Let $A$ be a $2times2$ matrix such that $A^TA = I_2$. [closed]











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How do I explain that the rows of $A$ are unit vectors orthogonal to each other in $mathbb{R}^2$? I genuinely don't know where to start. Thanks in advance for any help.










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closed as off-topic by user302797, GNUSupporter 8964民主女神 地下教會, amWhy, ncmathsadist, Davide Giraudo Nov 23 at 23:53


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user302797, GNUSupporter 8964民主女神 地下教會, amWhy, Davide Giraudo

If this question can be reworded to fit the rules in the help center, please edit the question.













  • that's called an orthogonal matrix (en.wikipedia.org/wiki/Orthogonal_matrix)
    – D...
    Nov 23 at 10:27










  • Hint: the elements of $AA^T$ are scalar products of the rows of $A$, and $AA^TA=A$, and $A$ is regular.
    – Jean-Claude Arbaut
    Nov 23 at 10:31












  • See math.stackexchange.com/questions/2028990/…
    – Widawensen
    Nov 23 at 10:46








  • 1




    Possible duplicate of Orthogonal matrix and orthonormal columns
    – ncmathsadist
    Nov 23 at 21:55















up vote
-1
down vote

favorite












How do I explain that the rows of $A$ are unit vectors orthogonal to each other in $mathbb{R}^2$? I genuinely don't know where to start. Thanks in advance for any help.










share|cite|improve this question















closed as off-topic by user302797, GNUSupporter 8964民主女神 地下教會, amWhy, ncmathsadist, Davide Giraudo Nov 23 at 23:53


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user302797, GNUSupporter 8964民主女神 地下教會, amWhy, Davide Giraudo

If this question can be reworded to fit the rules in the help center, please edit the question.













  • that's called an orthogonal matrix (en.wikipedia.org/wiki/Orthogonal_matrix)
    – D...
    Nov 23 at 10:27










  • Hint: the elements of $AA^T$ are scalar products of the rows of $A$, and $AA^TA=A$, and $A$ is regular.
    – Jean-Claude Arbaut
    Nov 23 at 10:31












  • See math.stackexchange.com/questions/2028990/…
    – Widawensen
    Nov 23 at 10:46








  • 1




    Possible duplicate of Orthogonal matrix and orthonormal columns
    – ncmathsadist
    Nov 23 at 21:55













up vote
-1
down vote

favorite









up vote
-1
down vote

favorite











How do I explain that the rows of $A$ are unit vectors orthogonal to each other in $mathbb{R}^2$? I genuinely don't know where to start. Thanks in advance for any help.










share|cite|improve this question















How do I explain that the rows of $A$ are unit vectors orthogonal to each other in $mathbb{R}^2$? I genuinely don't know where to start. Thanks in advance for any help.







linear-algebra matrices






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edited Nov 23 at 10:34









Yadati Kiran

1,444518




1,444518










asked Nov 23 at 10:23









DwightD

42




42




closed as off-topic by user302797, GNUSupporter 8964民主女神 地下教會, amWhy, ncmathsadist, Davide Giraudo Nov 23 at 23:53


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user302797, GNUSupporter 8964民主女神 地下教會, amWhy, Davide Giraudo

If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by user302797, GNUSupporter 8964民主女神 地下教會, amWhy, ncmathsadist, Davide Giraudo Nov 23 at 23:53


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user302797, GNUSupporter 8964民主女神 地下教會, amWhy, Davide Giraudo

If this question can be reworded to fit the rules in the help center, please edit the question.












  • that's called an orthogonal matrix (en.wikipedia.org/wiki/Orthogonal_matrix)
    – D...
    Nov 23 at 10:27










  • Hint: the elements of $AA^T$ are scalar products of the rows of $A$, and $AA^TA=A$, and $A$ is regular.
    – Jean-Claude Arbaut
    Nov 23 at 10:31












  • See math.stackexchange.com/questions/2028990/…
    – Widawensen
    Nov 23 at 10:46








  • 1




    Possible duplicate of Orthogonal matrix and orthonormal columns
    – ncmathsadist
    Nov 23 at 21:55


















  • that's called an orthogonal matrix (en.wikipedia.org/wiki/Orthogonal_matrix)
    – D...
    Nov 23 at 10:27










  • Hint: the elements of $AA^T$ are scalar products of the rows of $A$, and $AA^TA=A$, and $A$ is regular.
    – Jean-Claude Arbaut
    Nov 23 at 10:31












  • See math.stackexchange.com/questions/2028990/…
    – Widawensen
    Nov 23 at 10:46








  • 1




    Possible duplicate of Orthogonal matrix and orthonormal columns
    – ncmathsadist
    Nov 23 at 21:55
















that's called an orthogonal matrix (en.wikipedia.org/wiki/Orthogonal_matrix)
– D...
Nov 23 at 10:27




that's called an orthogonal matrix (en.wikipedia.org/wiki/Orthogonal_matrix)
– D...
Nov 23 at 10:27












Hint: the elements of $AA^T$ are scalar products of the rows of $A$, and $AA^TA=A$, and $A$ is regular.
– Jean-Claude Arbaut
Nov 23 at 10:31






Hint: the elements of $AA^T$ are scalar products of the rows of $A$, and $AA^TA=A$, and $A$ is regular.
– Jean-Claude Arbaut
Nov 23 at 10:31














See math.stackexchange.com/questions/2028990/…
– Widawensen
Nov 23 at 10:46






See math.stackexchange.com/questions/2028990/…
– Widawensen
Nov 23 at 10:46






1




1




Possible duplicate of Orthogonal matrix and orthonormal columns
– ncmathsadist
Nov 23 at 21:55




Possible duplicate of Orthogonal matrix and orthonormal columns
– ncmathsadist
Nov 23 at 21:55










3 Answers
3






active

oldest

votes

















up vote
1
down vote













Well, since A is a $2times 2$ matrix, one way to do this is brute force:



$$A=
begin{pmatrix}
a & b \
c & d \
end{pmatrix}$$



$$A^TA= begin{pmatrix}
a & b \
c & d \
end{pmatrix}
begin{pmatrix}
a & c \
b & d \
end{pmatrix} =
begin{pmatrix}
a^2 +b^2 & ac + bd \
ca +db & c^2 +d^2 \
end{pmatrix} = I_2$$



From there, you can conclude about $a^2+b^2$,$c^2+d^2$ and $ac+bd$






share|cite|improve this answer





















  • Actually you have written $AA^T$.
    – Jean-Claude Arbaut
    Nov 24 at 9:50


















up vote
1
down vote













Writing the matrix as two row vectors $r_0,r_1$, the elements of the matrix $A^TA$ are



$$begin{pmatrix}r_0r_0^T&r_0r_1^T\r_1r_0^T&r_1r_1^Tend{pmatrix},$$ which are dot products. Conclude.






share|cite|improve this answer























  • Actually, it's just slightly more complicated: one has to consider $AA^T$ to get scalar products of rows. But it's easy to prove first that $AA^T=I_2$.
    – Jean-Claude Arbaut
    Nov 23 at 10:30




















up vote
1
down vote













If $A=left(begin{smallmatrix}a&b\c&dend{smallmatrix}right)$, the first thing that you do is to compute $A^TA$; the second one is to see what it means to say that $A^TA=operatorname{Id}_2$. You will see that it means three things:




  1. The first column has norm $1$;

  2. The second column has norm $1$;

  3. The columns ar orthogonal.






share|cite|improve this answer






























    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    1
    down vote













    Well, since A is a $2times 2$ matrix, one way to do this is brute force:



    $$A=
    begin{pmatrix}
    a & b \
    c & d \
    end{pmatrix}$$



    $$A^TA= begin{pmatrix}
    a & b \
    c & d \
    end{pmatrix}
    begin{pmatrix}
    a & c \
    b & d \
    end{pmatrix} =
    begin{pmatrix}
    a^2 +b^2 & ac + bd \
    ca +db & c^2 +d^2 \
    end{pmatrix} = I_2$$



    From there, you can conclude about $a^2+b^2$,$c^2+d^2$ and $ac+bd$






    share|cite|improve this answer





















    • Actually you have written $AA^T$.
      – Jean-Claude Arbaut
      Nov 24 at 9:50















    up vote
    1
    down vote













    Well, since A is a $2times 2$ matrix, one way to do this is brute force:



    $$A=
    begin{pmatrix}
    a & b \
    c & d \
    end{pmatrix}$$



    $$A^TA= begin{pmatrix}
    a & b \
    c & d \
    end{pmatrix}
    begin{pmatrix}
    a & c \
    b & d \
    end{pmatrix} =
    begin{pmatrix}
    a^2 +b^2 & ac + bd \
    ca +db & c^2 +d^2 \
    end{pmatrix} = I_2$$



    From there, you can conclude about $a^2+b^2$,$c^2+d^2$ and $ac+bd$






    share|cite|improve this answer





















    • Actually you have written $AA^T$.
      – Jean-Claude Arbaut
      Nov 24 at 9:50













    up vote
    1
    down vote










    up vote
    1
    down vote









    Well, since A is a $2times 2$ matrix, one way to do this is brute force:



    $$A=
    begin{pmatrix}
    a & b \
    c & d \
    end{pmatrix}$$



    $$A^TA= begin{pmatrix}
    a & b \
    c & d \
    end{pmatrix}
    begin{pmatrix}
    a & c \
    b & d \
    end{pmatrix} =
    begin{pmatrix}
    a^2 +b^2 & ac + bd \
    ca +db & c^2 +d^2 \
    end{pmatrix} = I_2$$



    From there, you can conclude about $a^2+b^2$,$c^2+d^2$ and $ac+bd$






    share|cite|improve this answer












    Well, since A is a $2times 2$ matrix, one way to do this is brute force:



    $$A=
    begin{pmatrix}
    a & b \
    c & d \
    end{pmatrix}$$



    $$A^TA= begin{pmatrix}
    a & b \
    c & d \
    end{pmatrix}
    begin{pmatrix}
    a & c \
    b & d \
    end{pmatrix} =
    begin{pmatrix}
    a^2 +b^2 & ac + bd \
    ca +db & c^2 +d^2 \
    end{pmatrix} = I_2$$



    From there, you can conclude about $a^2+b^2$,$c^2+d^2$ and $ac+bd$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 23 at 10:31









    F.Carette

    1,20112




    1,20112












    • Actually you have written $AA^T$.
      – Jean-Claude Arbaut
      Nov 24 at 9:50


















    • Actually you have written $AA^T$.
      – Jean-Claude Arbaut
      Nov 24 at 9:50
















    Actually you have written $AA^T$.
    – Jean-Claude Arbaut
    Nov 24 at 9:50




    Actually you have written $AA^T$.
    – Jean-Claude Arbaut
    Nov 24 at 9:50










    up vote
    1
    down vote













    Writing the matrix as two row vectors $r_0,r_1$, the elements of the matrix $A^TA$ are



    $$begin{pmatrix}r_0r_0^T&r_0r_1^T\r_1r_0^T&r_1r_1^Tend{pmatrix},$$ which are dot products. Conclude.






    share|cite|improve this answer























    • Actually, it's just slightly more complicated: one has to consider $AA^T$ to get scalar products of rows. But it's easy to prove first that $AA^T=I_2$.
      – Jean-Claude Arbaut
      Nov 23 at 10:30

















    up vote
    1
    down vote













    Writing the matrix as two row vectors $r_0,r_1$, the elements of the matrix $A^TA$ are



    $$begin{pmatrix}r_0r_0^T&r_0r_1^T\r_1r_0^T&r_1r_1^Tend{pmatrix},$$ which are dot products. Conclude.






    share|cite|improve this answer























    • Actually, it's just slightly more complicated: one has to consider $AA^T$ to get scalar products of rows. But it's easy to prove first that $AA^T=I_2$.
      – Jean-Claude Arbaut
      Nov 23 at 10:30















    up vote
    1
    down vote










    up vote
    1
    down vote









    Writing the matrix as two row vectors $r_0,r_1$, the elements of the matrix $A^TA$ are



    $$begin{pmatrix}r_0r_0^T&r_0r_1^T\r_1r_0^T&r_1r_1^Tend{pmatrix},$$ which are dot products. Conclude.






    share|cite|improve this answer














    Writing the matrix as two row vectors $r_0,r_1$, the elements of the matrix $A^TA$ are



    $$begin{pmatrix}r_0r_0^T&r_0r_1^T\r_1r_0^T&r_1r_1^Tend{pmatrix},$$ which are dot products. Conclude.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Nov 23 at 10:31

























    answered Nov 23 at 10:28









    Yves Daoust

    123k668219




    123k668219












    • Actually, it's just slightly more complicated: one has to consider $AA^T$ to get scalar products of rows. But it's easy to prove first that $AA^T=I_2$.
      – Jean-Claude Arbaut
      Nov 23 at 10:30




















    • Actually, it's just slightly more complicated: one has to consider $AA^T$ to get scalar products of rows. But it's easy to prove first that $AA^T=I_2$.
      – Jean-Claude Arbaut
      Nov 23 at 10:30


















    Actually, it's just slightly more complicated: one has to consider $AA^T$ to get scalar products of rows. But it's easy to prove first that $AA^T=I_2$.
    – Jean-Claude Arbaut
    Nov 23 at 10:30






    Actually, it's just slightly more complicated: one has to consider $AA^T$ to get scalar products of rows. But it's easy to prove first that $AA^T=I_2$.
    – Jean-Claude Arbaut
    Nov 23 at 10:30












    up vote
    1
    down vote













    If $A=left(begin{smallmatrix}a&b\c&dend{smallmatrix}right)$, the first thing that you do is to compute $A^TA$; the second one is to see what it means to say that $A^TA=operatorname{Id}_2$. You will see that it means three things:




    1. The first column has norm $1$;

    2. The second column has norm $1$;

    3. The columns ar orthogonal.






    share|cite|improve this answer



























      up vote
      1
      down vote













      If $A=left(begin{smallmatrix}a&b\c&dend{smallmatrix}right)$, the first thing that you do is to compute $A^TA$; the second one is to see what it means to say that $A^TA=operatorname{Id}_2$. You will see that it means three things:




      1. The first column has norm $1$;

      2. The second column has norm $1$;

      3. The columns ar orthogonal.






      share|cite|improve this answer

























        up vote
        1
        down vote










        up vote
        1
        down vote









        If $A=left(begin{smallmatrix}a&b\c&dend{smallmatrix}right)$, the first thing that you do is to compute $A^TA$; the second one is to see what it means to say that $A^TA=operatorname{Id}_2$. You will see that it means three things:




        1. The first column has norm $1$;

        2. The second column has norm $1$;

        3. The columns ar orthogonal.






        share|cite|improve this answer














        If $A=left(begin{smallmatrix}a&b\c&dend{smallmatrix}right)$, the first thing that you do is to compute $A^TA$; the second one is to see what it means to say that $A^TA=operatorname{Id}_2$. You will see that it means three things:




        1. The first column has norm $1$;

        2. The second column has norm $1$;

        3. The columns ar orthogonal.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 23 at 10:46

























        answered Nov 23 at 10:29









        José Carlos Santos

        147k22117218




        147k22117218















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