Proving that $text{tr}^2(A) - 4det(A) geq 0$












-1












$begingroup$


I'm trying to figure out how to prove that when given a diagonal matrix



begin{align}
A=
begin{bmatrix}
lambda & 0\
0 & mu
end{bmatrix},
end{align}
with a positive determinant $det(A) > 0$, the following statement is true
$$text{tr}^2(A) - 4det(A) geq 0$$
Thank you in advance.






linear-algebra determinant







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Have you tried writing the trace and determinant in terms of $lambda$ and $mu$ and then doing a little algebra?
    $endgroup$
    – Gerry Myerson
    Nov 30 '18 at 11:36










  • $begingroup$
    I have, I get $frac{lambda^2+mu^2-2lambdamu}{4}>lambdamu.$... I can't really figure what to do from here.
    $endgroup$
    – Jens Kramer
    Nov 30 '18 at 11:39








  • 1




    $begingroup$
    I think you've slipped up in your algebra. Also, that $lambda^2+mu^2-2lambdamu$ looks familiar....
    $endgroup$
    – Gerry Myerson
    Nov 30 '18 at 11:41






  • 1




    $begingroup$
    Algebra, algebra! Please, please, please check your algebra!
    $endgroup$
    – Gerry Myerson
    Nov 30 '18 at 11:49






  • 2




    $begingroup$
    The expansion of the quadratic term tr(A)^2.. So $lambda^2+mu^2+2lambdamugeq 4lambdamu$ which the implies $lambda^2+mu^2-2lambdamugeq 0$. Simplifying further gets $(lambda-mu)^2geq0$. I'm not quite sure what this is telling me.
    $endgroup$
    – Jens Kramer
    Nov 30 '18 at 11:59


















-1












$begingroup$


I'm trying to figure out how to prove that when given a diagonal matrix



begin{align}
A=
begin{bmatrix}
lambda & 0\
0 & mu
end{bmatrix},
end{align}
with a positive determinant $det(A) > 0$, the following statement is true
$$text{tr}^2(A) - 4det(A) geq 0$$
Thank you in advance.






linear-algebra determinant







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Have you tried writing the trace and determinant in terms of $lambda$ and $mu$ and then doing a little algebra?
    $endgroup$
    – Gerry Myerson
    Nov 30 '18 at 11:36










  • $begingroup$
    I have, I get $frac{lambda^2+mu^2-2lambdamu}{4}>lambdamu.$... I can't really figure what to do from here.
    $endgroup$
    – Jens Kramer
    Nov 30 '18 at 11:39








  • 1




    $begingroup$
    I think you've slipped up in your algebra. Also, that $lambda^2+mu^2-2lambdamu$ looks familiar....
    $endgroup$
    – Gerry Myerson
    Nov 30 '18 at 11:41






  • 1




    $begingroup$
    Algebra, algebra! Please, please, please check your algebra!
    $endgroup$
    – Gerry Myerson
    Nov 30 '18 at 11:49






  • 2




    $begingroup$
    The expansion of the quadratic term tr(A)^2.. So $lambda^2+mu^2+2lambdamugeq 4lambdamu$ which the implies $lambda^2+mu^2-2lambdamugeq 0$. Simplifying further gets $(lambda-mu)^2geq0$. I'm not quite sure what this is telling me.
    $endgroup$
    – Jens Kramer
    Nov 30 '18 at 11:59
















-1












-1








-1





$begingroup$


I'm trying to figure out how to prove that when given a diagonal matrix



begin{align}
A=
begin{bmatrix}
lambda & 0\
0 & mu
end{bmatrix},
end{align}
with a positive determinant $det(A) > 0$, the following statement is true
$$text{tr}^2(A) - 4det(A) geq 0$$
Thank you in advance.






linear-algebra determinant







share|cite|improve this question











$endgroup$




I'm trying to figure out how to prove that when given a diagonal matrix



begin{align}
A=
begin{bmatrix}
lambda & 0\
0 & mu
end{bmatrix},
end{align}
with a positive determinant $det(A) > 0$, the following statement is true
$$text{tr}^2(A) - 4det(A) geq 0$$
Thank you in advance.






linear-algebra determinant




linear-algebra determinant






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 30 '18 at 11:43









Rebellos

14.6k31247




14.6k31247










asked Nov 30 '18 at 11:34









Jens KramerJens Kramer

557




557








  • 3




    $begingroup$
    Have you tried writing the trace and determinant in terms of $lambda$ and $mu$ and then doing a little algebra?
    $endgroup$
    – Gerry Myerson
    Nov 30 '18 at 11:36










  • $begingroup$
    I have, I get $frac{lambda^2+mu^2-2lambdamu}{4}>lambdamu.$... I can't really figure what to do from here.
    $endgroup$
    – Jens Kramer
    Nov 30 '18 at 11:39








  • 1




    $begingroup$
    I think you've slipped up in your algebra. Also, that $lambda^2+mu^2-2lambdamu$ looks familiar....
    $endgroup$
    – Gerry Myerson
    Nov 30 '18 at 11:41






  • 1




    $begingroup$
    Algebra, algebra! Please, please, please check your algebra!
    $endgroup$
    – Gerry Myerson
    Nov 30 '18 at 11:49






  • 2




    $begingroup$
    The expansion of the quadratic term tr(A)^2.. So $lambda^2+mu^2+2lambdamugeq 4lambdamu$ which the implies $lambda^2+mu^2-2lambdamugeq 0$. Simplifying further gets $(lambda-mu)^2geq0$. I'm not quite sure what this is telling me.
    $endgroup$
    – Jens Kramer
    Nov 30 '18 at 11:59
















  • 3




    $begingroup$
    Have you tried writing the trace and determinant in terms of $lambda$ and $mu$ and then doing a little algebra?
    $endgroup$
    – Gerry Myerson
    Nov 30 '18 at 11:36










  • $begingroup$
    I have, I get $frac{lambda^2+mu^2-2lambdamu}{4}>lambdamu.$... I can't really figure what to do from here.
    $endgroup$
    – Jens Kramer
    Nov 30 '18 at 11:39








  • 1




    $begingroup$
    I think you've slipped up in your algebra. Also, that $lambda^2+mu^2-2lambdamu$ looks familiar....
    $endgroup$
    – Gerry Myerson
    Nov 30 '18 at 11:41






  • 1




    $begingroup$
    Algebra, algebra! Please, please, please check your algebra!
    $endgroup$
    – Gerry Myerson
    Nov 30 '18 at 11:49






  • 2




    $begingroup$
    The expansion of the quadratic term tr(A)^2.. So $lambda^2+mu^2+2lambdamugeq 4lambdamu$ which the implies $lambda^2+mu^2-2lambdamugeq 0$. Simplifying further gets $(lambda-mu)^2geq0$. I'm not quite sure what this is telling me.
    $endgroup$
    – Jens Kramer
    Nov 30 '18 at 11:59










3




3




$begingroup$
Have you tried writing the trace and determinant in terms of $lambda$ and $mu$ and then doing a little algebra?
$endgroup$
– Gerry Myerson
Nov 30 '18 at 11:36




$begingroup$
Have you tried writing the trace and determinant in terms of $lambda$ and $mu$ and then doing a little algebra?
$endgroup$
– Gerry Myerson
Nov 30 '18 at 11:36












$begingroup$
I have, I get $frac{lambda^2+mu^2-2lambdamu}{4}>lambdamu.$... I can't really figure what to do from here.
$endgroup$
– Jens Kramer
Nov 30 '18 at 11:39






$begingroup$
I have, I get $frac{lambda^2+mu^2-2lambdamu}{4}>lambdamu.$... I can't really figure what to do from here.
$endgroup$
– Jens Kramer
Nov 30 '18 at 11:39






1




1




$begingroup$
I think you've slipped up in your algebra. Also, that $lambda^2+mu^2-2lambdamu$ looks familiar....
$endgroup$
– Gerry Myerson
Nov 30 '18 at 11:41




$begingroup$
I think you've slipped up in your algebra. Also, that $lambda^2+mu^2-2lambdamu$ looks familiar....
$endgroup$
– Gerry Myerson
Nov 30 '18 at 11:41




1




1




$begingroup$
Algebra, algebra! Please, please, please check your algebra!
$endgroup$
– Gerry Myerson
Nov 30 '18 at 11:49




$begingroup$
Algebra, algebra! Please, please, please check your algebra!
$endgroup$
– Gerry Myerson
Nov 30 '18 at 11:49




2




2




$begingroup$
The expansion of the quadratic term tr(A)^2.. So $lambda^2+mu^2+2lambdamugeq 4lambdamu$ which the implies $lambda^2+mu^2-2lambdamugeq 0$. Simplifying further gets $(lambda-mu)^2geq0$. I'm not quite sure what this is telling me.
$endgroup$
– Jens Kramer
Nov 30 '18 at 11:59






$begingroup$
The expansion of the quadratic term tr(A)^2.. So $lambda^2+mu^2+2lambdamugeq 4lambdamu$ which the implies $lambda^2+mu^2-2lambdamugeq 0$. Simplifying further gets $(lambda-mu)^2geq0$. I'm not quite sure what this is telling me.
$endgroup$
– Jens Kramer
Nov 30 '18 at 11:59












3 Answers
3






active

oldest

votes


















2












$begingroup$

More generally, the characteristic polynomial of a $2 times 2$ matrix $A$ is $x^2-operatorname{tr}(A)x + det(A)$. The roots of this polynomial are the eigenvalues of $A$ and so are real iff the discriminant $operatorname{tr}(A)^2 -4det(A)$ is nonnegative.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Hint :



    Simply using the definitions of $text{tr}$ and $det$ :



    The trace of the matrix $A$, is :



    $$text{tr}(A) = lambda + mu $$



    The determinant of $A$, simply is :



    $$det(A) = lambda mu$$



    Can you now follow a simple algebraic path to $text{tr}^2(A) - 4det(A)$ and conclude about its sign ? Use also the condition that as you stated, the determinant is positive.






    share|cite|improve this answer









    $endgroup$





















      -1












      $begingroup$

      $begin{align}
      tr(A)^2-4 det(A)geq 0
      end{align} iff (lambda + mu)^2 ge 4 lambda mu iff lambda^2 -2 lambda mu + mu^2 ge 0 iff (lambda- mu)^2 ge 0$      .






      share|cite|improve this answer









      $endgroup$









      • 2




        $begingroup$
        Simply elaborated, but why give a thorough solution to a simple problem of which the OP have showed zero attempts ?
        $endgroup$
        – Rebellos
        Nov 30 '18 at 11:39






      • 3




        $begingroup$
        Not leaving much for OP to do, Fred.
        $endgroup$
        – Gerry Myerson
        Nov 30 '18 at 11:42











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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      More generally, the characteristic polynomial of a $2 times 2$ matrix $A$ is $x^2-operatorname{tr}(A)x + det(A)$. The roots of this polynomial are the eigenvalues of $A$ and so are real iff the discriminant $operatorname{tr}(A)^2 -4det(A)$ is nonnegative.






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        More generally, the characteristic polynomial of a $2 times 2$ matrix $A$ is $x^2-operatorname{tr}(A)x + det(A)$. The roots of this polynomial are the eigenvalues of $A$ and so are real iff the discriminant $operatorname{tr}(A)^2 -4det(A)$ is nonnegative.






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          More generally, the characteristic polynomial of a $2 times 2$ matrix $A$ is $x^2-operatorname{tr}(A)x + det(A)$. The roots of this polynomial are the eigenvalues of $A$ and so are real iff the discriminant $operatorname{tr}(A)^2 -4det(A)$ is nonnegative.






          share|cite|improve this answer









          $endgroup$



          More generally, the characteristic polynomial of a $2 times 2$ matrix $A$ is $x^2-operatorname{tr}(A)x + det(A)$. The roots of this polynomial are the eigenvalues of $A$ and so are real iff the discriminant $operatorname{tr}(A)^2 -4det(A)$ is nonnegative.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 30 '18 at 12:07









          lhflhf

          165k10171396




          165k10171396























              0












              $begingroup$

              Hint :



              Simply using the definitions of $text{tr}$ and $det$ :



              The trace of the matrix $A$, is :



              $$text{tr}(A) = lambda + mu $$



              The determinant of $A$, simply is :



              $$det(A) = lambda mu$$



              Can you now follow a simple algebraic path to $text{tr}^2(A) - 4det(A)$ and conclude about its sign ? Use also the condition that as you stated, the determinant is positive.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Hint :



                Simply using the definitions of $text{tr}$ and $det$ :



                The trace of the matrix $A$, is :



                $$text{tr}(A) = lambda + mu $$



                The determinant of $A$, simply is :



                $$det(A) = lambda mu$$



                Can you now follow a simple algebraic path to $text{tr}^2(A) - 4det(A)$ and conclude about its sign ? Use also the condition that as you stated, the determinant is positive.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Hint :



                  Simply using the definitions of $text{tr}$ and $det$ :



                  The trace of the matrix $A$, is :



                  $$text{tr}(A) = lambda + mu $$



                  The determinant of $A$, simply is :



                  $$det(A) = lambda mu$$



                  Can you now follow a simple algebraic path to $text{tr}^2(A) - 4det(A)$ and conclude about its sign ? Use also the condition that as you stated, the determinant is positive.






                  share|cite|improve this answer









                  $endgroup$



                  Hint :



                  Simply using the definitions of $text{tr}$ and $det$ :



                  The trace of the matrix $A$, is :



                  $$text{tr}(A) = lambda + mu $$



                  The determinant of $A$, simply is :



                  $$det(A) = lambda mu$$



                  Can you now follow a simple algebraic path to $text{tr}^2(A) - 4det(A)$ and conclude about its sign ? Use also the condition that as you stated, the determinant is positive.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 30 '18 at 11:38









                  RebellosRebellos

                  14.6k31247




                  14.6k31247























                      -1












                      $begingroup$

                      $begin{align}
                      tr(A)^2-4 det(A)geq 0
                      end{align} iff (lambda + mu)^2 ge 4 lambda mu iff lambda^2 -2 lambda mu + mu^2 ge 0 iff (lambda- mu)^2 ge 0$                      .






                      share|cite|improve this answer









                      $endgroup$









                      • 2




                        $begingroup$
                        Simply elaborated, but why give a thorough solution to a simple problem of which the OP have showed zero attempts ?
                        $endgroup$
                        – Rebellos
                        Nov 30 '18 at 11:39






                      • 3




                        $begingroup$
                        Not leaving much for OP to do, Fred.
                        $endgroup$
                        – Gerry Myerson
                        Nov 30 '18 at 11:42
















                      -1












                      $begingroup$

                      $begin{align}
                      tr(A)^2-4 det(A)geq 0
                      end{align} iff (lambda + mu)^2 ge 4 lambda mu iff lambda^2 -2 lambda mu + mu^2 ge 0 iff (lambda- mu)^2 ge 0$                      .






                      share|cite|improve this answer









                      $endgroup$









                      • 2




                        $begingroup$
                        Simply elaborated, but why give a thorough solution to a simple problem of which the OP have showed zero attempts ?
                        $endgroup$
                        – Rebellos
                        Nov 30 '18 at 11:39






                      • 3




                        $begingroup$
                        Not leaving much for OP to do, Fred.
                        $endgroup$
                        – Gerry Myerson
                        Nov 30 '18 at 11:42














                      -1












                      -1








                      -1





                      $begingroup$

                      $begin{align}
                      tr(A)^2-4 det(A)geq 0
                      end{align} iff (lambda + mu)^2 ge 4 lambda mu iff lambda^2 -2 lambda mu + mu^2 ge 0 iff (lambda- mu)^2 ge 0$                      .






                      share|cite|improve this answer









                      $endgroup$



                      $begin{align}
                      tr(A)^2-4 det(A)geq 0
                      end{align} iff (lambda + mu)^2 ge 4 lambda mu iff lambda^2 -2 lambda mu + mu^2 ge 0 iff (lambda- mu)^2 ge 0$                      .







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Nov 30 '18 at 11:38









                      FredFred

                      46.1k1848




                      46.1k1848








                      • 2




                        $begingroup$
                        Simply elaborated, but why give a thorough solution to a simple problem of which the OP have showed zero attempts ?
                        $endgroup$
                        – Rebellos
                        Nov 30 '18 at 11:39






                      • 3




                        $begingroup$
                        Not leaving much for OP to do, Fred.
                        $endgroup$
                        – Gerry Myerson
                        Nov 30 '18 at 11:42














                      • 2




                        $begingroup$
                        Simply elaborated, but why give a thorough solution to a simple problem of which the OP have showed zero attempts ?
                        $endgroup$
                        – Rebellos
                        Nov 30 '18 at 11:39






                      • 3




                        $begingroup$
                        Not leaving much for OP to do, Fred.
                        $endgroup$
                        – Gerry Myerson
                        Nov 30 '18 at 11:42








                      2




                      2




                      $begingroup$
                      Simply elaborated, but why give a thorough solution to a simple problem of which the OP have showed zero attempts ?
                      $endgroup$
                      – Rebellos
                      Nov 30 '18 at 11:39




                      $begingroup$
                      Simply elaborated, but why give a thorough solution to a simple problem of which the OP have showed zero attempts ?
                      $endgroup$
                      – Rebellos
                      Nov 30 '18 at 11:39




                      3




                      3




                      $begingroup$
                      Not leaving much for OP to do, Fred.
                      $endgroup$
                      – Gerry Myerson
                      Nov 30 '18 at 11:42




                      $begingroup$
                      Not leaving much for OP to do, Fred.
                      $endgroup$
                      – Gerry Myerson
                      Nov 30 '18 at 11:42


















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