Bilinear form in finite dimension space
$begingroup$
How to prove that there exist a unique operator $T:Vto V$ such that
$$f(x,y)=g(Tx,y)$$
where $f$ is any bilinear form and $g$ is a non-degenerate bilinear form, and $V$ is a finite dimensional space.
I proved the uniqueness but I dont know how to prove the existence.
linear-algebra
asked Dec 21 '18 at 17:31
Bilal Jafar KarakiBilal Jafar Karaki
382112
$endgroup$
add a comment |
$begingroup$
How to prove that there exist a unique operator $T:Vto V$ such that
$$f(x,y)=g(Tx,y)$$
where $f$ is any bilinear form and $g$ is a non-degenerate bilinear form, and $V$ is a finite dimensional space.
I proved the uniqueness but I dont know how to prove the existence.
linear-algebra
asked Dec 21 '18 at 17:31
Bilal Jafar KarakiBilal Jafar Karaki
382112
$endgroup$
2
$begingroup$
We're missing some information here. Do we know, for instance, that $g$ is a non-degenerate bilinear form?
$endgroup$
– Omnomnomnom
Dec 21 '18 at 17:37
$begingroup$
yes, g is a non-degenerate bilinear form.
$endgroup$
– Bilal Jafar Karaki
Dec 21 '18 at 17:38
add a comment |
$begingroup$
How to prove that there exist a unique operator $T:Vto V$ such that
$$f(x,y)=g(Tx,y)$$
where $f$ is any bilinear form and $g$ is a non-degenerate bilinear form, and $V$ is a finite dimensional space.
I proved the uniqueness but I dont know how to prove the existence.
linear-algebra
asked Dec 21 '18 at 17:31
Bilal Jafar KarakiBilal Jafar Karaki
382112
$endgroup$
How to prove that there exist a unique operator $T:Vto V$ such that
$$f(x,y)=g(Tx,y)$$
where $f$ is any bilinear form and $g$ is a non-degenerate bilinear form, and $V$ is a finite dimensional space.
I proved the uniqueness but I dont know how to prove the existence.
linear-algebra
linear-algebra
asked Dec 21 '18 at 17:31
Bilal Jafar KarakiBilal Jafar Karaki
382112
asked Dec 21 '18 at 17:31
Bilal Jafar KarakiBilal Jafar Karaki
382112
edited Dec 21 '18 at 17:39
Bilal Jafar Karaki
asked Dec 21 '18 at 17:31
Bilal Jafar KarakiBilal Jafar Karaki
382112
asked Dec 21 '18 at 17:31
Bilal Jafar KarakiBilal Jafar Karaki
382112
asked Dec 21 '18 at 17:31
Bilal Jafar KarakiBilal Jafar Karaki
382112
382112
2
$begingroup$
We're missing some information here. Do we know, for instance, that $g$ is a non-degenerate bilinear form?
$endgroup$
– Omnomnomnom
Dec 21 '18 at 17:37
$begingroup$
yes, g is a non-degenerate bilinear form.
$endgroup$
– Bilal Jafar Karaki
Dec 21 '18 at 17:38
add a comment |
2
$begingroup$
We're missing some information here. Do we know, for instance, that $g$ is a non-degenerate bilinear form?
$endgroup$
– Omnomnomnom
Dec 21 '18 at 17:37
$begingroup$
yes, g is a non-degenerate bilinear form.
$endgroup$
– Bilal Jafar Karaki
Dec 21 '18 at 17:38
2
2
$begingroup$
We're missing some information here. Do we know, for instance, that $g$ is a non-degenerate bilinear form?
$endgroup$
– Omnomnomnom
Dec 21 '18 at 17:37
$begingroup$
We're missing some information here. Do we know, for instance, that $g$ is a non-degenerate bilinear form?
$endgroup$
– Omnomnomnom
Dec 21 '18 at 17:37
$begingroup$
yes, g is a non-degenerate bilinear form.
$endgroup$
– Bilal Jafar Karaki
Dec 21 '18 at 17:38
$begingroup$
yes, g is a non-degenerate bilinear form.
$endgroup$
– Bilal Jafar Karaki
Dec 21 '18 at 17:38
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint: fixing a basis for $V$, it suffices to work over $mathbb{R}^n$. Note that there exist matrices $A$ and $B$ such that
$$ f(x,y) = x^top cdot A cdot y qquad text{and} qquad g(x,y) = x^top B cdot y.$$
Now show there exists a matrix $T$ such that $T^top B = A$.
answered Dec 21 '18 at 17:40
SvanNSvanN
2,0661422
$endgroup$
add a comment |
$begingroup$
The form $g$ defines an isomorphism $hat{g}colon Vto V^*$ by setting
$$
hat{g}(x)colon ymapsto g(x,y)
$$
Similarly, $f$ defines a linear map $hat{f}colon Vto V^*$ by $hat{f}(x)(y)=f(x,y)$.
Then we get $Tcolon Vto V$, $T=hat{g}^{-1}circhat{f}$, which is the same as saying that $hat{g}circ T=hat{f}$. In particular, for every $xin V$, we have
$$
hat{f}(x)=hat{g}circ T(x)=hat{g}(T(x))
$$
and therefore, for every $yin V$,
$$
hat{f}(x)(y)=hat{g}circ T(x)=hat{g}(Tx)(y)
$$
which is the same as saying that $f(x,y)=g(Tx,y)$.
How do we prove that $hat{g}$ is an isomorphism? Since $V$ is finite dimensional, it is sufficient to show it is injective. If $hat{g}(x)=0$, then $g(x,y)=0$, for every $yin V$. Being $g$ nondegenerate, this implies $x=0$.
answered Dec 21 '18 at 18:47
egregegreg
183k1486205
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add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
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$begingroup$
Hint: fixing a basis for $V$, it suffices to work over $mathbb{R}^n$. Note that there exist matrices $A$ and $B$ such that
$$ f(x,y) = x^top cdot A cdot y qquad text{and} qquad g(x,y) = x^top B cdot y.$$
Now show there exists a matrix $T$ such that $T^top B = A$.
answered Dec 21 '18 at 17:40
SvanNSvanN
2,0661422
$endgroup$
add a comment |
$begingroup$
Hint: fixing a basis for $V$, it suffices to work over $mathbb{R}^n$. Note that there exist matrices $A$ and $B$ such that
$$ f(x,y) = x^top cdot A cdot y qquad text{and} qquad g(x,y) = x^top B cdot y.$$
Now show there exists a matrix $T$ such that $T^top B = A$.
answered Dec 21 '18 at 17:40
SvanNSvanN
2,0661422
$endgroup$
add a comment |
$begingroup$
Hint: fixing a basis for $V$, it suffices to work over $mathbb{R}^n$. Note that there exist matrices $A$ and $B$ such that
$$ f(x,y) = x^top cdot A cdot y qquad text{and} qquad g(x,y) = x^top B cdot y.$$
Now show there exists a matrix $T$ such that $T^top B = A$.
answered Dec 21 '18 at 17:40
SvanNSvanN
2,0661422
$endgroup$
Hint: fixing a basis for $V$, it suffices to work over $mathbb{R}^n$. Note that there exist matrices $A$ and $B$ such that
$$ f(x,y) = x^top cdot A cdot y qquad text{and} qquad g(x,y) = x^top B cdot y.$$
Now show there exists a matrix $T$ such that $T^top B = A$.
answered Dec 21 '18 at 17:40
SvanNSvanN
2,0661422
answered Dec 21 '18 at 17:40
SvanNSvanN
2,0661422
answered Dec 21 '18 at 17:40
SvanNSvanN
2,0661422
answered Dec 21 '18 at 17:40
SvanNSvanN
2,0661422
2,0661422
add a comment |
add a comment |
$begingroup$
The form $g$ defines an isomorphism $hat{g}colon Vto V^*$ by setting
$$
hat{g}(x)colon ymapsto g(x,y)
$$
Similarly, $f$ defines a linear map $hat{f}colon Vto V^*$ by $hat{f}(x)(y)=f(x,y)$.
Then we get $Tcolon Vto V$, $T=hat{g}^{-1}circhat{f}$, which is the same as saying that $hat{g}circ T=hat{f}$. In particular, for every $xin V$, we have
$$
hat{f}(x)=hat{g}circ T(x)=hat{g}(T(x))
$$
and therefore, for every $yin V$,
$$
hat{f}(x)(y)=hat{g}circ T(x)=hat{g}(Tx)(y)
$$
which is the same as saying that $f(x,y)=g(Tx,y)$.
How do we prove that $hat{g}$ is an isomorphism? Since $V$ is finite dimensional, it is sufficient to show it is injective. If $hat{g}(x)=0$, then $g(x,y)=0$, for every $yin V$. Being $g$ nondegenerate, this implies $x=0$.
answered Dec 21 '18 at 18:47
egregegreg
183k1486205
$endgroup$
add a comment |
$begingroup$
The form $g$ defines an isomorphism $hat{g}colon Vto V^*$ by setting
$$
hat{g}(x)colon ymapsto g(x,y)
$$
Similarly, $f$ defines a linear map $hat{f}colon Vto V^*$ by $hat{f}(x)(y)=f(x,y)$.
Then we get $Tcolon Vto V$, $T=hat{g}^{-1}circhat{f}$, which is the same as saying that $hat{g}circ T=hat{f}$. In particular, for every $xin V$, we have
$$
hat{f}(x)=hat{g}circ T(x)=hat{g}(T(x))
$$
and therefore, for every $yin V$,
$$
hat{f}(x)(y)=hat{g}circ T(x)=hat{g}(Tx)(y)
$$
which is the same as saying that $f(x,y)=g(Tx,y)$.
How do we prove that $hat{g}$ is an isomorphism? Since $V$ is finite dimensional, it is sufficient to show it is injective. If $hat{g}(x)=0$, then $g(x,y)=0$, for every $yin V$. Being $g$ nondegenerate, this implies $x=0$.
answered Dec 21 '18 at 18:47
egregegreg
183k1486205
$endgroup$
add a comment |
$begingroup$
The form $g$ defines an isomorphism $hat{g}colon Vto V^*$ by setting
$$
hat{g}(x)colon ymapsto g(x,y)
$$
Similarly, $f$ defines a linear map $hat{f}colon Vto V^*$ by $hat{f}(x)(y)=f(x,y)$.
Then we get $Tcolon Vto V$, $T=hat{g}^{-1}circhat{f}$, which is the same as saying that $hat{g}circ T=hat{f}$. In particular, for every $xin V$, we have
$$
hat{f}(x)=hat{g}circ T(x)=hat{g}(T(x))
$$
and therefore, for every $yin V$,
$$
hat{f}(x)(y)=hat{g}circ T(x)=hat{g}(Tx)(y)
$$
which is the same as saying that $f(x,y)=g(Tx,y)$.
How do we prove that $hat{g}$ is an isomorphism? Since $V$ is finite dimensional, it is sufficient to show it is injective. If $hat{g}(x)=0$, then $g(x,y)=0$, for every $yin V$. Being $g$ nondegenerate, this implies $x=0$.
answered Dec 21 '18 at 18:47
egregegreg
183k1486205
$endgroup$
The form $g$ defines an isomorphism $hat{g}colon Vto V^*$ by setting
$$
hat{g}(x)colon ymapsto g(x,y)
$$
Similarly, $f$ defines a linear map $hat{f}colon Vto V^*$ by $hat{f}(x)(y)=f(x,y)$.
Then we get $Tcolon Vto V$, $T=hat{g}^{-1}circhat{f}$, which is the same as saying that $hat{g}circ T=hat{f}$. In particular, for every $xin V$, we have
$$
hat{f}(x)=hat{g}circ T(x)=hat{g}(T(x))
$$
and therefore, for every $yin V$,
$$
hat{f}(x)(y)=hat{g}circ T(x)=hat{g}(Tx)(y)
$$
which is the same as saying that $f(x,y)=g(Tx,y)$.
How do we prove that $hat{g}$ is an isomorphism? Since $V$ is finite dimensional, it is sufficient to show it is injective. If $hat{g}(x)=0$, then $g(x,y)=0$, for every $yin V$. Being $g$ nondegenerate, this implies $x=0$.
answered Dec 21 '18 at 18:47
egregegreg
183k1486205
answered Dec 21 '18 at 18:47
egregegreg
183k1486205
answered Dec 21 '18 at 18:47
egregegreg
183k1486205
answered Dec 21 '18 at 18:47
egregegreg
183k1486205
183k1486205
add a comment |
add a comment |
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2
$begingroup$
We're missing some information here. Do we know, for instance, that $g$ is a non-degenerate bilinear form?
$endgroup$
– Omnomnomnom
Dec 21 '18 at 17:37
$begingroup$
yes, g is a non-degenerate bilinear form.
$endgroup$
– Bilal Jafar Karaki
Dec 21 '18 at 17:38