Which of the following statements about determinants are correct?












-1












$begingroup$



Which of the following statements about determinants are correct?




  1. $det(A^2)>0$, for all invertible matrices $A$


  2. $det(A+A^{-1})=det(A)+dfrac{1}{det(A)}$, for all invertible matrices $A$


  3. $det(vv^T)>0$, for all column vectors $v ≠ 0$


  4. $det(AB^T)=det((A^T)B)$, for all square matrices $A$ and $B$





Which of the statements are correct? I do not feel secure about which of them that is true.



My answer:
My calculations have given me that $2$ and $4$ are true. Am I correct?






matrices symmetric-matrices







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    kindly edit your post to include your attempts using mathjax.
    $endgroup$
    – Siong Thye Goh
    Dec 19 '18 at 6:54












  • $begingroup$
    Okey. Sorry I forgot
    $endgroup$
    – Jacob Andreasson
    Dec 19 '18 at 6:55










  • $begingroup$
    What do you know about determinants?
    $endgroup$
    – bubba
    Dec 19 '18 at 6:58






  • 1




    $begingroup$
    Do you know, for example, that $det(AB) = det(A)cdotdet(B)$ ?
    $endgroup$
    – bubba
    Dec 19 '18 at 6:59












  • $begingroup$
    Sorry I have calculate totally wrong
    $endgroup$
    – Jacob Andreasson
    Dec 19 '18 at 7:00
















-1












$begingroup$



Which of the following statements about determinants are correct?




  1. $det(A^2)>0$, for all invertible matrices $A$


  2. $det(A+A^{-1})=det(A)+dfrac{1}{det(A)}$, for all invertible matrices $A$


  3. $det(vv^T)>0$, for all column vectors $v ≠ 0$


  4. $det(AB^T)=det((A^T)B)$, for all square matrices $A$ and $B$





Which of the statements are correct? I do not feel secure about which of them that is true.



My answer:
My calculations have given me that $2$ and $4$ are true. Am I correct?






matrices symmetric-matrices







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    kindly edit your post to include your attempts using mathjax.
    $endgroup$
    – Siong Thye Goh
    Dec 19 '18 at 6:54












  • $begingroup$
    Okey. Sorry I forgot
    $endgroup$
    – Jacob Andreasson
    Dec 19 '18 at 6:55










  • $begingroup$
    What do you know about determinants?
    $endgroup$
    – bubba
    Dec 19 '18 at 6:58






  • 1




    $begingroup$
    Do you know, for example, that $det(AB) = det(A)cdotdet(B)$ ?
    $endgroup$
    – bubba
    Dec 19 '18 at 6:59












  • $begingroup$
    Sorry I have calculate totally wrong
    $endgroup$
    – Jacob Andreasson
    Dec 19 '18 at 7:00














-1












-1








-1





$begingroup$



Which of the following statements about determinants are correct?




  1. $det(A^2)>0$, for all invertible matrices $A$


  2. $det(A+A^{-1})=det(A)+dfrac{1}{det(A)}$, for all invertible matrices $A$


  3. $det(vv^T)>0$, for all column vectors $v ≠ 0$


  4. $det(AB^T)=det((A^T)B)$, for all square matrices $A$ and $B$





Which of the statements are correct? I do not feel secure about which of them that is true.



My answer:
My calculations have given me that $2$ and $4$ are true. Am I correct?






matrices symmetric-matrices







share|cite|improve this question











$endgroup$





Which of the following statements about determinants are correct?




  1. $det(A^2)>0$, for all invertible matrices $A$


  2. $det(A+A^{-1})=det(A)+dfrac{1}{det(A)}$, for all invertible matrices $A$


  3. $det(vv^T)>0$, for all column vectors $v ≠ 0$


  4. $det(AB^T)=det((A^T)B)$, for all square matrices $A$ and $B$





Which of the statements are correct? I do not feel secure about which of them that is true.



My answer:
My calculations have given me that $2$ and $4$ are true. Am I correct?






matrices symmetric-matrices




matrices symmetric-matrices






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 19 '18 at 7:12









Henno Brandsma

113k348122




113k348122










asked Dec 19 '18 at 6:53









Jacob AndreassonJacob Andreasson

35




35








  • 1




    $begingroup$
    kindly edit your post to include your attempts using mathjax.
    $endgroup$
    – Siong Thye Goh
    Dec 19 '18 at 6:54












  • $begingroup$
    Okey. Sorry I forgot
    $endgroup$
    – Jacob Andreasson
    Dec 19 '18 at 6:55










  • $begingroup$
    What do you know about determinants?
    $endgroup$
    – bubba
    Dec 19 '18 at 6:58






  • 1




    $begingroup$
    Do you know, for example, that $det(AB) = det(A)cdotdet(B)$ ?
    $endgroup$
    – bubba
    Dec 19 '18 at 6:59












  • $begingroup$
    Sorry I have calculate totally wrong
    $endgroup$
    – Jacob Andreasson
    Dec 19 '18 at 7:00














  • 1




    $begingroup$
    kindly edit your post to include your attempts using mathjax.
    $endgroup$
    – Siong Thye Goh
    Dec 19 '18 at 6:54












  • $begingroup$
    Okey. Sorry I forgot
    $endgroup$
    – Jacob Andreasson
    Dec 19 '18 at 6:55










  • $begingroup$
    What do you know about determinants?
    $endgroup$
    – bubba
    Dec 19 '18 at 6:58






  • 1




    $begingroup$
    Do you know, for example, that $det(AB) = det(A)cdotdet(B)$ ?
    $endgroup$
    – bubba
    Dec 19 '18 at 6:59












  • $begingroup$
    Sorry I have calculate totally wrong
    $endgroup$
    – Jacob Andreasson
    Dec 19 '18 at 7:00








1




1




$begingroup$
kindly edit your post to include your attempts using mathjax.
$endgroup$
– Siong Thye Goh
Dec 19 '18 at 6:54






$begingroup$
kindly edit your post to include your attempts using mathjax.
$endgroup$
– Siong Thye Goh
Dec 19 '18 at 6:54














$begingroup$
Okey. Sorry I forgot
$endgroup$
– Jacob Andreasson
Dec 19 '18 at 6:55




$begingroup$
Okey. Sorry I forgot
$endgroup$
– Jacob Andreasson
Dec 19 '18 at 6:55












$begingroup$
What do you know about determinants?
$endgroup$
– bubba
Dec 19 '18 at 6:58




$begingroup$
What do you know about determinants?
$endgroup$
– bubba
Dec 19 '18 at 6:58




1




1




$begingroup$
Do you know, for example, that $det(AB) = det(A)cdotdet(B)$ ?
$endgroup$
– bubba
Dec 19 '18 at 6:59






$begingroup$
Do you know, for example, that $det(AB) = det(A)cdotdet(B)$ ?
$endgroup$
– bubba
Dec 19 '18 at 6:59














$begingroup$
Sorry I have calculate totally wrong
$endgroup$
– Jacob Andreasson
Dec 19 '18 at 7:00




$begingroup$
Sorry I have calculate totally wrong
$endgroup$
– Jacob Andreasson
Dec 19 '18 at 7:00










2 Answers
2






active

oldest

votes


















3












$begingroup$


  • $det (A^2)=det A cdot det A=(det A)^2>0$


  • $det (I+I)=det (2I)=2^n .det I=2^n$ whereas $det I+frac{1}{det I}=2$


  • $vv^T$ has rank $1$, so its eigenvalues are zero [$(n-1)$ times] and its trace and hence $det vv^T=0$

  • $det (AB^T)=det A cdot det B^T=det A cdot det B=det A^T.det B=det(A^TB)$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    so the only one that is true is 4?
    $endgroup$
    – Jacob Andreasson
    Dec 19 '18 at 7:27










  • $begingroup$
    it is A*B$^T$ not (AB)$^T$
    $endgroup$
    – Jacob Andreasson
    Dec 19 '18 at 7:30










  • $begingroup$
    see my answer again. use $det A=det A^T$
    $endgroup$
    – Chinnapparaj R
    Dec 19 '18 at 7:33










  • $begingroup$
    1 and 4 is true then?
    $endgroup$
    – Jacob Andreasson
    Dec 19 '18 at 7:36










  • $begingroup$
    yes...................
    $endgroup$
    – Chinnapparaj R
    Dec 19 '18 at 7:36



















1












$begingroup$

For 1, use that $det(AB)= det(A)det(B)$ to find that $det(A^2) = det(A)^2 > 0$.



For 2, take $A = I $.



For 3, note that the rows of every matrix $vv^T$ are scalar multiples of $v$. Thus, the determinant will be zero for all vectors with more thant $1$ entry.



For 4, use that $det(A^T) = det(A)$, so $det(AB^T) = det((AB^T)^T) = det(BA^T) = det(A^TB)$.






share|cite|improve this answer









$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$


    • $det (A^2)=det A cdot det A=(det A)^2>0$


    • $det (I+I)=det (2I)=2^n .det I=2^n$ whereas $det I+frac{1}{det I}=2$


    • $vv^T$ has rank $1$, so its eigenvalues are zero [$(n-1)$ times] and its trace and hence $det vv^T=0$

    • $det (AB^T)=det A cdot det B^T=det A cdot det B=det A^T.det B=det(A^TB)$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      so the only one that is true is 4?
      $endgroup$
      – Jacob Andreasson
      Dec 19 '18 at 7:27










    • $begingroup$
      it is A*B$^T$ not (AB)$^T$
      $endgroup$
      – Jacob Andreasson
      Dec 19 '18 at 7:30










    • $begingroup$
      see my answer again. use $det A=det A^T$
      $endgroup$
      – Chinnapparaj R
      Dec 19 '18 at 7:33










    • $begingroup$
      1 and 4 is true then?
      $endgroup$
      – Jacob Andreasson
      Dec 19 '18 at 7:36










    • $begingroup$
      yes...................
      $endgroup$
      – Chinnapparaj R
      Dec 19 '18 at 7:36
















    3












    $begingroup$


    • $det (A^2)=det A cdot det A=(det A)^2>0$


    • $det (I+I)=det (2I)=2^n .det I=2^n$ whereas $det I+frac{1}{det I}=2$


    • $vv^T$ has rank $1$, so its eigenvalues are zero [$(n-1)$ times] and its trace and hence $det vv^T=0$

    • $det (AB^T)=det A cdot det B^T=det A cdot det B=det A^T.det B=det(A^TB)$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      so the only one that is true is 4?
      $endgroup$
      – Jacob Andreasson
      Dec 19 '18 at 7:27










    • $begingroup$
      it is A*B$^T$ not (AB)$^T$
      $endgroup$
      – Jacob Andreasson
      Dec 19 '18 at 7:30










    • $begingroup$
      see my answer again. use $det A=det A^T$
      $endgroup$
      – Chinnapparaj R
      Dec 19 '18 at 7:33










    • $begingroup$
      1 and 4 is true then?
      $endgroup$
      – Jacob Andreasson
      Dec 19 '18 at 7:36










    • $begingroup$
      yes...................
      $endgroup$
      – Chinnapparaj R
      Dec 19 '18 at 7:36














    3












    3








    3





    $begingroup$


    • $det (A^2)=det A cdot det A=(det A)^2>0$


    • $det (I+I)=det (2I)=2^n .det I=2^n$ whereas $det I+frac{1}{det I}=2$


    • $vv^T$ has rank $1$, so its eigenvalues are zero [$(n-1)$ times] and its trace and hence $det vv^T=0$

    • $det (AB^T)=det A cdot det B^T=det A cdot det B=det A^T.det B=det(A^TB)$






    share|cite|improve this answer











    $endgroup$




    • $det (A^2)=det A cdot det A=(det A)^2>0$


    • $det (I+I)=det (2I)=2^n .det I=2^n$ whereas $det I+frac{1}{det I}=2$


    • $vv^T$ has rank $1$, so its eigenvalues are zero [$(n-1)$ times] and its trace and hence $det vv^T=0$

    • $det (AB^T)=det A cdot det B^T=det A cdot det B=det A^T.det B=det(A^TB)$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 19 '18 at 7:32

























    answered Dec 19 '18 at 7:25









    Chinnapparaj RChinnapparaj R

    5,7412928




    5,7412928












    • $begingroup$
      so the only one that is true is 4?
      $endgroup$
      – Jacob Andreasson
      Dec 19 '18 at 7:27










    • $begingroup$
      it is A*B$^T$ not (AB)$^T$
      $endgroup$
      – Jacob Andreasson
      Dec 19 '18 at 7:30










    • $begingroup$
      see my answer again. use $det A=det A^T$
      $endgroup$
      – Chinnapparaj R
      Dec 19 '18 at 7:33










    • $begingroup$
      1 and 4 is true then?
      $endgroup$
      – Jacob Andreasson
      Dec 19 '18 at 7:36










    • $begingroup$
      yes...................
      $endgroup$
      – Chinnapparaj R
      Dec 19 '18 at 7:36


















    • $begingroup$
      so the only one that is true is 4?
      $endgroup$
      – Jacob Andreasson
      Dec 19 '18 at 7:27










    • $begingroup$
      it is A*B$^T$ not (AB)$^T$
      $endgroup$
      – Jacob Andreasson
      Dec 19 '18 at 7:30










    • $begingroup$
      see my answer again. use $det A=det A^T$
      $endgroup$
      – Chinnapparaj R
      Dec 19 '18 at 7:33










    • $begingroup$
      1 and 4 is true then?
      $endgroup$
      – Jacob Andreasson
      Dec 19 '18 at 7:36










    • $begingroup$
      yes...................
      $endgroup$
      – Chinnapparaj R
      Dec 19 '18 at 7:36
















    $begingroup$
    so the only one that is true is 4?
    $endgroup$
    – Jacob Andreasson
    Dec 19 '18 at 7:27




    $begingroup$
    so the only one that is true is 4?
    $endgroup$
    – Jacob Andreasson
    Dec 19 '18 at 7:27












    $begingroup$
    it is A*B$^T$ not (AB)$^T$
    $endgroup$
    – Jacob Andreasson
    Dec 19 '18 at 7:30




    $begingroup$
    it is A*B$^T$ not (AB)$^T$
    $endgroup$
    – Jacob Andreasson
    Dec 19 '18 at 7:30












    $begingroup$
    see my answer again. use $det A=det A^T$
    $endgroup$
    – Chinnapparaj R
    Dec 19 '18 at 7:33




    $begingroup$
    see my answer again. use $det A=det A^T$
    $endgroup$
    – Chinnapparaj R
    Dec 19 '18 at 7:33












    $begingroup$
    1 and 4 is true then?
    $endgroup$
    – Jacob Andreasson
    Dec 19 '18 at 7:36




    $begingroup$
    1 and 4 is true then?
    $endgroup$
    – Jacob Andreasson
    Dec 19 '18 at 7:36












    $begingroup$
    yes...................
    $endgroup$
    – Chinnapparaj R
    Dec 19 '18 at 7:36




    $begingroup$
    yes...................
    $endgroup$
    – Chinnapparaj R
    Dec 19 '18 at 7:36











    1












    $begingroup$

    For 1, use that $det(AB)= det(A)det(B)$ to find that $det(A^2) = det(A)^2 > 0$.



    For 2, take $A = I $.



    For 3, note that the rows of every matrix $vv^T$ are scalar multiples of $v$. Thus, the determinant will be zero for all vectors with more thant $1$ entry.



    For 4, use that $det(A^T) = det(A)$, so $det(AB^T) = det((AB^T)^T) = det(BA^T) = det(A^TB)$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      For 1, use that $det(AB)= det(A)det(B)$ to find that $det(A^2) = det(A)^2 > 0$.



      For 2, take $A = I $.



      For 3, note that the rows of every matrix $vv^T$ are scalar multiples of $v$. Thus, the determinant will be zero for all vectors with more thant $1$ entry.



      For 4, use that $det(A^T) = det(A)$, so $det(AB^T) = det((AB^T)^T) = det(BA^T) = det(A^TB)$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        For 1, use that $det(AB)= det(A)det(B)$ to find that $det(A^2) = det(A)^2 > 0$.



        For 2, take $A = I $.



        For 3, note that the rows of every matrix $vv^T$ are scalar multiples of $v$. Thus, the determinant will be zero for all vectors with more thant $1$ entry.



        For 4, use that $det(A^T) = det(A)$, so $det(AB^T) = det((AB^T)^T) = det(BA^T) = det(A^TB)$.






        share|cite|improve this answer









        $endgroup$



        For 1, use that $det(AB)= det(A)det(B)$ to find that $det(A^2) = det(A)^2 > 0$.



        For 2, take $A = I $.



        For 3, note that the rows of every matrix $vv^T$ are scalar multiples of $v$. Thus, the determinant will be zero for all vectors with more thant $1$ entry.



        For 4, use that $det(A^T) = det(A)$, so $det(AB^T) = det((AB^T)^T) = det(BA^T) = det(A^TB)$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 19 '18 at 7:31









        Anthony TerAnthony Ter

        37116




        37116






























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            Ida-Boy-Ed-Garten

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            Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich ...
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