Transpose rules and algebra












0












$begingroup$


A, B and P are all squared matrices of the same order ($nXn$).



It is given that:



$PP^{T}=I$



$B^{T}B=I$



$A=P^{T}BP$



Which of the following is correct:



$(1) A=B$



$(2) AA^{T}=I$



$(3) (PB)^{-1}=(PB)^{T}$



I have started with (2). This is what I've done:



$(P^{T}BP)(P^{T}BP)$



$P^{T}BB^{T}P$



Something is not right. Can I say that



$B^{T}B=BB^{T}$ ?



According to the book, (2) and (3) are correct while (1) is not. I can't figure out why.






linear-algebra transpose







share|cite|improve this question











$endgroup$












  • $begingroup$
    Your item (2) just reads "$AA^T$"; but what about $AA^T$? There's no proposition there . . .
    $endgroup$
    – Robert Lewis
    Jan 14 at 8:10










  • $begingroup$
    $(P^mathrm TBP)(P^mathrm TBP)=P^mathrm TBBPne P^mathrm TBB^mathrm TP$ unless $B$ is symmetric.
    $endgroup$
    – W. Zhu
    Jan 14 at 8:26


















0












$begingroup$


A, B and P are all squared matrices of the same order ($nXn$).



It is given that:



$PP^{T}=I$



$B^{T}B=I$



$A=P^{T}BP$



Which of the following is correct:



$(1) A=B$



$(2) AA^{T}=I$



$(3) (PB)^{-1}=(PB)^{T}$



I have started with (2). This is what I've done:



$(P^{T}BP)(P^{T}BP)$



$P^{T}BB^{T}P$



Something is not right. Can I say that



$B^{T}B=BB^{T}$ ?



According to the book, (2) and (3) are correct while (1) is not. I can't figure out why.






linear-algebra transpose







share|cite|improve this question











$endgroup$












  • $begingroup$
    Your item (2) just reads "$AA^T$"; but what about $AA^T$? There's no proposition there . . .
    $endgroup$
    – Robert Lewis
    Jan 14 at 8:10










  • $begingroup$
    $(P^mathrm TBP)(P^mathrm TBP)=P^mathrm TBBPne P^mathrm TBB^mathrm TP$ unless $B$ is symmetric.
    $endgroup$
    – W. Zhu
    Jan 14 at 8:26
















0












0








0





$begingroup$


A, B and P are all squared matrices of the same order ($nXn$).



It is given that:



$PP^{T}=I$



$B^{T}B=I$



$A=P^{T}BP$



Which of the following is correct:



$(1) A=B$



$(2) AA^{T}=I$



$(3) (PB)^{-1}=(PB)^{T}$



I have started with (2). This is what I've done:



$(P^{T}BP)(P^{T}BP)$



$P^{T}BB^{T}P$



Something is not right. Can I say that



$B^{T}B=BB^{T}$ ?



According to the book, (2) and (3) are correct while (1) is not. I can't figure out why.






linear-algebra transpose







share|cite|improve this question











$endgroup$




A, B and P are all squared matrices of the same order ($nXn$).



It is given that:



$PP^{T}=I$



$B^{T}B=I$



$A=P^{T}BP$



Which of the following is correct:



$(1) A=B$



$(2) AA^{T}=I$



$(3) (PB)^{-1}=(PB)^{T}$



I have started with (2). This is what I've done:



$(P^{T}BP)(P^{T}BP)$



$P^{T}BB^{T}P$



Something is not right. Can I say that



$B^{T}B=BB^{T}$ ?



According to the book, (2) and (3) are correct while (1) is not. I can't figure out why.






linear-algebra transpose




linear-algebra transpose






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 14 at 8:18







user3275222

















asked Jan 14 at 8:02









user3275222user3275222

31829




31829












  • $begingroup$
    Your item (2) just reads "$AA^T$"; but what about $AA^T$? There's no proposition there . . .
    $endgroup$
    – Robert Lewis
    Jan 14 at 8:10










  • $begingroup$
    $(P^mathrm TBP)(P^mathrm TBP)=P^mathrm TBBPne P^mathrm TBB^mathrm TP$ unless $B$ is symmetric.
    $endgroup$
    – W. Zhu
    Jan 14 at 8:26




















  • $begingroup$
    Your item (2) just reads "$AA^T$"; but what about $AA^T$? There's no proposition there . . .
    $endgroup$
    – Robert Lewis
    Jan 14 at 8:10










  • $begingroup$
    $(P^mathrm TBP)(P^mathrm TBP)=P^mathrm TBBPne P^mathrm TBB^mathrm TP$ unless $B$ is symmetric.
    $endgroup$
    – W. Zhu
    Jan 14 at 8:26


















$begingroup$
Your item (2) just reads "$AA^T$"; but what about $AA^T$? There's no proposition there . . .
$endgroup$
– Robert Lewis
Jan 14 at 8:10




$begingroup$
Your item (2) just reads "$AA^T$"; but what about $AA^T$? There's no proposition there . . .
$endgroup$
– Robert Lewis
Jan 14 at 8:10












$begingroup$
$(P^mathrm TBP)(P^mathrm TBP)=P^mathrm TBBPne P^mathrm TBB^mathrm TP$ unless $B$ is symmetric.
$endgroup$
– W. Zhu
Jan 14 at 8:26






$begingroup$
$(P^mathrm TBP)(P^mathrm TBP)=P^mathrm TBBPne P^mathrm TBB^mathrm TP$ unless $B$ is symmetric.
$endgroup$
– W. Zhu
Jan 14 at 8:26












2 Answers
2






active

oldest

votes


















1












$begingroup$

To prove that (1) is false I will write a counterexample in terms of linear transformations instead of matrices. Let $P(x_1,x_2,cdots,x_n)=(x_1,0,cdots,0)$, $A=0$ and $B=I-P$. Then $A neq B$ but $A=P^{T}BP$.



(2) follows from $AA^{T}=(P^{T}BP)(P^{T}BP)^{T}=I$ because $(MN)^{T}=N^{T}M^{T}$. [Observe that $P^{T}P=I$ and $BB^{T}=I$]. To prove (3) it is enough to show that $(PB)(PB)^{T}=I$ which is true since LHS $=B^{T}P^{T}PB=B^{T}B=I$.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Hint: If $U$ and $V$ are square matrices (of the same size) , then



    $UV=I iff VU=I.$






    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









      1












      $begingroup$

      To prove that (1) is false I will write a counterexample in terms of linear transformations instead of matrices. Let $P(x_1,x_2,cdots,x_n)=(x_1,0,cdots,0)$, $A=0$ and $B=I-P$. Then $A neq B$ but $A=P^{T}BP$.



      (2) follows from $AA^{T}=(P^{T}BP)(P^{T}BP)^{T}=I$ because $(MN)^{T}=N^{T}M^{T}$. [Observe that $P^{T}P=I$ and $BB^{T}=I$]. To prove (3) it is enough to show that $(PB)(PB)^{T}=I$ which is true since LHS $=B^{T}P^{T}PB=B^{T}B=I$.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        To prove that (1) is false I will write a counterexample in terms of linear transformations instead of matrices. Let $P(x_1,x_2,cdots,x_n)=(x_1,0,cdots,0)$, $A=0$ and $B=I-P$. Then $A neq B$ but $A=P^{T}BP$.



        (2) follows from $AA^{T}=(P^{T}BP)(P^{T}BP)^{T}=I$ because $(MN)^{T}=N^{T}M^{T}$. [Observe that $P^{T}P=I$ and $BB^{T}=I$]. To prove (3) it is enough to show that $(PB)(PB)^{T}=I$ which is true since LHS $=B^{T}P^{T}PB=B^{T}B=I$.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          To prove that (1) is false I will write a counterexample in terms of linear transformations instead of matrices. Let $P(x_1,x_2,cdots,x_n)=(x_1,0,cdots,0)$, $A=0$ and $B=I-P$. Then $A neq B$ but $A=P^{T}BP$.



          (2) follows from $AA^{T}=(P^{T}BP)(P^{T}BP)^{T}=I$ because $(MN)^{T}=N^{T}M^{T}$. [Observe that $P^{T}P=I$ and $BB^{T}=I$]. To prove (3) it is enough to show that $(PB)(PB)^{T}=I$ which is true since LHS $=B^{T}P^{T}PB=B^{T}B=I$.






          share|cite|improve this answer









          $endgroup$



          To prove that (1) is false I will write a counterexample in terms of linear transformations instead of matrices. Let $P(x_1,x_2,cdots,x_n)=(x_1,0,cdots,0)$, $A=0$ and $B=I-P$. Then $A neq B$ but $A=P^{T}BP$.



          (2) follows from $AA^{T}=(P^{T}BP)(P^{T}BP)^{T}=I$ because $(MN)^{T}=N^{T}M^{T}$. [Observe that $P^{T}P=I$ and $BB^{T}=I$]. To prove (3) it is enough to show that $(PB)(PB)^{T}=I$ which is true since LHS $=B^{T}P^{T}PB=B^{T}B=I$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 14 at 8:19









          Kavi Rama MurthyKavi Rama Murthy

          69.1k53169




          69.1k53169























              0












              $begingroup$

              Hint: If $U$ and $V$ are square matrices (of the same size) , then



              $UV=I iff VU=I.$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Hint: If $U$ and $V$ are square matrices (of the same size) , then



                $UV=I iff VU=I.$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Hint: If $U$ and $V$ are square matrices (of the same size) , then



                  $UV=I iff VU=I.$






                  share|cite|improve this answer









                  $endgroup$



                  Hint: If $U$ and $V$ are square matrices (of the same size) , then



                  $UV=I iff VU=I.$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 14 at 8:07









                  FredFred

                  48.6k11849




                  48.6k11849






























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