























    6














    If I have a symmetric and positive definite $ntimes n$ matrix $Q$ and a full row-rank totally unimodular $mtimes n$, where $m<n$, matrix $A$, is it posible to show that the matrix
    $$Q-A^T(AQ^{-1}A^T)^{-1}A$$
    is also positive definite?



    I have show that it is true when the matrix $Q$ has dimention $2times 2$, by doing all the posible situations. Also, If $m=n$, I have that the matrix $Q-A^T(AQ^{-1}A^T)^{-1}A$ can be zero.






    positive-definite positive-semidefinite







    share|cite|improve this question



























      6












      6








      6


      1





      If I have a symmetric and positive definite $ntimes n$ matrix $Q$ and a full row-rank totally unimodular $mtimes n$, where $m<n$, matrix $A$, is it posible to show that the matrix
      $$Q-A^T(AQ^{-1}A^T)^{-1}A$$
      is also positive definite?



      I have show that it is true when the matrix $Q$ has dimention $2times 2$, by doing all the posible situations. Also, If $m=n$, I have that the matrix $Q-A^T(AQ^{-1}A^T)^{-1}A$ can be zero.






      positive-definite positive-semidefinite







      share|cite|improve this question















      If I have a symmetric and positive definite $ntimes n$ matrix $Q$ and a full row-rank totally unimodular $mtimes n$, where $m<n$, matrix $A$, is it posible to show that the matrix
      $$Q-A^T(AQ^{-1}A^T)^{-1}A$$
      is also positive definite?



      I have show that it is true when the matrix $Q$ has dimention $2times 2$, by doing all the posible situations. Also, If $m=n$, I have that the matrix $Q-A^T(AQ^{-1}A^T)^{-1}A$ can be zero.






      positive-definite positive-semidefinite




      positive-definite positive-semidefinite






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 26 at 21:32

























      asked Nov 26 at 19:50









      GuadalupeAnimation

      30719




      30719






















          1 Answer
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          3














          The matrix $Q - A^T(AQ^{-1}A^T)^{-1}A$ will generally be positive semidefinite. Let



          $$M/Q : = Q - A^T(AQ^{-1}A^T)^{-1}A$$



          and note that $M/Q$ is the lower Schur complement of the block matrix



          $$
          M=
          begin{bmatrix}
          AQ^{-1}A^T & A \
          A^T & Q\
          end{bmatrix}.
          $$



          Since $Q$ is positive definite, it follows that $Q^{-1}$ is positive definite and that $Q^{1/2}$ exists. Furthermore $(Q^{-1})^{1/2}$ exists, hence let $Q^{-1/2} := (Q^{-1})^{1/2}.$ Note that



          $$
          begin{bmatrix}
          AQ^{-1}A^T & A \
          A^T & Q\
          end{bmatrix}
          =
          begin{bmatrix}
          AQ^{-1/2} \
          Q^{1/2}\
          end{bmatrix}
          begin{bmatrix}
          Q^{-1/2}A^T & Q^{1/2} \
          end{bmatrix}.
          $$



          Therefore $M$ is positive semidefinite. Since $M$ is positive semidefinite, so too is its lower Schur complement.






          share|cite|improve this answer























          • Why does $Q^{-1}$ being positive definite imply that $M$ is positive definite?
            – Omnomnomnom
            Nov 26 at 20:03










          • @Omnomnomnom you're right; it should be positive SEMIdefinite.
            – SZN
            Nov 26 at 20:05










          • then why does $Q^{-1}$ being positive definite imply that $M$ is positive semi-definite? Still don't see it.
            – Omnomnomnom
            Nov 26 at 20:07










          • @Omnomnomnom thank you for your comment. I've tried to add more detail.
            – SZN
            Nov 26 at 20:27










          • for non-symmetric matrices, non-negative eigenvalues alone do not imply that the matrix is positive definite.
            – Omnomnomnom
            Nov 26 at 20:44











          Your Answer





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          1 Answer
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          1 Answer
          1






          active

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          active

          oldest

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          active

          oldest

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          3














          The matrix $Q - A^T(AQ^{-1}A^T)^{-1}A$ will generally be positive semidefinite. Let



          $$M/Q : = Q - A^T(AQ^{-1}A^T)^{-1}A$$



          and note that $M/Q$ is the lower Schur complement of the block matrix



          $$
          M=
          begin{bmatrix}
          AQ^{-1}A^T & A \
          A^T & Q\
          end{bmatrix}.
          $$



          Since $Q$ is positive definite, it follows that $Q^{-1}$ is positive definite and that $Q^{1/2}$ exists. Furthermore $(Q^{-1})^{1/2}$ exists, hence let $Q^{-1/2} := (Q^{-1})^{1/2}.$ Note that



          $$
          begin{bmatrix}
          AQ^{-1}A^T & A \
          A^T & Q\
          end{bmatrix}
          =
          begin{bmatrix}
          AQ^{-1/2} \
          Q^{1/2}\
          end{bmatrix}
          begin{bmatrix}
          Q^{-1/2}A^T & Q^{1/2} \
          end{bmatrix}.
          $$



          Therefore $M$ is positive semidefinite. Since $M$ is positive semidefinite, so too is its lower Schur complement.






          share|cite|improve this answer























          • Why does $Q^{-1}$ being positive definite imply that $M$ is positive definite?
            – Omnomnomnom
            Nov 26 at 20:03










          • @Omnomnomnom you're right; it should be positive SEMIdefinite.
            – SZN
            Nov 26 at 20:05










          • then why does $Q^{-1}$ being positive definite imply that $M$ is positive semi-definite? Still don't see it.
            – Omnomnomnom
            Nov 26 at 20:07










          • @Omnomnomnom thank you for your comment. I've tried to add more detail.
            – SZN
            Nov 26 at 20:27










          • for non-symmetric matrices, non-negative eigenvalues alone do not imply that the matrix is positive definite.
            – Omnomnomnom
            Nov 26 at 20:44
















          3














          The matrix $Q - A^T(AQ^{-1}A^T)^{-1}A$ will generally be positive semidefinite. Let



          $$M/Q : = Q - A^T(AQ^{-1}A^T)^{-1}A$$



          and note that $M/Q$ is the lower Schur complement of the block matrix



          $$
          M=
          begin{bmatrix}
          AQ^{-1}A^T & A \
          A^T & Q\
          end{bmatrix}.
          $$



          Since $Q$ is positive definite, it follows that $Q^{-1}$ is positive definite and that $Q^{1/2}$ exists. Furthermore $(Q^{-1})^{1/2}$ exists, hence let $Q^{-1/2} := (Q^{-1})^{1/2}.$ Note that



          $$
          begin{bmatrix}
          AQ^{-1}A^T & A \
          A^T & Q\
          end{bmatrix}
          =
          begin{bmatrix}
          AQ^{-1/2} \
          Q^{1/2}\
          end{bmatrix}
          begin{bmatrix}
          Q^{-1/2}A^T & Q^{1/2} \
          end{bmatrix}.
          $$



          Therefore $M$ is positive semidefinite. Since $M$ is positive semidefinite, so too is its lower Schur complement.






          share|cite|improve this answer























          • Why does $Q^{-1}$ being positive definite imply that $M$ is positive definite?
            – Omnomnomnom
            Nov 26 at 20:03










          • @Omnomnomnom you're right; it should be positive SEMIdefinite.
            – SZN
            Nov 26 at 20:05










          • then why does $Q^{-1}$ being positive definite imply that $M$ is positive semi-definite? Still don't see it.
            – Omnomnomnom
            Nov 26 at 20:07










          • @Omnomnomnom thank you for your comment. I've tried to add more detail.
            – SZN
            Nov 26 at 20:27










          • for non-symmetric matrices, non-negative eigenvalues alone do not imply that the matrix is positive definite.
            – Omnomnomnom
            Nov 26 at 20:44














          3












          3








          3






          The matrix $Q - A^T(AQ^{-1}A^T)^{-1}A$ will generally be positive semidefinite. Let



          $$M/Q : = Q - A^T(AQ^{-1}A^T)^{-1}A$$



          and note that $M/Q$ is the lower Schur complement of the block matrix



          $$
          M=
          begin{bmatrix}
          AQ^{-1}A^T & A \
          A^T & Q\
          end{bmatrix}.
          $$



          Since $Q$ is positive definite, it follows that $Q^{-1}$ is positive definite and that $Q^{1/2}$ exists. Furthermore $(Q^{-1})^{1/2}$ exists, hence let $Q^{-1/2} := (Q^{-1})^{1/2}.$ Note that



          $$
          begin{bmatrix}
          AQ^{-1}A^T & A \
          A^T & Q\
          end{bmatrix}
          =
          begin{bmatrix}
          AQ^{-1/2} \
          Q^{1/2}\
          end{bmatrix}
          begin{bmatrix}
          Q^{-1/2}A^T & Q^{1/2} \
          end{bmatrix}.
          $$



          Therefore $M$ is positive semidefinite. Since $M$ is positive semidefinite, so too is its lower Schur complement.






          share|cite|improve this answer














          The matrix $Q - A^T(AQ^{-1}A^T)^{-1}A$ will generally be positive semidefinite. Let



          $$M/Q : = Q - A^T(AQ^{-1}A^T)^{-1}A$$



          and note that $M/Q$ is the lower Schur complement of the block matrix



          $$
          M=
          begin{bmatrix}
          AQ^{-1}A^T & A \
          A^T & Q\
          end{bmatrix}.
          $$



          Since $Q$ is positive definite, it follows that $Q^{-1}$ is positive definite and that $Q^{1/2}$ exists. Furthermore $(Q^{-1})^{1/2}$ exists, hence let $Q^{-1/2} := (Q^{-1})^{1/2}.$ Note that



          $$
          begin{bmatrix}
          AQ^{-1}A^T & A \
          A^T & Q\
          end{bmatrix}
          =
          begin{bmatrix}
          AQ^{-1/2} \
          Q^{1/2}\
          end{bmatrix}
          begin{bmatrix}
          Q^{-1/2}A^T & Q^{1/2} \
          end{bmatrix}.
          $$



          Therefore $M$ is positive semidefinite. Since $M$ is positive semidefinite, so too is its lower Schur complement.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 26 at 21:09

























          answered Nov 26 at 19:55









          SZN

          2,713720




          2,713720












          • Why does $Q^{-1}$ being positive definite imply that $M$ is positive definite?
            – Omnomnomnom
            Nov 26 at 20:03










          • @Omnomnomnom you're right; it should be positive SEMIdefinite.
            – SZN
            Nov 26 at 20:05










          • then why does $Q^{-1}$ being positive definite imply that $M$ is positive semi-definite? Still don't see it.
            – Omnomnomnom
            Nov 26 at 20:07










          • @Omnomnomnom thank you for your comment. I've tried to add more detail.
            – SZN
            Nov 26 at 20:27










          • for non-symmetric matrices, non-negative eigenvalues alone do not imply that the matrix is positive definite.
            – Omnomnomnom
            Nov 26 at 20:44


















          • Why does $Q^{-1}$ being positive definite imply that $M$ is positive definite?
            – Omnomnomnom
            Nov 26 at 20:03










          • @Omnomnomnom you're right; it should be positive SEMIdefinite.
            – SZN
            Nov 26 at 20:05










          • then why does $Q^{-1}$ being positive definite imply that $M$ is positive semi-definite? Still don't see it.
            – Omnomnomnom
            Nov 26 at 20:07










          • @Omnomnomnom thank you for your comment. I've tried to add more detail.
            – SZN
            Nov 26 at 20:27










          • for non-symmetric matrices, non-negative eigenvalues alone do not imply that the matrix is positive definite.
            – Omnomnomnom
            Nov 26 at 20:44
















          Why does $Q^{-1}$ being positive definite imply that $M$ is positive definite?
          – Omnomnomnom
          Nov 26 at 20:03




          Why does $Q^{-1}$ being positive definite imply that $M$ is positive definite?
          – Omnomnomnom
          Nov 26 at 20:03












          @Omnomnomnom you're right; it should be positive SEMIdefinite.
          – SZN
          Nov 26 at 20:05




          @Omnomnomnom you're right; it should be positive SEMIdefinite.
          – SZN
          Nov 26 at 20:05












          then why does $Q^{-1}$ being positive definite imply that $M$ is positive semi-definite? Still don't see it.
          – Omnomnomnom
          Nov 26 at 20:07




          then why does $Q^{-1}$ being positive definite imply that $M$ is positive semi-definite? Still don't see it.
          – Omnomnomnom
          Nov 26 at 20:07












          @Omnomnomnom thank you for your comment. I've tried to add more detail.
          – SZN
          Nov 26 at 20:27




          @Omnomnomnom thank you for your comment. I've tried to add more detail.
          – SZN
          Nov 26 at 20:27












          for non-symmetric matrices, non-negative eigenvalues alone do not imply that the matrix is positive definite.
          – Omnomnomnom
          Nov 26 at 20:44




          for non-symmetric matrices, non-negative eigenvalues alone do not imply that the matrix is positive definite.
          – Omnomnomnom
          Nov 26 at 20:44


















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