Is a square zero matrix positive semidefinite?












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Does the fact that a square zero matrix contains non-negative eigenvalues (zeros) make it proper to say it is positive semidefinite?






linear-algebra matrices positive-semidefinite







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    $begingroup$


    Does the fact that a square zero matrix contains non-negative eigenvalues (zeros) make it proper to say it is positive semidefinite?






    linear-algebra matrices positive-semidefinite







    share|cite|improve this question











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      2












      2








      2





      $begingroup$


      Does the fact that a square zero matrix contains non-negative eigenvalues (zeros) make it proper to say it is positive semidefinite?






      linear-algebra matrices positive-semidefinite







      share|cite|improve this question











      $endgroup$




      Does the fact that a square zero matrix contains non-negative eigenvalues (zeros) make it proper to say it is positive semidefinite?






      linear-algebra matrices positive-semidefinite




      linear-algebra matrices positive-semidefinite






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      share|cite|improve this question













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      edited yesterday







      Kay

















      asked yesterday









      KayKay

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          The $n times n$ zero matrix is positive semidefinite and negative semidefinite.






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            $begingroup$

            "When in doubt, go back to the basic definitions"! The definition of "positive semi-definite" is "all eigen-values are non-negative". The eigenvalues or the zero matrix are all 0 so, yes, the zero matrix is positive semi-definite. And, as Gary Moon said, it is also negative semi-definite.






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            • 2




              $begingroup$
              Nitpicking, the definition of "positive-semidefinite" is "a symmetric matrix $A$ with $vec x^{mathsf T}!Avec x ge 0$ for all $vec x in mathbb R^n$." The eigenvalue test is equivalent to this, but is not the definition.
              $endgroup$
              – Misha Lavrov
              21 hours ago













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            2 Answers
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            2 Answers
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            active

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            active

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            active

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            6












            $begingroup$

            The $n times n$ zero matrix is positive semidefinite and negative semidefinite.






            share|cite|improve this answer









            $endgroup$


















              6












              $begingroup$

              The $n times n$ zero matrix is positive semidefinite and negative semidefinite.






              share|cite|improve this answer









              $endgroup$
















                6












                6








                6





                $begingroup$

                The $n times n$ zero matrix is positive semidefinite and negative semidefinite.






                share|cite|improve this answer









                $endgroup$



                The $n times n$ zero matrix is positive semidefinite and negative semidefinite.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered yesterday









                Gary MoonGary Moon

                42613




                42613























                    4












                    $begingroup$

                    "When in doubt, go back to the basic definitions"! The definition of "positive semi-definite" is "all eigen-values are non-negative". The eigenvalues or the zero matrix are all 0 so, yes, the zero matrix is positive semi-definite. And, as Gary Moon said, it is also negative semi-definite.






                    share|cite|improve this answer









                    $endgroup$









                    • 2




                      $begingroup$
                      Nitpicking, the definition of "positive-semidefinite" is "a symmetric matrix $A$ with $vec x^{mathsf T}!Avec x ge 0$ for all $vec x in mathbb R^n$." The eigenvalue test is equivalent to this, but is not the definition.
                      $endgroup$
                      – Misha Lavrov
                      21 hours ago


















                    4












                    $begingroup$

                    "When in doubt, go back to the basic definitions"! The definition of "positive semi-definite" is "all eigen-values are non-negative". The eigenvalues or the zero matrix are all 0 so, yes, the zero matrix is positive semi-definite. And, as Gary Moon said, it is also negative semi-definite.






                    share|cite|improve this answer









                    $endgroup$









                    • 2




                      $begingroup$
                      Nitpicking, the definition of "positive-semidefinite" is "a symmetric matrix $A$ with $vec x^{mathsf T}!Avec x ge 0$ for all $vec x in mathbb R^n$." The eigenvalue test is equivalent to this, but is not the definition.
                      $endgroup$
                      – Misha Lavrov
                      21 hours ago
















                    4












                    4








                    4





                    $begingroup$

                    "When in doubt, go back to the basic definitions"! The definition of "positive semi-definite" is "all eigen-values are non-negative". The eigenvalues or the zero matrix are all 0 so, yes, the zero matrix is positive semi-definite. And, as Gary Moon said, it is also negative semi-definite.






                    share|cite|improve this answer









                    $endgroup$



                    "When in doubt, go back to the basic definitions"! The definition of "positive semi-definite" is "all eigen-values are non-negative". The eigenvalues or the zero matrix are all 0 so, yes, the zero matrix is positive semi-definite. And, as Gary Moon said, it is also negative semi-definite.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered yesterday









                    user247327user247327

                    11.5k1516




                    11.5k1516








                    • 2




                      $begingroup$
                      Nitpicking, the definition of "positive-semidefinite" is "a symmetric matrix $A$ with $vec x^{mathsf T}!Avec x ge 0$ for all $vec x in mathbb R^n$." The eigenvalue test is equivalent to this, but is not the definition.
                      $endgroup$
                      – Misha Lavrov
                      21 hours ago
















                    • 2




                      $begingroup$
                      Nitpicking, the definition of "positive-semidefinite" is "a symmetric matrix $A$ with $vec x^{mathsf T}!Avec x ge 0$ for all $vec x in mathbb R^n$." The eigenvalue test is equivalent to this, but is not the definition.
                      $endgroup$
                      – Misha Lavrov
                      21 hours ago










                    2




                    2




                    $begingroup$
                    Nitpicking, the definition of "positive-semidefinite" is "a symmetric matrix $A$ with $vec x^{mathsf T}!Avec x ge 0$ for all $vec x in mathbb R^n$." The eigenvalue test is equivalent to this, but is not the definition.
                    $endgroup$
                    – Misha Lavrov
                    21 hours ago






                    $begingroup$
                    Nitpicking, the definition of "positive-semidefinite" is "a symmetric matrix $A$ with $vec x^{mathsf T}!Avec x ge 0$ for all $vec x in mathbb R^n$." The eigenvalue test is equivalent to this, but is not the definition.
                    $endgroup$
                    – Misha Lavrov
                    21 hours ago




















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