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Sum of symmetric, positive semidefinite matrices


Does this conic combination generate all $ntimes n$ real symmetric positive-semidefinite matrices?A is symmetric iff A=P-Q, where P,Q are positive definite matricesCriterion for positive semidefinite matricesAnother property of symmetric positive semidefinite matricesSum of rank-$k$ positive semidefinite matrices is at least of rank $k$Simultaneous Diagonalization of Symmetric Positive Semidefinite matricesA question about Hermitian and positive semidefinite matricesIs this matrix product positive semidefinite?Sum of rank 1 positive semidefinite and negative semidefinite matricesPositive semidefinite matrix using Schur Complement













0












$begingroup$



Let $A in mathbbR^m times n, B in mathbbR^p times n$. Show that $A^TA+ B^TB$ is invertible if and only if $ker A cap ker B =lbrace 0 rbrace$.




I could show that if it's invertible, then $ker A cap ker B= lbrace 0 rbrace$. Any help for the converse?










share|cite|improve this question











$endgroup$











  • $begingroup$
    How did you show the forward direction? Possibly the converse can be shown similarly.
    $endgroup$
    – Minus One-Twelfth
    yesterday















0












$begingroup$



Let $A in mathbbR^m times n, B in mathbbR^p times n$. Show that $A^TA+ B^TB$ is invertible if and only if $ker A cap ker B =lbrace 0 rbrace$.




I could show that if it's invertible, then $ker A cap ker B= lbrace 0 rbrace$. Any help for the converse?










share|cite|improve this question











$endgroup$











  • $begingroup$
    How did you show the forward direction? Possibly the converse can be shown similarly.
    $endgroup$
    – Minus One-Twelfth
    yesterday













0












0








0





$begingroup$



Let $A in mathbbR^m times n, B in mathbbR^p times n$. Show that $A^TA+ B^TB$ is invertible if and only if $ker A cap ker B =lbrace 0 rbrace$.




I could show that if it's invertible, then $ker A cap ker B= lbrace 0 rbrace$. Any help for the converse?










share|cite|improve this question











$endgroup$





Let $A in mathbbR^m times n, B in mathbbR^p times n$. Show that $A^TA+ B^TB$ is invertible if and only if $ker A cap ker B =lbrace 0 rbrace$.




I could show that if it's invertible, then $ker A cap ker B= lbrace 0 rbrace$. Any help for the converse?







linear-algebra matrices symmetric-matrices positive-semidefinite






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited yesterday









Rodrigo de Azevedo

13k41960




13k41960










asked yesterday









mich95mich95

6,97211126




6,97211126











  • $begingroup$
    How did you show the forward direction? Possibly the converse can be shown similarly.
    $endgroup$
    – Minus One-Twelfth
    yesterday
















  • $begingroup$
    How did you show the forward direction? Possibly the converse can be shown similarly.
    $endgroup$
    – Minus One-Twelfth
    yesterday















$begingroup$
How did you show the forward direction? Possibly the converse can be shown similarly.
$endgroup$
– Minus One-Twelfth
yesterday




$begingroup$
How did you show the forward direction? Possibly the converse can be shown similarly.
$endgroup$
– Minus One-Twelfth
yesterday










2 Answers
2






active

oldest

votes


















0












$begingroup$

$newcommandmathbf0newcommandxmathbfx$Hints: Recall or try and show that $colorbluex^T M^T Mx = 0text iff x in ker M$ for any matrix $M$, and recall that for any square matrix, it is invertible iff its kernel is $$. Then to show the converse, your goal is to show that if $ker A cap ker B = $, then $A^T A + B^T B$ has kernel $ $. To show this, suppose that $x in kerleft(A^T A + B^T Bright)$ and try and deduce that $x = $. Note also that $x^T left(A^T A + B^T Bright) x = x^T A^T A x + x^T B^T Bx$.






share|cite|improve this answer











$endgroup$




















    1












    $begingroup$

    Here's an approach: suppose that $A^TA + B^TB$ is not invertible. Then, there exists a non-zero vector $x$ such that $(A^TA + B^TB)x = 0$. It follows that
    $$
    0 = x^T(A^TA + B^TB)x = (Ax)^T(Ax) + (Bx)^T(Bx) = |Ax|^2 + |Bx|^2
    $$

    conclude that $x in ker(A) cap ker(B)$.






    share|cite|improve this answer









    $endgroup$












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      2 Answers
      2






      active

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      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      0












      $begingroup$

      $newcommandmathbf0newcommandxmathbfx$Hints: Recall or try and show that $colorbluex^T M^T Mx = 0text iff x in ker M$ for any matrix $M$, and recall that for any square matrix, it is invertible iff its kernel is $$. Then to show the converse, your goal is to show that if $ker A cap ker B = $, then $A^T A + B^T B$ has kernel $ $. To show this, suppose that $x in kerleft(A^T A + B^T Bright)$ and try and deduce that $x = $. Note also that $x^T left(A^T A + B^T Bright) x = x^T A^T A x + x^T B^T Bx$.






      share|cite|improve this answer











      $endgroup$

















        0












        $begingroup$

        $newcommandmathbf0newcommandxmathbfx$Hints: Recall or try and show that $colorbluex^T M^T Mx = 0text iff x in ker M$ for any matrix $M$, and recall that for any square matrix, it is invertible iff its kernel is $$. Then to show the converse, your goal is to show that if $ker A cap ker B = $, then $A^T A + B^T B$ has kernel $ $. To show this, suppose that $x in kerleft(A^T A + B^T Bright)$ and try and deduce that $x = $. Note also that $x^T left(A^T A + B^T Bright) x = x^T A^T A x + x^T B^T Bx$.






        share|cite|improve this answer











        $endgroup$















          0












          0








          0





          $begingroup$

          $newcommandmathbf0newcommandxmathbfx$Hints: Recall or try and show that $colorbluex^T M^T Mx = 0text iff x in ker M$ for any matrix $M$, and recall that for any square matrix, it is invertible iff its kernel is $$. Then to show the converse, your goal is to show that if $ker A cap ker B = $, then $A^T A + B^T B$ has kernel $ $. To show this, suppose that $x in kerleft(A^T A + B^T Bright)$ and try and deduce that $x = $. Note also that $x^T left(A^T A + B^T Bright) x = x^T A^T A x + x^T B^T Bx$.






          share|cite|improve this answer











          $endgroup$



          $newcommandmathbf0newcommandxmathbfx$Hints: Recall or try and show that $colorbluex^T M^T Mx = 0text iff x in ker M$ for any matrix $M$, and recall that for any square matrix, it is invertible iff its kernel is $$. Then to show the converse, your goal is to show that if $ker A cap ker B = $, then $A^T A + B^T B$ has kernel $ $. To show this, suppose that $x in kerleft(A^T A + B^T Bright)$ and try and deduce that $x = $. Note also that $x^T left(A^T A + B^T Bright) x = x^T A^T A x + x^T B^T Bx$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited yesterday

























          answered yesterday









          Minus One-TwelfthMinus One-Twelfth

          2,07219




          2,07219





















              1












              $begingroup$

              Here's an approach: suppose that $A^TA + B^TB$ is not invertible. Then, there exists a non-zero vector $x$ such that $(A^TA + B^TB)x = 0$. It follows that
              $$
              0 = x^T(A^TA + B^TB)x = (Ax)^T(Ax) + (Bx)^T(Bx) = |Ax|^2 + |Bx|^2
              $$

              conclude that $x in ker(A) cap ker(B)$.






              share|cite|improve this answer









              $endgroup$

















                1












                $begingroup$

                Here's an approach: suppose that $A^TA + B^TB$ is not invertible. Then, there exists a non-zero vector $x$ such that $(A^TA + B^TB)x = 0$. It follows that
                $$
                0 = x^T(A^TA + B^TB)x = (Ax)^T(Ax) + (Bx)^T(Bx) = |Ax|^2 + |Bx|^2
                $$

                conclude that $x in ker(A) cap ker(B)$.






                share|cite|improve this answer









                $endgroup$















                  1












                  1








                  1





                  $begingroup$

                  Here's an approach: suppose that $A^TA + B^TB$ is not invertible. Then, there exists a non-zero vector $x$ such that $(A^TA + B^TB)x = 0$. It follows that
                  $$
                  0 = x^T(A^TA + B^TB)x = (Ax)^T(Ax) + (Bx)^T(Bx) = |Ax|^2 + |Bx|^2
                  $$

                  conclude that $x in ker(A) cap ker(B)$.






                  share|cite|improve this answer









                  $endgroup$



                  Here's an approach: suppose that $A^TA + B^TB$ is not invertible. Then, there exists a non-zero vector $x$ such that $(A^TA + B^TB)x = 0$. It follows that
                  $$
                  0 = x^T(A^TA + B^TB)x = (Ax)^T(Ax) + (Bx)^T(Bx) = |Ax|^2 + |Bx|^2
                  $$

                  conclude that $x in ker(A) cap ker(B)$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered yesterday









                  OmnomnomnomOmnomnomnom

                  128k791186




                  128k791186



























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