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

114k348124




114k348124










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

    5,8682928




    5,8682928












    • $begingroup$
      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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