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Proof: Every reflexive $sigma$-sesquilinear form, $sigma ne mathbb1$, is proportional with a hermitian form

0

$begingroup$

$V$ is a $K$-vector field and $sigma$ a field automorphism of $K$.

(Definition: Two $sigma$-sesquilinear forms $f$ and $f'$ are said to be proportional if there exists an $ell in K^*$ such that $f'(v,w) = ell f(v,w), forall v,w in V$.)

Proof:

Suppose that there exists $v in V$ such that $f(v,v) ne 0$. Then we have $f(u,w)=fracf(v,v)f(v,v)^sigma f(w,u)^sigma$ for all $u,w in V$. Define $f':V times V to K: (x,y) mapsto f(v,v)^sigma f(x,y)$, this a $sigma$-sesquilinear form proportional with $f$. We can easily check that $f'$ is a hermitian form.

This proof is all over the place and results are stated without proper explanation.

---

"Suppose that there exists $v in V$ such that $f(v,v) ne 0$." We know that if $f$ is a $sigma$-sesquilinear form ($sigma ne 1$) for which $f(u,u)=0, forall u in V$, then $f$ is the null form. We'd like to prove the theorem for non-trivial sesquilinear forms, therefore such a $v$ exists.

Then we have $f(u,w)=fracf(v,v)f(v,v)^sigma f(w,u)^sigma$ for all $u,w in V$. Where does this expression come from? This is what I wrote next to it: $exists g = af Rightarrow a = ka^sigma Rightarrow a=f(x,x).$ I don't understand what this has to do with the derivation of that expression. Any ideas?

We can easily check that $f'$ is a hermitian form. I have to prove that $f(v,w) = f(w,v)^sigma$. Well, $$ f(v,w) = f(v,v)^sigmaf(v,w)$$ and $$ f(w,v)^sigma = f(v,v)^sigma^2f(w,v)= f(v,v)f(w,v) quad (ftext is reflexive)$$ I don't see the equality here. Did I miss something?

The last to paragraphs are the parts of the proof I don't fully understand. Could somebody help me out. Thanks.

linear-algebra

share|cite|improve this question

asked 2 days ago

ZacharyZachary

1939

$endgroup$

add a comment | 

0

$begingroup$

$V$ is a $K$-vector field and $sigma$ a field automorphism of $K$.

(Definition: Two $sigma$-sesquilinear forms $f$ and $f'$ are said to be proportional if there exists an $ell in K^*$ such that $f'(v,w) = ell f(v,w), forall v,w in V$.)

Proof:

Suppose that there exists $v in V$ such that $f(v,v) ne 0$. Then we have $f(u,w)=fracf(v,v)f(v,v)^sigma f(w,u)^sigma$ for all $u,w in V$. Define $f':V times V to K: (x,y) mapsto f(v,v)^sigma f(x,y)$, this a $sigma$-sesquilinear form proportional with $f$. We can easily check that $f'$ is a hermitian form.

This proof is all over the place and results are stated without proper explanation.

---

"Suppose that there exists $v in V$ such that $f(v,v) ne 0$." We know that if $f$ is a $sigma$-sesquilinear form ($sigma ne 1$) for which $f(u,u)=0, forall u in V$, then $f$ is the null form. We'd like to prove the theorem for non-trivial sesquilinear forms, therefore such a $v$ exists.

Then we have $f(u,w)=fracf(v,v)f(v,v)^sigma f(w,u)^sigma$ for all $u,w in V$. Where does this expression come from? This is what I wrote next to it: $exists g = af Rightarrow a = ka^sigma Rightarrow a=f(x,x).$ I don't understand what this has to do with the derivation of that expression. Any ideas?

We can easily check that $f'$ is a hermitian form. I have to prove that $f(v,w) = f(w,v)^sigma$. Well, $$ f(v,w) = f(v,v)^sigmaf(v,w)$$ and $$ f(w,v)^sigma = f(v,v)^sigma^2f(w,v)= f(v,v)f(w,v) quad (ftext is reflexive)$$ I don't see the equality here. Did I miss something?

The last to paragraphs are the parts of the proof I don't fully understand. Could somebody help me out. Thanks.

linear-algebra

share|cite|improve this question

asked 2 days ago

ZacharyZachary

1939

$endgroup$

add a comment | 

$begingroup$

$V$ is a $K$-vector field and $sigma$ a field automorphism of $K$.

(Definition: Two $sigma$-sesquilinear forms $f$ and $f'$ are said to be proportional if there exists an $ell in K^*$ such that $f'(v,w) = ell f(v,w), forall v,w in V$.)

Proof:

Suppose that there exists $v in V$ such that $f(v,v) ne 0$. Then we have $f(u,w)=fracf(v,v)f(v,v)^sigma f(w,u)^sigma$ for all $u,w in V$. Define $f':V times V to K: (x,y) mapsto f(v,v)^sigma f(x,y)$, this a $sigma$-sesquilinear form proportional with $f$. We can easily check that $f'$ is a hermitian form.

This proof is all over the place and results are stated without proper explanation.

---

"Suppose that there exists $v in V$ such that $f(v,v) ne 0$." We know that if $f$ is a $sigma$-sesquilinear form ($sigma ne 1$) for which $f(u,u)=0, forall u in V$, then $f$ is the null form. We'd like to prove the theorem for non-trivial sesquilinear forms, therefore such a $v$ exists.

Then we have $f(u,w)=fracf(v,v)f(v,v)^sigma f(w,u)^sigma$ for all $u,w in V$. Where does this expression come from? This is what I wrote next to it: $exists g = af Rightarrow a = ka^sigma Rightarrow a=f(x,x).$ I don't understand what this has to do with the derivation of that expression. Any ideas?

We can easily check that $f'$ is a hermitian form. I have to prove that $f(v,w) = f(w,v)^sigma$. Well, $$ f(v,w) = f(v,v)^sigmaf(v,w)$$ and $$ f(w,v)^sigma = f(v,v)^sigma^2f(w,v)= f(v,v)f(w,v) quad (ftext is reflexive)$$ I don't see the equality here. Did I miss something?

The last to paragraphs are the parts of the proof I don't fully understand. Could somebody help me out. Thanks.

linear-algebra

share|cite|improve this question

asked 2 days ago

ZacharyZachary

1939

$endgroup$

share|cite|improve this question

asked 2 days ago

ZacharyZachary

1939

share|cite|improve this question

asked 2 days ago

ZacharyZachary

1939

asked 2 days ago

ZacharyZachary

1939

asked 2 days ago

ZacharyZachary

1939