Show that $L_A$ acts on by orthogonal transformation and in particular rotation.
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
edited Dec 9 at 15:08
José Carlos Santos
150k22120221
asked Dec 9 at 14:56
Join_PhD
1968
add a comment |
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
edited Dec 9 at 15:08
José Carlos Santos
150k22120221
asked Dec 9 at 14:56
Join_PhD
1968
add a comment |
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
edited Dec 9 at 15:08
José Carlos Santos
150k22120221
asked Dec 9 at 14:56
Join_PhD
1968
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
edited Dec 9 at 15:08
José Carlos Santos
150k22120221
asked Dec 9 at 14:56
Join_PhD
1968
edited Dec 9 at 15:08
José Carlos Santos
150k22120221
asked Dec 9 at 14:56
Join_PhD
1968
edited Dec 9 at 15:08
José Carlos Santos
150k22120221
edited Dec 9 at 15:08
José Carlos Santos
150k22120221
edited Dec 9 at 15:08
José Carlos Santos
150k22120221
150k22120221
asked Dec 9 at 14:56
Join_PhD
1968
asked Dec 9 at 14:56
Join_PhD
1968
asked Dec 9 at 14:56
Join_PhD
1968
1968
add a comment |
add a comment |
1 Answer
1
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oldest
votes
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 $det A=1$.
answered Dec 9 at 15:07
José Carlos Santos
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
|
show 4 more comments
Your Answer
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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 $det A=1$.
answered Dec 9 at 15:07
José Carlos Santos
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
|
show 4 more comments
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 $det A=1$.
answered Dec 9 at 15:07
José Carlos Santos
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
|
show 4 more comments
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 $det A=1$.
answered Dec 9 at 15:07
José Carlos Santos
150k22120221
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 $det A=1$.
answered Dec 9 at 15:07
José Carlos Santos
150k22120221
answered Dec 9 at 15:07
José Carlos Santos
150k22120221
answered Dec 9 at 15:07
José Carlos Santos
150k22120221
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
|
show 4 more comments
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
|
show 4 more comments
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