If $exists v$ such that $v,T(v),dots,T^n-1(v)$ is linearly independent then $T$ has $n$ different eigenvalues
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$T:mathbbR^nto mathbbR^n$ is a normal linear operator.
Assume that the inner product between two vectors $u,v in mathbbR^n$ is of the form:
$(u,v)=sum _i=1^n u_icdot overlinev_i$
If the vector space were unitary then I know how to prove the claim, because if the vector space were unitary , given that $T$ is normal and $v,T(v),dots,T^n-1(v)$ is linearly independent, then I can show that the minimal polynomial of $T$ is equal to the characteristic polynomial of $T$ and also that $T$ diagonalizable, so this solves the problem.
But since the vector space is $mathbbR^n$ given that $T$ is normal doesn't imply that $T$ is diagonalizable.
I need some help, please.
linear-algebra
edited Aug 13 at 10:34
egreg
165k1180187
asked Aug 13 at 9:52
idan di
372110
 |Â
show 6 more comments
up vote
3
down vote
favorite
$T:mathbbR^nto mathbbR^n$ is a normal linear operator.
Assume that the inner product between two vectors $u,v in mathbbR^n$ is of the form:
$(u,v)=sum _i=1^n u_icdot overlinev_i$
If the vector space were unitary then I know how to prove the claim, because if the vector space were unitary , given that $T$ is normal and $v,T(v),dots,T^n-1(v)$ is linearly independent, then I can show that the minimal polynomial of $T$ is equal to the characteristic polynomial of $T$ and also that $T$ diagonalizable, so this solves the problem.
But since the vector space is $mathbbR^n$ given that $T$ is normal doesn't imply that $T$ is diagonalizable.
I need some help, please.
linear-algebra
edited Aug 13 at 10:34
egreg
165k1180187
asked Aug 13 at 9:52
idan di
372110
Why does this solve the problem in the unitary case? How do you know that the diagonal contains different elements?
– uniquesolution
Aug 13 at 9:54
Is this an exercise you have been assigned? Are there no other hypotheses?
– Arnaud D.
Aug 13 at 9:57
If $V$ is a finite dimensional unitary vector space, and $T$ is normal, there exists a basis of $V$ containing eigenvectors of $T$ and therefore $T$ is diagonalizable .
– idan di
Aug 13 at 9:57
@ArnaudD. just that $T$ is normal and the set $leftv,Tleft(vright),...,T^n-1left(vright)right$ is linear independent
– idan di
Aug 13 at 9:59
1
@uniquesolution and If $V$ is a finite dimensional unitary vector space, and $T$ is normal, there exists a basis of $V$ containing eigenvectors of $T$ and therefore $T$ is diagonalizable . so the minimal poylnomial of $T$ is of the form $left(t-lambda _1right)cdot ...cdot left(t-lambda _kright)$ such that $::kle n$, and all the roots are different. If we assume by contradiction that $k<n$ then we will get a contradiction that the set $leftv,Tleft(vright),...,T^n-1left(vright)right$ is linear independent
– idan di
Aug 13 at 10:05
 |Â
show 6 more comments
up vote
3
down vote
favorite
up vote
3
down vote
favorite
$T:mathbbR^nto mathbbR^n$ is a normal linear operator.
Assume that the inner product between two vectors $u,v in mathbbR^n$ is of the form:
$(u,v)=sum _i=1^n u_icdot overlinev_i$
If the vector space were unitary then I know how to prove the claim, because if the vector space were unitary , given that $T$ is normal and $v,T(v),dots,T^n-1(v)$ is linearly independent, then I can show that the minimal polynomial of $T$ is equal to the characteristic polynomial of $T$ and also that $T$ diagonalizable, so this solves the problem.
But since the vector space is $mathbbR^n$ given that $T$ is normal doesn't imply that $T$ is diagonalizable.
I need some help, please.
linear-algebra
edited Aug 13 at 10:34
egreg
165k1180187
asked Aug 13 at 9:52
idan di
372110
$T:mathbbR^nto mathbbR^n$ is a normal linear operator.
Assume that the inner product between two vectors $u,v in mathbbR^n$ is of the form:
$(u,v)=sum _i=1^n u_icdot overlinev_i$
If the vector space were unitary then I know how to prove the claim, because if the vector space were unitary , given that $T$ is normal and $v,T(v),dots,T^n-1(v)$ is linearly independent, then I can show that the minimal polynomial of $T$ is equal to the characteristic polynomial of $T$ and also that $T$ diagonalizable, so this solves the problem.
But since the vector space is $mathbbR^n$ given that $T$ is normal doesn't imply that $T$ is diagonalizable.
I need some help, please.
linear-algebra
edited Aug 13 at 10:34
egreg
165k1180187
asked Aug 13 at 9:52
idan di
372110
edited Aug 13 at 10:34
egreg
165k1180187
edited Aug 13 at 10:34
egreg
165k1180187
edited Aug 13 at 10:34
egreg
165k1180187
165k1180187
asked Aug 13 at 9:52
idan di
372110
asked Aug 13 at 9:52
idan di
372110
asked Aug 13 at 9:52
idan di
372110
372110
Why does this solve the problem in the unitary case? How do you know that the diagonal contains different elements?
– uniquesolution
Aug 13 at 9:54
Is this an exercise you have been assigned? Are there no other hypotheses?
– Arnaud D.
Aug 13 at 9:57
If $V$ is a finite dimensional unitary vector space, and $T$ is normal, there exists a basis of $V$ containing eigenvectors of $T$ and therefore $T$ is diagonalizable .
– idan di
Aug 13 at 9:57
@ArnaudD. just that $T$ is normal and the set $leftv,Tleft(vright),...,T^n-1left(vright)right$ is linear independent
– idan di
Aug 13 at 9:59
1
@uniquesolution and If $V$ is a finite dimensional unitary vector space, and $T$ is normal, there exists a basis of $V$ containing eigenvectors of $T$ and therefore $T$ is diagonalizable . so the minimal poylnomial of $T$ is of the form $left(t-lambda _1right)cdot ...cdot left(t-lambda _kright)$ such that $::kle n$, and all the roots are different. If we assume by contradiction that $k<n$ then we will get a contradiction that the set $leftv,Tleft(vright),...,T^n-1left(vright)right$ is linear independent
– idan di
Aug 13 at 10:05
 |Â
show 6 more comments
Why does this solve the problem in the unitary case? How do you know that the diagonal contains different elements?
– uniquesolution
Aug 13 at 9:54
Is this an exercise you have been assigned? Are there no other hypotheses?
– Arnaud D.
Aug 13 at 9:57
If $V$ is a finite dimensional unitary vector space, and $T$ is normal, there exists a basis of $V$ containing eigenvectors of $T$ and therefore $T$ is diagonalizable .
– idan di
Aug 13 at 9:57
@ArnaudD. just that $T$ is normal and the set $leftv,Tleft(vright),...,T^n-1left(vright)right$ is linear independent
– idan di
Aug 13 at 9:59
1
@uniquesolution and If $V$ is a finite dimensional unitary vector space, and $T$ is normal, there exists a basis of $V$ containing eigenvectors of $T$ and therefore $T$ is diagonalizable . so the minimal poylnomial of $T$ is of the form $left(t-lambda _1right)cdot ...cdot left(t-lambda _kright)$ such that $::kle n$, and all the roots are different. If we assume by contradiction that $k<n$ then we will get a contradiction that the set $leftv,Tleft(vright),...,T^n-1left(vright)right$ is linear independent
– idan di
Aug 13 at 10:05
Why does this solve the problem in the unitary case? How do you know that the diagonal contains different elements?
– uniquesolution
Aug 13 at 9:54
Why does this solve the problem in the unitary case? How do you know that the diagonal contains different elements?
– uniquesolution
Aug 13 at 9:54
Is this an exercise you have been assigned? Are there no other hypotheses?
– Arnaud D.
Aug 13 at 9:57
Is this an exercise you have been assigned? Are there no other hypotheses?
– Arnaud D.
Aug 13 at 9:57
If $V$ is a finite dimensional unitary vector space, and $T$ is normal, there exists a basis of $V$ containing eigenvectors of $T$ and therefore $T$ is diagonalizable .
– idan di
Aug 13 at 9:57
If $V$ is a finite dimensional unitary vector space, and $T$ is normal, there exists a basis of $V$ containing eigenvectors of $T$ and therefore $T$ is diagonalizable .
– idan di
Aug 13 at 9:57
@ArnaudD. just that $T$ is normal and the set $leftv,Tleft(vright),...,T^n-1left(vright)right$ is linear independent
– idan di
Aug 13 at 9:59
@ArnaudD. just that $T$ is normal and the set $leftv,Tleft(vright),...,T^n-1left(vright)right$ is linear independent
– idan di
Aug 13 at 9:59
1
1
@uniquesolution and If $V$ is a finite dimensional unitary vector space, and $T$ is normal, there exists a basis of $V$ containing eigenvectors of $T$ and therefore $T$ is diagonalizable . so the minimal poylnomial of $T$ is of the form $left(t-lambda _1right)cdot ...cdot left(t-lambda _kright)$ such that $::kle n$, and all the roots are different. If we assume by contradiction that $k<n$ then we will get a contradiction that the set $leftv,Tleft(vright),...,T^n-1left(vright)right$ is linear independent
– idan di
Aug 13 at 10:05
@uniquesolution and If $V$ is a finite dimensional unitary vector space, and $T$ is normal, there exists a basis of $V$ containing eigenvectors of $T$ and therefore $T$ is diagonalizable . so the minimal poylnomial of $T$ is of the form $left(t-lambda _1right)cdot ...cdot left(t-lambda _kright)$ such that $::kle n$, and all the roots are different. If we assume by contradiction that $k<n$ then we will get a contradiction that the set $leftv,Tleft(vright),...,T^n-1left(vright)right$ is linear independent
– idan di
Aug 13 at 10:05
 |Â
show 6 more comments
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Why does this solve the problem in the unitary case? How do you know that the diagonal contains different elements?
– uniquesolution
Aug 13 at 9:54
Is this an exercise you have been assigned? Are there no other hypotheses?
– Arnaud D.
Aug 13 at 9:57
If $V$ is a finite dimensional unitary vector space, and $T$ is normal, there exists a basis of $V$ containing eigenvectors of $T$ and therefore $T$ is diagonalizable .
– idan di
Aug 13 at 9:57
@ArnaudD. just that $T$ is normal and the set $leftv,Tleft(vright),...,T^n-1left(vright)right$ is linear independent
– idan di
Aug 13 at 9:59
1
@uniquesolution and If $V$ is a finite dimensional unitary vector space, and $T$ is normal, there exists a basis of $V$ containing eigenvectors of $T$ and therefore $T$ is diagonalizable . so the minimal poylnomial of $T$ is of the form $left(t-lambda _1right)cdot ...cdot left(t-lambda _kright)$ such that $::kle n$, and all the roots are different. If we assume by contradiction that $k<n$ then we will get a contradiction that the set $leftv,Tleft(vright),...,T^n-1left(vright)right$ is linear independent
– idan di
Aug 13 at 10:05