Mixing
About change of basis
$begingroup$
This is proof Lemma 1.2.1 from Scharlau's book Quadratic and Hermitian Forms.
Lemma 2.1. $$B_{b,mathfrak E'}=T^TB_{b,mathfrak E}T'.$$
Proof. We compute
$$b(e_i',e_j')=b(sum e_kt_{ki}, sum e_lt_{lj})=sum_{k,l} t_{ki} b(e_k,e_l)t_{lj}$$
yielding the $(i,j)$ entry of $T^TB_{b,mathfrak E}T'$.
I don't know why bilinear form splits like this. In first step they just take new basis as linear combination of standard basis. Then next step is unclear to me. Can someone please tell in explicit way assuming n basis
linear-algebra linear-transformations bilinear-form
edited Jul 2 '18 at 12:54
Martin Sleziak
45k10122277
asked Jun 22 '18 at 5:54
maths studentmaths student
6321521
$endgroup$
add a comment |
$begingroup$
This is proof Lemma 1.2.1 from Scharlau's book Quadratic and Hermitian Forms.
Lemma 2.1. $$B_{b,mathfrak E'}=T^TB_{b,mathfrak E}T'.$$
Proof. We compute
$$b(e_i',e_j')=b(sum e_kt_{ki}, sum e_lt_{lj})=sum_{k,l} t_{ki} b(e_k,e_l)t_{lj}$$
yielding the $(i,j)$ entry of $T^TB_{b,mathfrak E}T'$.
I don't know why bilinear form splits like this. In first step they just take new basis as linear combination of standard basis. Then next step is unclear to me. Can someone please tell in explicit way assuming n basis
linear-algebra linear-transformations bilinear-form
edited Jul 2 '18 at 12:54
Martin Sleziak
45k10122277
asked Jun 22 '18 at 5:54
maths studentmaths student
6321521
$endgroup$
add a comment |
$begingroup$
This is proof Lemma 1.2.1 from Scharlau's book Quadratic and Hermitian Forms.
Lemma 2.1. $$B_{b,mathfrak E'}=T^TB_{b,mathfrak E}T'.$$
Proof. We compute
$$b(e_i',e_j')=b(sum e_kt_{ki}, sum e_lt_{lj})=sum_{k,l} t_{ki} b(e_k,e_l)t_{lj}$$
yielding the $(i,j)$ entry of $T^TB_{b,mathfrak E}T'$.
I don't know why bilinear form splits like this. In first step they just take new basis as linear combination of standard basis. Then next step is unclear to me. Can someone please tell in explicit way assuming n basis
linear-algebra linear-transformations bilinear-form
edited Jul 2 '18 at 12:54
Martin Sleziak
45k10122277
asked Jun 22 '18 at 5:54
maths studentmaths student
6321521
$endgroup$
This is proof Lemma 1.2.1 from Scharlau's book Quadratic and Hermitian Forms.
Lemma 2.1. $$B_{b,mathfrak E'}=T^TB_{b,mathfrak E}T'.$$
Proof. We compute
$$b(e_i',e_j')=b(sum e_kt_{ki}, sum e_lt_{lj})=sum_{k,l} t_{ki} b(e_k,e_l)t_{lj}$$
yielding the $(i,j)$ entry of $T^TB_{b,mathfrak E}T'$.
I don't know why bilinear form splits like this. In first step they just take new basis as linear combination of standard basis. Then next step is unclear to me. Can someone please tell in explicit way assuming n basis
linear-algebra linear-transformations bilinear-form
linear-algebra linear-transformations bilinear-form
edited Jul 2 '18 at 12:54
Martin Sleziak
45k10122277
asked Jun 22 '18 at 5:54
maths studentmaths student
6321521
edited Jul 2 '18 at 12:54
Martin Sleziak
45k10122277
asked Jun 22 '18 at 5:54
maths studentmaths student
6321521
edited Jul 2 '18 at 12:54
Martin Sleziak
45k10122277
edited Jul 2 '18 at 12:54
Martin Sleziak
45k10122277
edited Jul 2 '18 at 12:54
Martin Sleziak
45k10122277
45k10122277
asked Jun 22 '18 at 5:54
maths studentmaths student
6321521
asked Jun 22 '18 at 5:54
maths studentmaths student
6321521
asked Jun 22 '18 at 5:54
maths studentmaths student
6321521
6321521
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
you simply use the definition of bilinear functions, i.e. linearity in every component: $b(sum_{i=1}^{n} t_ie_i,y)=sum_{i=1}^{n}t_ib(e_i,y)$ for a fixed $y$ in the second component and similarly $b(x,sum_{j=1}^{n} t_je_j)=sum_{j=1}^{n}t_jb(x,e_j)$. Then just use them both together to get above result.
answered Jun 22 '18 at 6:29
SimonsaysSimonsays
521210
$endgroup$
add a comment |
$begingroup$
Given a bilinear form $B$ in the standard basis
$$x^tBy$$
by a change of basis with matrix $T$ we have
- $x=Tu$
- $y=Tv$
and therefore
$$x^tBy=(Tu)^tBTv=u^tT^tBTv$$
answered Jun 22 '18 at 6:30
gimusigimusi
92.9k84594
$endgroup$
add a comment |
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2 Answers
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2 Answers
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active
oldest
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$begingroup$
you simply use the definition of bilinear functions, i.e. linearity in every component: $b(sum_{i=1}^{n} t_ie_i,y)=sum_{i=1}^{n}t_ib(e_i,y)$ for a fixed $y$ in the second component and similarly $b(x,sum_{j=1}^{n} t_je_j)=sum_{j=1}^{n}t_jb(x,e_j)$. Then just use them both together to get above result.
answered Jun 22 '18 at 6:29
SimonsaysSimonsays
521210
$endgroup$
add a comment |
$begingroup$
you simply use the definition of bilinear functions, i.e. linearity in every component: $b(sum_{i=1}^{n} t_ie_i,y)=sum_{i=1}^{n}t_ib(e_i,y)$ for a fixed $y$ in the second component and similarly $b(x,sum_{j=1}^{n} t_je_j)=sum_{j=1}^{n}t_jb(x,e_j)$. Then just use them both together to get above result.
answered Jun 22 '18 at 6:29
SimonsaysSimonsays
521210
$endgroup$
add a comment |
$begingroup$
you simply use the definition of bilinear functions, i.e. linearity in every component: $b(sum_{i=1}^{n} t_ie_i,y)=sum_{i=1}^{n}t_ib(e_i,y)$ for a fixed $y$ in the second component and similarly $b(x,sum_{j=1}^{n} t_je_j)=sum_{j=1}^{n}t_jb(x,e_j)$. Then just use them both together to get above result.
answered Jun 22 '18 at 6:29
SimonsaysSimonsays
521210
$endgroup$
you simply use the definition of bilinear functions, i.e. linearity in every component: $b(sum_{i=1}^{n} t_ie_i,y)=sum_{i=1}^{n}t_ib(e_i,y)$ for a fixed $y$ in the second component and similarly $b(x,sum_{j=1}^{n} t_je_j)=sum_{j=1}^{n}t_jb(x,e_j)$. Then just use them both together to get above result.
answered Jun 22 '18 at 6:29
SimonsaysSimonsays
521210
answered Jun 22 '18 at 6:29
SimonsaysSimonsays
521210
answered Jun 22 '18 at 6:29
SimonsaysSimonsays
521210
answered Jun 22 '18 at 6:29
SimonsaysSimonsays
521210
521210
add a comment |
add a comment |
$begingroup$
Given a bilinear form $B$ in the standard basis
$$x^tBy$$
by a change of basis with matrix $T$ we have
- $x=Tu$
- $y=Tv$
and therefore
$$x^tBy=(Tu)^tBTv=u^tT^tBTv$$
answered Jun 22 '18 at 6:30
gimusigimusi
92.9k84594
$endgroup$
add a comment |
$begingroup$
Given a bilinear form $B$ in the standard basis
$$x^tBy$$
by a change of basis with matrix $T$ we have
- $x=Tu$
- $y=Tv$
and therefore
$$x^tBy=(Tu)^tBTv=u^tT^tBTv$$
answered Jun 22 '18 at 6:30
gimusigimusi
92.9k84594
$endgroup$
add a comment |
$begingroup$
Given a bilinear form $B$ in the standard basis
$$x^tBy$$
by a change of basis with matrix $T$ we have
- $x=Tu$
- $y=Tv$
and therefore
$$x^tBy=(Tu)^tBTv=u^tT^tBTv$$
answered Jun 22 '18 at 6:30
gimusigimusi
92.9k84594
$endgroup$
Given a bilinear form $B$ in the standard basis
$$x^tBy$$
by a change of basis with matrix $T$ we have
- $x=Tu$
- $y=Tv$
and therefore
$$x^tBy=(Tu)^tBTv=u^tT^tBTv$$
answered Jun 22 '18 at 6:30
gimusigimusi
92.9k84594
answered Jun 22 '18 at 6:30
gimusigimusi
92.9k84594
answered Jun 22 '18 at 6:30
gimusigimusi
92.9k84594
answered Jun 22 '18 at 6:30
gimusigimusi
92.9k84594
92.9k84594
add a comment |
add a comment |
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