Show that $L_A$ acts on by orthogonal transformation and in particular rotation.












0















Let $A$ be a $3times 3$ orthogonal matrix with determinant $=1$.
Let $v$ be an eigen vector corresponding to $1$ of $A$.Let $W=text{span}{v}$.
Show that $L_A$ preserves $W^perp$ and it acts on it by orthogonal transformation and in particular rotation.




MY TRY::



Given $Av=vimplies v=A^{-1}v$



Also $L_A:Bbb R^3to Bbb R^3$ defined by $L_A(x)=Ax$.



Let $win W^perpimplies langle w,vrangle =0$



To show $L_A(w)in W^perp$.



Now $L_A(w)=Aw$



Also $langle Aw,vrangle =langle w,A^Tvrangle= langle
w,A^{-1}vrangle=langle w,vrangle =0$



Thus $L_A(w)in W^perp$.-------------(Proved)



hence we can consider the restriction $L_A:W^perpto W^perp$



But how can I show that it is an orthogonal transformation and in aprticular a rotation?






linear-algebra matrices geometry eigenvalues-eigenvectors orthogonal-matrices







share|cite|improve this question





























    0















    Let $A$ be a $3times 3$ orthogonal matrix with determinant $=1$.
    Let $v$ be an eigen vector corresponding to $1$ of $A$.Let $W=text{span}{v}$.
    Show that $L_A$ preserves $W^perp$ and it acts on it by orthogonal transformation and in particular rotation.




    MY TRY::



    Given $Av=vimplies v=A^{-1}v$



    Also $L_A:Bbb R^3to Bbb R^3$ defined by $L_A(x)=Ax$.



    Let $win W^perpimplies langle w,vrangle =0$



    To show $L_A(w)in W^perp$.



    Now $L_A(w)=Aw$



    Also $langle Aw,vrangle =langle w,A^Tvrangle= langle
    w,A^{-1}vrangle=langle w,vrangle =0$



    Thus $L_A(w)in W^perp$.-------------(Proved)



    hence we can consider the restriction $L_A:W^perpto W^perp$



    But how can I show that it is an orthogonal transformation and in aprticular a rotation?






    linear-algebra matrices geometry eigenvalues-eigenvectors orthogonal-matrices







    share|cite|improve this question



























      0












      0








      0








      Let $A$ be a $3times 3$ orthogonal matrix with determinant $=1$.
      Let $v$ be an eigen vector corresponding to $1$ of $A$.Let $W=text{span}{v}$.
      Show that $L_A$ preserves $W^perp$ and it acts on it by orthogonal transformation and in particular rotation.




      MY TRY::



      Given $Av=vimplies v=A^{-1}v$



      Also $L_A:Bbb R^3to Bbb R^3$ defined by $L_A(x)=Ax$.



      Let $win W^perpimplies langle w,vrangle =0$



      To show $L_A(w)in W^perp$.



      Now $L_A(w)=Aw$



      Also $langle Aw,vrangle =langle w,A^Tvrangle= langle
      w,A^{-1}vrangle=langle w,vrangle =0$



      Thus $L_A(w)in W^perp$.-------------(Proved)



      hence we can consider the restriction $L_A:W^perpto W^perp$



      But how can I show that it is an orthogonal transformation and in aprticular a rotation?






      linear-algebra matrices geometry eigenvalues-eigenvectors orthogonal-matrices







      share|cite|improve this question
















      Let $A$ be a $3times 3$ orthogonal matrix with determinant $=1$.
      Let $v$ be an eigen vector corresponding to $1$ of $A$.Let $W=text{span}{v}$.
      Show that $L_A$ preserves $W^perp$ and it acts on it by orthogonal transformation and in particular rotation.




      MY TRY::



      Given $Av=vimplies v=A^{-1}v$



      Also $L_A:Bbb R^3to Bbb R^3$ defined by $L_A(x)=Ax$.



      Let $win W^perpimplies langle w,vrangle =0$



      To show $L_A(w)in W^perp$.



      Now $L_A(w)=Aw$



      Also $langle Aw,vrangle =langle w,A^Tvrangle= langle
      w,A^{-1}vrangle=langle w,vrangle =0$



      Thus $L_A(w)in W^perp$.-------------(Proved)



      hence we can consider the restriction $L_A:W^perpto W^perp$



      But how can I show that it is an orthogonal transformation and in aprticular a rotation?






      linear-algebra matrices geometry eigenvalues-eigenvectors orthogonal-matrices




      linear-algebra matrices geometry eigenvalues-eigenvectors orthogonal-matrices






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 9 at 15:08









      José Carlos Santos

      150k22120221




      150k22120221










      asked Dec 9 at 14:56









      Join_PhD

      1968




      1968






















          1 Answer
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          Since $A.W^perpsubset W^perp$ and since $A$ is orthogonal, $A$ induces an orthogonal map from $W^perp$ into itself. Suppose that it is not a rotation. Fix an orthonormal basis $B$ of $W^perp$. Then the matrix of $A|_{W^perp}$ with respect to $B$ is an orthogonal matrix with determinant $-1$. But then $det A=1times(-1)=-1$. This is impossible, since we're assuming that $det A=1$.






          share|cite|improve this answer





















          • My problem is I dont understand how to show that restriction of A to $W^perp$ is orthogonal,is restriction of an orthogonal transformation an orthogonal transformation?if yes why?
            – Join_PhD
            Dec 9 at 15:19












          • How to prove it?
            – Join_PhD
            Dec 9 at 15:19










          • Since $A$ is orthogonal, then$$(forall v,winmathbb{R}^3):langle A.v,A.wrangle=langle v,wrangle.$$In particular,$$(forall v,win W^perp):langle A.v,A.wrangle=langle v,wrangle,$$which means that $A|_{W^perp}$ is orthogonal.
            – José Carlos Santos
            Dec 9 at 15:22












          • Are you including the vector $v$ along with the basis B to form a basis of $L_A$
            – Join_PhD
            Dec 9 at 15:33










          • Otherwise how can you include the eigen value -1
            – Join_PhD
            Dec 9 at 15:33











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          1 Answer
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          Since $A.W^perpsubset W^perp$ and since $A$ is orthogonal, $A$ induces an orthogonal map from $W^perp$ into itself. Suppose that it is not a rotation. Fix an orthonormal basis $B$ of $W^perp$. Then the matrix of $A|_{W^perp}$ with respect to $B$ is an orthogonal matrix with determinant $-1$. But then $det A=1times(-1)=-1$. This is impossible, since we're assuming that $det A=1$.






          share|cite|improve this answer





















          • My problem is I dont understand how to show that restriction of A to $W^perp$ is orthogonal,is restriction of an orthogonal transformation an orthogonal transformation?if yes why?
            – Join_PhD
            Dec 9 at 15:19












          • How to prove it?
            – Join_PhD
            Dec 9 at 15:19










          • Since $A$ is orthogonal, then$$(forall v,winmathbb{R}^3):langle A.v,A.wrangle=langle v,wrangle.$$In particular,$$(forall v,win W^perp):langle A.v,A.wrangle=langle v,wrangle,$$which means that $A|_{W^perp}$ is orthogonal.
            – José Carlos Santos
            Dec 9 at 15:22












          • Are you including the vector $v$ along with the basis B to form a basis of $L_A$
            – Join_PhD
            Dec 9 at 15:33










          • Otherwise how can you include the eigen value -1
            – Join_PhD
            Dec 9 at 15:33
















          1














          Since $A.W^perpsubset W^perp$ and since $A$ is orthogonal, $A$ induces an orthogonal map from $W^perp$ into itself. Suppose that it is not a rotation. Fix an orthonormal basis $B$ of $W^perp$. Then the matrix of $A|_{W^perp}$ with respect to $B$ is an orthogonal matrix with determinant $-1$. But then $det A=1times(-1)=-1$. This is impossible, since we're assuming that $det A=1$.






          share|cite|improve this answer





















          • My problem is I dont understand how to show that restriction of A to $W^perp$ is orthogonal,is restriction of an orthogonal transformation an orthogonal transformation?if yes why?
            – Join_PhD
            Dec 9 at 15:19












          • How to prove it?
            – Join_PhD
            Dec 9 at 15:19










          • Since $A$ is orthogonal, then$$(forall v,winmathbb{R}^3):langle A.v,A.wrangle=langle v,wrangle.$$In particular,$$(forall v,win W^perp):langle A.v,A.wrangle=langle v,wrangle,$$which means that $A|_{W^perp}$ is orthogonal.
            – José Carlos Santos
            Dec 9 at 15:22












          • Are you including the vector $v$ along with the basis B to form a basis of $L_A$
            – Join_PhD
            Dec 9 at 15:33










          • Otherwise how can you include the eigen value -1
            – Join_PhD
            Dec 9 at 15:33














          1












          1








          1






          Since $A.W^perpsubset W^perp$ and since $A$ is orthogonal, $A$ induces an orthogonal map from $W^perp$ into itself. Suppose that it is not a rotation. Fix an orthonormal basis $B$ of $W^perp$. Then the matrix of $A|_{W^perp}$ with respect to $B$ is an orthogonal matrix with determinant $-1$. But then $det A=1times(-1)=-1$. This is impossible, since we're assuming that $det A=1$.






          share|cite|improve this answer












          Since $A.W^perpsubset W^perp$ and since $A$ is orthogonal, $A$ induces an orthogonal map from $W^perp$ into itself. Suppose that it is not a rotation. Fix an orthonormal basis $B$ of $W^perp$. Then the matrix of $A|_{W^perp}$ with respect to $B$ is an orthogonal matrix with determinant $-1$. But then $det A=1times(-1)=-1$. This is impossible, since we're assuming that $det A=1$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 9 at 15:07









          José Carlos Santos

          150k22120221




          150k22120221












          • My problem is I dont understand how to show that restriction of A to $W^perp$ is orthogonal,is restriction of an orthogonal transformation an orthogonal transformation?if yes why?
            – Join_PhD
            Dec 9 at 15:19












          • How to prove it?
            – Join_PhD
            Dec 9 at 15:19










          • Since $A$ is orthogonal, then$$(forall v,winmathbb{R}^3):langle A.v,A.wrangle=langle v,wrangle.$$In particular,$$(forall v,win W^perp):langle A.v,A.wrangle=langle v,wrangle,$$which means that $A|_{W^perp}$ is orthogonal.
            – José Carlos Santos
            Dec 9 at 15:22












          • Are you including the vector $v$ along with the basis B to form a basis of $L_A$
            – Join_PhD
            Dec 9 at 15:33










          • Otherwise how can you include the eigen value -1
            – Join_PhD
            Dec 9 at 15:33


















          • My problem is I dont understand how to show that restriction of A to $W^perp$ is orthogonal,is restriction of an orthogonal transformation an orthogonal transformation?if yes why?
            – Join_PhD
            Dec 9 at 15:19












          • How to prove it?
            – Join_PhD
            Dec 9 at 15:19










          • Since $A$ is orthogonal, then$$(forall v,winmathbb{R}^3):langle A.v,A.wrangle=langle v,wrangle.$$In particular,$$(forall v,win W^perp):langle A.v,A.wrangle=langle v,wrangle,$$which means that $A|_{W^perp}$ is orthogonal.
            – José Carlos Santos
            Dec 9 at 15:22












          • Are you including the vector $v$ along with the basis B to form a basis of $L_A$
            – Join_PhD
            Dec 9 at 15:33










          • Otherwise how can you include the eigen value -1
            – Join_PhD
            Dec 9 at 15:33
















          My problem is I dont understand how to show that restriction of A to $W^perp$ is orthogonal,is restriction of an orthogonal transformation an orthogonal transformation?if yes why?
          – Join_PhD
          Dec 9 at 15:19






          My problem is I dont understand how to show that restriction of A to $W^perp$ is orthogonal,is restriction of an orthogonal transformation an orthogonal transformation?if yes why?
          – Join_PhD
          Dec 9 at 15:19














          How to prove it?
          – Join_PhD
          Dec 9 at 15:19




          How to prove it?
          – Join_PhD
          Dec 9 at 15:19












          Since $A$ is orthogonal, then$$(forall v,winmathbb{R}^3):langle A.v,A.wrangle=langle v,wrangle.$$In particular,$$(forall v,win W^perp):langle A.v,A.wrangle=langle v,wrangle,$$which means that $A|_{W^perp}$ is orthogonal.
          – José Carlos Santos
          Dec 9 at 15:22






          Since $A$ is orthogonal, then$$(forall v,winmathbb{R}^3):langle A.v,A.wrangle=langle v,wrangle.$$In particular,$$(forall v,win W^perp):langle A.v,A.wrangle=langle v,wrangle,$$which means that $A|_{W^perp}$ is orthogonal.
          – José Carlos Santos
          Dec 9 at 15:22














          Are you including the vector $v$ along with the basis B to form a basis of $L_A$
          – Join_PhD
          Dec 9 at 15:33




          Are you including the vector $v$ along with the basis B to form a basis of $L_A$
          – Join_PhD
          Dec 9 at 15:33












          Otherwise how can you include the eigen value -1
          – Join_PhD
          Dec 9 at 15:33




          Otherwise how can you include the eigen value -1
          – Join_PhD
          Dec 9 at 15:33


















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