Mixing
Let ${^ts}:Eto F$ be the transpose of $s:Fto E$. Show that $text{Im}(s)cap ker (,^ts)={0_E}$.
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Let $sin mathcal{L}(F,E)$
$$displaystyle F overset{s}{longrightarrow} Eoverset{^ts}{longrightarrow} F$$
I spent two days to show that :$$text{Im}(s)cap ker (,^ts)={0_E}qquad tag{1}$$
I'm not sure that is right, I tried with specific matrix (3x3), and it works.
But when I want to provide a general proof, I struggle.
I used the Annihilator but no way to find a solution, may be this statement is wrong...?
I add more information about $E$ and $F$ regarding the comments
$F=mathbb{R}^p$ and $E=mathbb{R}^n$ with $ple n$
Or $F=(mathbb{R}^p,langlecdot,cdotrangle)$ and $E=(mathbb{R}^n,langlecdot,cdotrangle)$
The matrix associated at $u$ is $M$ and $Min mathcal{M}_{n,p}(mathbb{R})$
linear-algebra vector-spaces linear-transformations bilinear-form transpose
edited Nov 16 at 18:31
Batominovski
31.3k23187
asked Nov 16 at 15:57
Stu
1,1011313
add a comment |
up vote
3
down vote
favorite
Let $sin mathcal{L}(F,E)$
$$displaystyle F overset{s}{longrightarrow} Eoverset{^ts}{longrightarrow} F$$
I spent two days to show that :$$text{Im}(s)cap ker (,^ts)={0_E}qquad tag{1}$$
I'm not sure that is right, I tried with specific matrix (3x3), and it works.
But when I want to provide a general proof, I struggle.
I used the Annihilator but no way to find a solution, may be this statement is wrong...?
I add more information about $E$ and $F$ regarding the comments
$F=mathbb{R}^p$ and $E=mathbb{R}^n$ with $ple n$
Or $F=(mathbb{R}^p,langlecdot,cdotrangle)$ and $E=(mathbb{R}^n,langlecdot,cdotrangle)$
The matrix associated at $u$ is $M$ and $Min mathcal{M}_{n,p}(mathbb{R})$
linear-algebra vector-spaces linear-transformations bilinear-form transpose
edited Nov 16 at 18:31
Batominovski
31.3k23187
asked Nov 16 at 15:57
Stu
1,1011313
2
I think the problem is not clear. What is the base field here? Presumably, you have nondegenerate bilinear forms on $E$ and $F$. However, not all bilinear forms work. If the base field is of a positive characteristic, then this is false. Even when the characteristic of the base field is $0$, you can come up with a nondegenerate symmetric bilinear form that contradicts the claim.
– Batominovski
Nov 16 at 17:03
1
So, I would like the OP to confirm whether: (1) the base field is $mathbb{R}$, (2) $E$ and $F$ are finite-dimensional vector spaces, and (3) the vector spaces are equipped with positive-definite symmetric bilinear forms.
– Batominovski
Nov 16 at 17:05
I added more informations, but I didn't want to use the inner product at first...but it seems I must do.
– Stu
Nov 16 at 17:23
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Let $sin mathcal{L}(F,E)$
$$displaystyle F overset{s}{longrightarrow} Eoverset{^ts}{longrightarrow} F$$
I spent two days to show that :$$text{Im}(s)cap ker (,^ts)={0_E}qquad tag{1}$$
I'm not sure that is right, I tried with specific matrix (3x3), and it works.
But when I want to provide a general proof, I struggle.
I used the Annihilator but no way to find a solution, may be this statement is wrong...?
I add more information about $E$ and $F$ regarding the comments
$F=mathbb{R}^p$ and $E=mathbb{R}^n$ with $ple n$
Or $F=(mathbb{R}^p,langlecdot,cdotrangle)$ and $E=(mathbb{R}^n,langlecdot,cdotrangle)$
The matrix associated at $u$ is $M$ and $Min mathcal{M}_{n,p}(mathbb{R})$
linear-algebra vector-spaces linear-transformations bilinear-form transpose
edited Nov 16 at 18:31
Batominovski
31.3k23187
asked Nov 16 at 15:57
Stu
1,1011313
Let $sin mathcal{L}(F,E)$
$$displaystyle F overset{s}{longrightarrow} Eoverset{^ts}{longrightarrow} F$$
I spent two days to show that :$$text{Im}(s)cap ker (,^ts)={0_E}qquad tag{1}$$
I'm not sure that is right, I tried with specific matrix (3x3), and it works.
But when I want to provide a general proof, I struggle.
I used the Annihilator but no way to find a solution, may be this statement is wrong...?
I add more information about $E$ and $F$ regarding the comments
$F=mathbb{R}^p$ and $E=mathbb{R}^n$ with $ple n$
Or $F=(mathbb{R}^p,langlecdot,cdotrangle)$ and $E=(mathbb{R}^n,langlecdot,cdotrangle)$
The matrix associated at $u$ is $M$ and $Min mathcal{M}_{n,p}(mathbb{R})$
linear-algebra vector-spaces linear-transformations bilinear-form transpose
linear-algebra vector-spaces linear-transformations bilinear-form transpose
edited Nov 16 at 18:31
Batominovski
31.3k23187
asked Nov 16 at 15:57
Stu
1,1011313
edited Nov 16 at 18:31
Batominovski
31.3k23187
asked Nov 16 at 15:57
Stu
1,1011313
edited Nov 16 at 18:31
Batominovski
31.3k23187
edited Nov 16 at 18:31
Batominovski
31.3k23187
edited Nov 16 at 18:31
Batominovski
31.3k23187
31.3k23187
asked Nov 16 at 15:57
Stu
1,1011313
asked Nov 16 at 15:57
Stu
1,1011313
asked Nov 16 at 15:57
Stu
1,1011313
1,1011313
2
I think the problem is not clear. What is the base field here? Presumably, you have nondegenerate bilinear forms on $E$ and $F$. However, not all bilinear forms work. If the base field is of a positive characteristic, then this is false. Even when the characteristic of the base field is $0$, you can come up with a nondegenerate symmetric bilinear form that contradicts the claim.
– Batominovski
Nov 16 at 17:03
1
So, I would like the OP to confirm whether: (1) the base field is $mathbb{R}$, (2) $E$ and $F$ are finite-dimensional vector spaces, and (3) the vector spaces are equipped with positive-definite symmetric bilinear forms.
– Batominovski
Nov 16 at 17:05
I added more informations, but I didn't want to use the inner product at first...but it seems I must do.
– Stu
Nov 16 at 17:23
add a comment |
2
I think the problem is not clear. What is the base field here? Presumably, you have nondegenerate bilinear forms on $E$ and $F$. However, not all bilinear forms work. If the base field is of a positive characteristic, then this is false. Even when the characteristic of the base field is $0$, you can come up with a nondegenerate symmetric bilinear form that contradicts the claim.
– Batominovski
Nov 16 at 17:03
1
So, I would like the OP to confirm whether: (1) the base field is $mathbb{R}$, (2) $E$ and $F$ are finite-dimensional vector spaces, and (3) the vector spaces are equipped with positive-definite symmetric bilinear forms.
– Batominovski
Nov 16 at 17:05
I added more informations, but I didn't want to use the inner product at first...but it seems I must do.
– Stu
Nov 16 at 17:23
2
2
I think the problem is not clear. What is the base field here? Presumably, you have nondegenerate bilinear forms on $E$ and $F$. However, not all bilinear forms work. If the base field is of a positive characteristic, then this is false. Even when the characteristic of the base field is $0$, you can come up with a nondegenerate symmetric bilinear form that contradicts the claim.
– Batominovski
Nov 16 at 17:03
I think the problem is not clear. What is the base field here? Presumably, you have nondegenerate bilinear forms on $E$ and $F$. However, not all bilinear forms work. If the base field is of a positive characteristic, then this is false. Even when the characteristic of the base field is $0$, you can come up with a nondegenerate symmetric bilinear form that contradicts the claim.
– Batominovski
Nov 16 at 17:03
1
1
So, I would like the OP to confirm whether: (1) the base field is $mathbb{R}$, (2) $E$ and $F$ are finite-dimensional vector spaces, and (3) the vector spaces are equipped with positive-definite symmetric bilinear forms.
– Batominovski
Nov 16 at 17:05
So, I would like the OP to confirm whether: (1) the base field is $mathbb{R}$, (2) $E$ and $F$ are finite-dimensional vector spaces, and (3) the vector spaces are equipped with positive-definite symmetric bilinear forms.
– Batominovski
Nov 16 at 17:05
I added more informations, but I didn't want to use the inner product at first...but it seems I must do.
– Stu
Nov 16 at 17:23
I added more informations, but I didn't want to use the inner product at first...but it seems I must do.
– Stu
Nov 16 at 17:23
add a comment |
4 Answers
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2
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Let $s:Fto E$ be a linear transformation from a vector space $F$ to another vector space $E$ over the base field $mathbb{R}$. Both $E$ and $F$ are equipped with nondegenerate symmetric bilinear forms $langle_,_rangle_E$ and $langle_,_rangle_F$, respectively. Suppose further that $langle_,_rangle_E$ is positive-definite. Write $s^top:Eto F$ for the transpose of $s$. We claim that
$$text{im}(s)capker(s^top)={0_E},.$$
Recall that $s^top$ is the unique linear transformation from $E$ to $F$ such that $$biglangle s(x),ybigrangle_E=biglangle x,s^top(y)rangle_F$$
for all $xin F$ and $yin E$. Suppose that $yintext{im}(s)capker(s^top)$. Then, $yintext{im}(s)$, so $y=s(x)$ for some $xin F$. Sinc $yinker(s^top)$, we get $$s^topbig(s(x)big)=s^top(y)=0_F,.$$
Therefore,
$$Biglangle x,s^topbig(s(x)big)Bigrangle_F=biglangle x,0_Fbigrangle_F=0,.$$
By the definition of $s^top$, we have
$$langle y,yrangle_E=biglangle s(x),s(x)bigrangle_E=Biglangle x,s^topbig(s(x)big)Bigrangle_F=0,.$$
Since $langle_,_rangle_E$ is positive-definite, we have $y=0_E$. This proves the claim.
P.S.: If the ground field is $mathbb{C}$, we instead assume that $langle_,_rangle_E$ and $langle_,_rangle_F$ are nondegenerate Hermitian forms. The transpose map has to be replaced by the Hermitian conjugate. For the claim to hold, we need to assume further that $langle_,_rangle_E$ is positive-definite.
If the ground field $mathbb{K}$ is a field of characteristic $0$, but there is no positive-definiteness assumption on $langle_,_rangle_E$ (which makes no sense when $mathbb{K}$ is not a subfied of $mathbb{R}$ anyhow), then the claim is not true. For example, let $mathbb{K}:=mathbb{R}$, $E:=mathbb{R}^2$, and $F:=mathbb{R}^2$. Then we set $e_1:=(1,0)$, $e_2:=(0,1)$. Let $langle_,_rangle$ be the nondegenerate symmetric bilinear form on both $E$ and $F$ given by $$langle a_1e_1+a_2e_2,b_1e_1+b_2e_2rangle:=a_1b_1-a_2b_2$$
for all $a_1,a_2,b_1,b_2inmathbb{R}$. Take $s:Fto E$ to be the linear map sending both $e_1$ and $e_2$ to $e_1+e_2$. That is, $s$ is given by the matrix
$$begin{bmatrix}1&1\1&1end{bmatrix}$$
with respect to the ordered basis $(e_1,e_2)$ of both $F$ and $E$. Then, $s^top:Eto F$ is given by the matrix
$$begin{bmatrix}1&-1\-1&1end{bmatrix}$$
with respect to the ordered basis $(e_1,e_2)$ of both $E$ and $F$. This shows that $$text{im}(s)=text{span}_mathbb{R}{e_1+e_2}=ker(s^top),.$$
The claim fails in positive characteristics. If the ground field $mathbb{K}$ is of a positive characteristric $p$, then take $E$ and $F$ to be both $mathbb{K}^p$. Fix a basis ${e_1,e_2,ldots,e_p}$ of both $E$ and $F$. Equip $E$ and $F$ with the nondegenerate symmetric bilinear form $langle_,_rangle$ given by
$$leftlanglesum_{i=1}^p,a_ie_i,sum_{i=1}^p,b_ie_irightrangle:=sum_{i=1}^p,a_ib_i$$
for all $a_1,a_2,ldots,a_p,b_1,b_2,ldots,b_pinmathbb{K}$. If the map $s:Fto E$ is the $mathbb{K}$-linear map sending $e_imapsto e_1+e_2+ldots+e_p$ for all $i=1,2,ldots,p$, then its transpose $s^top:Eto F$ also sends $e_imapsto e_1+e_2+ldots+e_p$ for all $i=1,2,ldots,p$. In particular, this shows that $$text{span}_mathbb{K}{e_1+e_2+ldots+e_p}=text{im}(s)capker(s^top),.$$
answered Nov 16 at 17:49
Batominovski
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It is sufficient to show that
$$
ker s^tsubset(text{Im } s)^perp
$$
because we can then conclude by:
$$
ker s^t cap text{Im} s subset (text{Im} s)^perp cap text{Im} s ={0_E}
$$
The claim $ker s^tsubset(text{Im } s)^perp$ can be proven as follows:
begin{align*}
einker s^t &Rightarrow s^t(e)=0_F \
&Rightarrow forall fin F, langle f,s^t(e) rangle_F=0 \ &Rightarrow forall fin F, langle s(f),erangle_E=0 \
&Rightarrow forall fin F, s(f) perp e \
&Rightarrow e in (text{Im} s)^perp
end{align*}
answered Nov 16 at 18:20
Picaud Vincent
67815
1
I would like to note that the equality $ker (s^top)=big(text{im}(s)big)^perp$ is true for any nondegenerate symmetric bilinear form $langle_,_rangle_E$. However, the last equality $big(text{im}(s)big)^perpcaptext{im}(s)={0_E}$ must rely on some strong conditions on $langle_,_rangle_E$ such as positive-definiteness.
– Batominovski
Nov 16 at 18:27
thanks a lot I need some time to understand everything, and I'll be back
– Stu
Nov 16 at 18:28
@Batominovski you are right, and that is the reason why I simply wrote positive definite right at the beginning. Also a more general result would also hold only using duality bracket $<. ,.>_{Etimes E^*}$ and $<. ,.>_{Ftimes F^*}$ (without introducing scalar products $<. ,.>_E$ and $<. ,.>_F$
– Picaud Vincent
Nov 16 at 18:30
Shorter proof (sorry it is nearly a complete rewrite, but the idea is the same)
– Picaud Vincent
Nov 16 at 21:39
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In order that the transpose is a map $Fto E$ you need some assumptions: both vector spaces need to be equipped with a nondegenerate bilinear form, so that we can define isomorphisms $Eto E^*$ and $Fto F^*$ (the dual spaces), assuming finite dimensionality.
In particular, if the base field is $mathbb{R}$, the two space might be inner product spaces.
If $langle{cdot},{cdot}rangle_E$ is the form on $E$, then for each $vin E$, the map
$$
e_vcolon Eto K,qquad e_v(x)=langle v,xrangle_E
$$
is an element of $E^*$. If $vne0$, then nondegeneracy implies there exists $xin E$ with $langle v,xrangle_Ene0$, so $ecolon Eto E^*$, $vmapsto e_v$ is an isomorphism (it is clear from bilinearity that $e_v$ is linear, for every $vin E$, and $e$ is linear as well).
Then the transpose ${}^{t!}s$ is the dual map composed with these isomorphisms:
$$
{}^{t!}s=e^{-1}circ s^*circ e
$$
(I denote by $e$ both maps $Eto E^*$ and $Fto F^*$, no confusion should arise).
If $w=s(v)$ and ${}^{t!}s(w)=0$, then also $s^*circ e_w=0$, which means $e_wcirc s=0$, that is,
$$
e_w(s(x))=0quadtext{for all $xin V$}
$$
hence
$$
langle w,s(x)rangle_F=0
$$
In particular, $langle s(v),s(v)rangle_F=0$, so by nondegeneracy, $s(v)=0$ and $w=0$.
answered Nov 16 at 23:16
egreg
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If you represent a linear map by a matrix then the image of the map is the sub-space spanned of the columns of the matrix. And the kernel of the map is the sub-space that is perpendicular to the rows of the matrix (because the inner product of each row with any vector in the kernel must be zero).
So the image of $s$ is the sub-space that is spanned by the columns of the matrix representing $s$ (relative to specific bases in $E$ and $F$).
And the kernel of $^ts$ is the sub-space that is perpendicular to the sub-space spanned by the rows of the matrix representing $^ts$.
But the columns of the first matrix are the rows of the second matrix. The result follows.
answered Nov 16 at 16:48
gandalf61
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4 Answers
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4 Answers
4
active
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active
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active
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up vote
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Let $s:Fto E$ be a linear transformation from a vector space $F$ to another vector space $E$ over the base field $mathbb{R}$. Both $E$ and $F$ are equipped with nondegenerate symmetric bilinear forms $langle_,_rangle_E$ and $langle_,_rangle_F$, respectively. Suppose further that $langle_,_rangle_E$ is positive-definite. Write $s^top:Eto F$ for the transpose of $s$. We claim that
$$text{im}(s)capker(s^top)={0_E},.$$
Recall that $s^top$ is the unique linear transformation from $E$ to $F$ such that $$biglangle s(x),ybigrangle_E=biglangle x,s^top(y)rangle_F$$
for all $xin F$ and $yin E$. Suppose that $yintext{im}(s)capker(s^top)$. Then, $yintext{im}(s)$, so $y=s(x)$ for some $xin F$. Sinc $yinker(s^top)$, we get $$s^topbig(s(x)big)=s^top(y)=0_F,.$$
Therefore,
$$Biglangle x,s^topbig(s(x)big)Bigrangle_F=biglangle x,0_Fbigrangle_F=0,.$$
By the definition of $s^top$, we have
$$langle y,yrangle_E=biglangle s(x),s(x)bigrangle_E=Biglangle x,s^topbig(s(x)big)Bigrangle_F=0,.$$
Since $langle_,_rangle_E$ is positive-definite, we have $y=0_E$. This proves the claim.
P.S.: If the ground field is $mathbb{C}$, we instead assume that $langle_,_rangle_E$ and $langle_,_rangle_F$ are nondegenerate Hermitian forms. The transpose map has to be replaced by the Hermitian conjugate. For the claim to hold, we need to assume further that $langle_,_rangle_E$ is positive-definite.
If the ground field $mathbb{K}$ is a field of characteristic $0$, but there is no positive-definiteness assumption on $langle_,_rangle_E$ (which makes no sense when $mathbb{K}$ is not a subfied of $mathbb{R}$ anyhow), then the claim is not true. For example, let $mathbb{K}:=mathbb{R}$, $E:=mathbb{R}^2$, and $F:=mathbb{R}^2$. Then we set $e_1:=(1,0)$, $e_2:=(0,1)$. Let $langle_,_rangle$ be the nondegenerate symmetric bilinear form on both $E$ and $F$ given by $$langle a_1e_1+a_2e_2,b_1e_1+b_2e_2rangle:=a_1b_1-a_2b_2$$
for all $a_1,a_2,b_1,b_2inmathbb{R}$. Take $s:Fto E$ to be the linear map sending both $e_1$ and $e_2$ to $e_1+e_2$. That is, $s$ is given by the matrix
$$begin{bmatrix}1&1\1&1end{bmatrix}$$
with respect to the ordered basis $(e_1,e_2)$ of both $F$ and $E$. Then, $s^top:Eto F$ is given by the matrix
$$begin{bmatrix}1&-1\-1&1end{bmatrix}$$
with respect to the ordered basis $(e_1,e_2)$ of both $E$ and $F$. This shows that $$text{im}(s)=text{span}_mathbb{R}{e_1+e_2}=ker(s^top),.$$
The claim fails in positive characteristics. If the ground field $mathbb{K}$ is of a positive characteristric $p$, then take $E$ and $F$ to be both $mathbb{K}^p$. Fix a basis ${e_1,e_2,ldots,e_p}$ of both $E$ and $F$. Equip $E$ and $F$ with the nondegenerate symmetric bilinear form $langle_,_rangle$ given by
$$leftlanglesum_{i=1}^p,a_ie_i,sum_{i=1}^p,b_ie_irightrangle:=sum_{i=1}^p,a_ib_i$$
for all $a_1,a_2,ldots,a_p,b_1,b_2,ldots,b_pinmathbb{K}$. If the map $s:Fto E$ is the $mathbb{K}$-linear map sending $e_imapsto e_1+e_2+ldots+e_p$ for all $i=1,2,ldots,p$, then its transpose $s^top:Eto F$ also sends $e_imapsto e_1+e_2+ldots+e_p$ for all $i=1,2,ldots,p$. In particular, this shows that $$text{span}_mathbb{K}{e_1+e_2+ldots+e_p}=text{im}(s)capker(s^top),.$$
answered Nov 16 at 17:49
Batominovski
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Let $s:Fto E$ be a linear transformation from a vector space $F$ to another vector space $E$ over the base field $mathbb{R}$. Both $E$ and $F$ are equipped with nondegenerate symmetric bilinear forms $langle_,_rangle_E$ and $langle_,_rangle_F$, respectively. Suppose further that $langle_,_rangle_E$ is positive-definite. Write $s^top:Eto F$ for the transpose of $s$. We claim that
$$text{im}(s)capker(s^top)={0_E},.$$
Recall that $s^top$ is the unique linear transformation from $E$ to $F$ such that $$biglangle s(x),ybigrangle_E=biglangle x,s^top(y)rangle_F$$
for all $xin F$ and $yin E$. Suppose that $yintext{im}(s)capker(s^top)$. Then, $yintext{im}(s)$, so $y=s(x)$ for some $xin F$. Sinc $yinker(s^top)$, we get $$s^topbig(s(x)big)=s^top(y)=0_F,.$$
Therefore,
$$Biglangle x,s^topbig(s(x)big)Bigrangle_F=biglangle x,0_Fbigrangle_F=0,.$$
By the definition of $s^top$, we have
$$langle y,yrangle_E=biglangle s(x),s(x)bigrangle_E=Biglangle x,s^topbig(s(x)big)Bigrangle_F=0,.$$
Since $langle_,_rangle_E$ is positive-definite, we have $y=0_E$. This proves the claim.
P.S.: If the ground field is $mathbb{C}$, we instead assume that $langle_,_rangle_E$ and $langle_,_rangle_F$ are nondegenerate Hermitian forms. The transpose map has to be replaced by the Hermitian conjugate. For the claim to hold, we need to assume further that $langle_,_rangle_E$ is positive-definite.
If the ground field $mathbb{K}$ is a field of characteristic $0$, but there is no positive-definiteness assumption on $langle_,_rangle_E$ (which makes no sense when $mathbb{K}$ is not a subfied of $mathbb{R}$ anyhow), then the claim is not true. For example, let $mathbb{K}:=mathbb{R}$, $E:=mathbb{R}^2$, and $F:=mathbb{R}^2$. Then we set $e_1:=(1,0)$, $e_2:=(0,1)$. Let $langle_,_rangle$ be the nondegenerate symmetric bilinear form on both $E$ and $F$ given by $$langle a_1e_1+a_2e_2,b_1e_1+b_2e_2rangle:=a_1b_1-a_2b_2$$
for all $a_1,a_2,b_1,b_2inmathbb{R}$. Take $s:Fto E$ to be the linear map sending both $e_1$ and $e_2$ to $e_1+e_2$. That is, $s$ is given by the matrix
$$begin{bmatrix}1&1\1&1end{bmatrix}$$
with respect to the ordered basis $(e_1,e_2)$ of both $F$ and $E$. Then, $s^top:Eto F$ is given by the matrix
$$begin{bmatrix}1&-1\-1&1end{bmatrix}$$
with respect to the ordered basis $(e_1,e_2)$ of both $E$ and $F$. This shows that $$text{im}(s)=text{span}_mathbb{R}{e_1+e_2}=ker(s^top),.$$
The claim fails in positive characteristics. If the ground field $mathbb{K}$ is of a positive characteristric $p$, then take $E$ and $F$ to be both $mathbb{K}^p$. Fix a basis ${e_1,e_2,ldots,e_p}$ of both $E$ and $F$. Equip $E$ and $F$ with the nondegenerate symmetric bilinear form $langle_,_rangle$ given by
$$leftlanglesum_{i=1}^p,a_ie_i,sum_{i=1}^p,b_ie_irightrangle:=sum_{i=1}^p,a_ib_i$$
for all $a_1,a_2,ldots,a_p,b_1,b_2,ldots,b_pinmathbb{K}$. If the map $s:Fto E$ is the $mathbb{K}$-linear map sending $e_imapsto e_1+e_2+ldots+e_p$ for all $i=1,2,ldots,p$, then its transpose $s^top:Eto F$ also sends $e_imapsto e_1+e_2+ldots+e_p$ for all $i=1,2,ldots,p$. In particular, this shows that $$text{span}_mathbb{K}{e_1+e_2+ldots+e_p}=text{im}(s)capker(s^top),.$$
answered Nov 16 at 17:49
Batominovski
31.3k23187
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Let $s:Fto E$ be a linear transformation from a vector space $F$ to another vector space $E$ over the base field $mathbb{R}$. Both $E$ and $F$ are equipped with nondegenerate symmetric bilinear forms $langle_,_rangle_E$ and $langle_,_rangle_F$, respectively. Suppose further that $langle_,_rangle_E$ is positive-definite. Write $s^top:Eto F$ for the transpose of $s$. We claim that
$$text{im}(s)capker(s^top)={0_E},.$$
Recall that $s^top$ is the unique linear transformation from $E$ to $F$ such that $$biglangle s(x),ybigrangle_E=biglangle x,s^top(y)rangle_F$$
for all $xin F$ and $yin E$. Suppose that $yintext{im}(s)capker(s^top)$. Then, $yintext{im}(s)$, so $y=s(x)$ for some $xin F$. Sinc $yinker(s^top)$, we get $$s^topbig(s(x)big)=s^top(y)=0_F,.$$
Therefore,
$$Biglangle x,s^topbig(s(x)big)Bigrangle_F=biglangle x,0_Fbigrangle_F=0,.$$
By the definition of $s^top$, we have
$$langle y,yrangle_E=biglangle s(x),s(x)bigrangle_E=Biglangle x,s^topbig(s(x)big)Bigrangle_F=0,.$$
Since $langle_,_rangle_E$ is positive-definite, we have $y=0_E$. This proves the claim.
P.S.: If the ground field is $mathbb{C}$, we instead assume that $langle_,_rangle_E$ and $langle_,_rangle_F$ are nondegenerate Hermitian forms. The transpose map has to be replaced by the Hermitian conjugate. For the claim to hold, we need to assume further that $langle_,_rangle_E$ is positive-definite.
If the ground field $mathbb{K}$ is a field of characteristic $0$, but there is no positive-definiteness assumption on $langle_,_rangle_E$ (which makes no sense when $mathbb{K}$ is not a subfied of $mathbb{R}$ anyhow), then the claim is not true. For example, let $mathbb{K}:=mathbb{R}$, $E:=mathbb{R}^2$, and $F:=mathbb{R}^2$. Then we set $e_1:=(1,0)$, $e_2:=(0,1)$. Let $langle_,_rangle$ be the nondegenerate symmetric bilinear form on both $E$ and $F$ given by $$langle a_1e_1+a_2e_2,b_1e_1+b_2e_2rangle:=a_1b_1-a_2b_2$$
for all $a_1,a_2,b_1,b_2inmathbb{R}$. Take $s:Fto E$ to be the linear map sending both $e_1$ and $e_2$ to $e_1+e_2$. That is, $s$ is given by the matrix
$$begin{bmatrix}1&1\1&1end{bmatrix}$$
with respect to the ordered basis $(e_1,e_2)$ of both $F$ and $E$. Then, $s^top:Eto F$ is given by the matrix
$$begin{bmatrix}1&-1\-1&1end{bmatrix}$$
with respect to the ordered basis $(e_1,e_2)$ of both $E$ and $F$. This shows that $$text{im}(s)=text{span}_mathbb{R}{e_1+e_2}=ker(s^top),.$$
The claim fails in positive characteristics. If the ground field $mathbb{K}$ is of a positive characteristric $p$, then take $E$ and $F$ to be both $mathbb{K}^p$. Fix a basis ${e_1,e_2,ldots,e_p}$ of both $E$ and $F$. Equip $E$ and $F$ with the nondegenerate symmetric bilinear form $langle_,_rangle$ given by
$$leftlanglesum_{i=1}^p,a_ie_i,sum_{i=1}^p,b_ie_irightrangle:=sum_{i=1}^p,a_ib_i$$
for all $a_1,a_2,ldots,a_p,b_1,b_2,ldots,b_pinmathbb{K}$. If the map $s:Fto E$ is the $mathbb{K}$-linear map sending $e_imapsto e_1+e_2+ldots+e_p$ for all $i=1,2,ldots,p$, then its transpose $s^top:Eto F$ also sends $e_imapsto e_1+e_2+ldots+e_p$ for all $i=1,2,ldots,p$. In particular, this shows that $$text{span}_mathbb{K}{e_1+e_2+ldots+e_p}=text{im}(s)capker(s^top),.$$
answered Nov 16 at 17:49
Batominovski
31.3k23187
Let $s:Fto E$ be a linear transformation from a vector space $F$ to another vector space $E$ over the base field $mathbb{R}$. Both $E$ and $F$ are equipped with nondegenerate symmetric bilinear forms $langle_,_rangle_E$ and $langle_,_rangle_F$, respectively. Suppose further that $langle_,_rangle_E$ is positive-definite. Write $s^top:Eto F$ for the transpose of $s$. We claim that
$$text{im}(s)capker(s^top)={0_E},.$$
Recall that $s^top$ is the unique linear transformation from $E$ to $F$ such that $$biglangle s(x),ybigrangle_E=biglangle x,s^top(y)rangle_F$$
for all $xin F$ and $yin E$. Suppose that $yintext{im}(s)capker(s^top)$. Then, $yintext{im}(s)$, so $y=s(x)$ for some $xin F$. Sinc $yinker(s^top)$, we get $$s^topbig(s(x)big)=s^top(y)=0_F,.$$
Therefore,
$$Biglangle x,s^topbig(s(x)big)Bigrangle_F=biglangle x,0_Fbigrangle_F=0,.$$
By the definition of $s^top$, we have
$$langle y,yrangle_E=biglangle s(x),s(x)bigrangle_E=Biglangle x,s^topbig(s(x)big)Bigrangle_F=0,.$$
Since $langle_,_rangle_E$ is positive-definite, we have $y=0_E$. This proves the claim.
P.S.: If the ground field is $mathbb{C}$, we instead assume that $langle_,_rangle_E$ and $langle_,_rangle_F$ are nondegenerate Hermitian forms. The transpose map has to be replaced by the Hermitian conjugate. For the claim to hold, we need to assume further that $langle_,_rangle_E$ is positive-definite.
If the ground field $mathbb{K}$ is a field of characteristic $0$, but there is no positive-definiteness assumption on $langle_,_rangle_E$ (which makes no sense when $mathbb{K}$ is not a subfied of $mathbb{R}$ anyhow), then the claim is not true. For example, let $mathbb{K}:=mathbb{R}$, $E:=mathbb{R}^2$, and $F:=mathbb{R}^2$. Then we set $e_1:=(1,0)$, $e_2:=(0,1)$. Let $langle_,_rangle$ be the nondegenerate symmetric bilinear form on both $E$ and $F$ given by $$langle a_1e_1+a_2e_2,b_1e_1+b_2e_2rangle:=a_1b_1-a_2b_2$$
for all $a_1,a_2,b_1,b_2inmathbb{R}$. Take $s:Fto E$ to be the linear map sending both $e_1$ and $e_2$ to $e_1+e_2$. That is, $s$ is given by the matrix
$$begin{bmatrix}1&1\1&1end{bmatrix}$$
with respect to the ordered basis $(e_1,e_2)$ of both $F$ and $E$. Then, $s^top:Eto F$ is given by the matrix
$$begin{bmatrix}1&-1\-1&1end{bmatrix}$$
with respect to the ordered basis $(e_1,e_2)$ of both $E$ and $F$. This shows that $$text{im}(s)=text{span}_mathbb{R}{e_1+e_2}=ker(s^top),.$$
The claim fails in positive characteristics. If the ground field $mathbb{K}$ is of a positive characteristric $p$, then take $E$ and $F$ to be both $mathbb{K}^p$. Fix a basis ${e_1,e_2,ldots,e_p}$ of both $E$ and $F$. Equip $E$ and $F$ with the nondegenerate symmetric bilinear form $langle_,_rangle$ given by
$$leftlanglesum_{i=1}^p,a_ie_i,sum_{i=1}^p,b_ie_irightrangle:=sum_{i=1}^p,a_ib_i$$
for all $a_1,a_2,ldots,a_p,b_1,b_2,ldots,b_pinmathbb{K}$. If the map $s:Fto E$ is the $mathbb{K}$-linear map sending $e_imapsto e_1+e_2+ldots+e_p$ for all $i=1,2,ldots,p$, then its transpose $s^top:Eto F$ also sends $e_imapsto e_1+e_2+ldots+e_p$ for all $i=1,2,ldots,p$. In particular, this shows that $$text{span}_mathbb{K}{e_1+e_2+ldots+e_p}=text{im}(s)capker(s^top),.$$
answered Nov 16 at 17:49
Batominovski
31.3k23187
edited Nov 16 at 18:40
answered Nov 16 at 17:49
Batominovski
31.3k23187
answered Nov 16 at 17:49
Batominovski
31.3k23187
answered Nov 16 at 17:49
Batominovski
31.3k23187
31.3k23187
add a comment |
add a comment |
up vote
2
down vote
It is sufficient to show that
$$
ker s^tsubset(text{Im } s)^perp
$$
because we can then conclude by:
$$
ker s^t cap text{Im} s subset (text{Im} s)^perp cap text{Im} s ={0_E}
$$
The claim $ker s^tsubset(text{Im } s)^perp$ can be proven as follows:
begin{align*}
einker s^t &Rightarrow s^t(e)=0_F \
&Rightarrow forall fin F, langle f,s^t(e) rangle_F=0 \ &Rightarrow forall fin F, langle s(f),erangle_E=0 \
&Rightarrow forall fin F, s(f) perp e \
&Rightarrow e in (text{Im} s)^perp
end{align*}
answered Nov 16 at 18:20
Picaud Vincent
67815
1
I would like to note that the equality $ker (s^top)=big(text{im}(s)big)^perp$ is true for any nondegenerate symmetric bilinear form $langle_,_rangle_E$. However, the last equality $big(text{im}(s)big)^perpcaptext{im}(s)={0_E}$ must rely on some strong conditions on $langle_,_rangle_E$ such as positive-definiteness.
– Batominovski
Nov 16 at 18:27
thanks a lot I need some time to understand everything, and I'll be back
– Stu
Nov 16 at 18:28
@Batominovski you are right, and that is the reason why I simply wrote positive definite right at the beginning. Also a more general result would also hold only using duality bracket $<. ,.>_{Etimes E^*}$ and $<. ,.>_{Ftimes F^*}$ (without introducing scalar products $<. ,.>_E$ and $<. ,.>_F$
– Picaud Vincent
Nov 16 at 18:30
Shorter proof (sorry it is nearly a complete rewrite, but the idea is the same)
– Picaud Vincent
Nov 16 at 21:39
add a comment |
up vote
2
down vote
It is sufficient to show that
$$
ker s^tsubset(text{Im } s)^perp
$$
because we can then conclude by:
$$
ker s^t cap text{Im} s subset (text{Im} s)^perp cap text{Im} s ={0_E}
$$
The claim $ker s^tsubset(text{Im } s)^perp$ can be proven as follows:
begin{align*}
einker s^t &Rightarrow s^t(e)=0_F \
&Rightarrow forall fin F, langle f,s^t(e) rangle_F=0 \ &Rightarrow forall fin F, langle s(f),erangle_E=0 \
&Rightarrow forall fin F, s(f) perp e \
&Rightarrow e in (text{Im} s)^perp
end{align*}
answered Nov 16 at 18:20
Picaud Vincent
67815
1
I would like to note that the equality $ker (s^top)=big(text{im}(s)big)^perp$ is true for any nondegenerate symmetric bilinear form $langle_,_rangle_E$. However, the last equality $big(text{im}(s)big)^perpcaptext{im}(s)={0_E}$ must rely on some strong conditions on $langle_,_rangle_E$ such as positive-definiteness.
– Batominovski
Nov 16 at 18:27
thanks a lot I need some time to understand everything, and I'll be back
– Stu
Nov 16 at 18:28
@Batominovski you are right, and that is the reason why I simply wrote positive definite right at the beginning. Also a more general result would also hold only using duality bracket $<. ,.>_{Etimes E^*}$ and $<. ,.>_{Ftimes F^*}$ (without introducing scalar products $<. ,.>_E$ and $<. ,.>_F$
– Picaud Vincent
Nov 16 at 18:30
Shorter proof (sorry it is nearly a complete rewrite, but the idea is the same)
– Picaud Vincent
Nov 16 at 21:39
add a comment |
up vote
2
down vote
up vote
2
down vote
It is sufficient to show that
$$
ker s^tsubset(text{Im } s)^perp
$$
because we can then conclude by:
$$
ker s^t cap text{Im} s subset (text{Im} s)^perp cap text{Im} s ={0_E}
$$
The claim $ker s^tsubset(text{Im } s)^perp$ can be proven as follows:
begin{align*}
einker s^t &Rightarrow s^t(e)=0_F \
&Rightarrow forall fin F, langle f,s^t(e) rangle_F=0 \ &Rightarrow forall fin F, langle s(f),erangle_E=0 \
&Rightarrow forall fin F, s(f) perp e \
&Rightarrow e in (text{Im} s)^perp
end{align*}
answered Nov 16 at 18:20
Picaud Vincent
67815
It is sufficient to show that
$$
ker s^tsubset(text{Im } s)^perp
$$
because we can then conclude by:
$$
ker s^t cap text{Im} s subset (text{Im} s)^perp cap text{Im} s ={0_E}
$$
The claim $ker s^tsubset(text{Im } s)^perp$ can be proven as follows:
begin{align*}
einker s^t &Rightarrow s^t(e)=0_F \
&Rightarrow forall fin F, langle f,s^t(e) rangle_F=0 \ &Rightarrow forall fin F, langle s(f),erangle_E=0 \
&Rightarrow forall fin F, s(f) perp e \
&Rightarrow e in (text{Im} s)^perp
end{align*}
answered Nov 16 at 18:20
Picaud Vincent
67815
edited Nov 16 at 22:13
answered Nov 16 at 18:20
Picaud Vincent
67815
answered Nov 16 at 18:20
Picaud Vincent
67815
answered Nov 16 at 18:20
Picaud Vincent
67815
67815
1
I would like to note that the equality $ker (s^top)=big(text{im}(s)big)^perp$ is true for any nondegenerate symmetric bilinear form $langle_,_rangle_E$. However, the last equality $big(text{im}(s)big)^perpcaptext{im}(s)={0_E}$ must rely on some strong conditions on $langle_,_rangle_E$ such as positive-definiteness.
– Batominovski
Nov 16 at 18:27
thanks a lot I need some time to understand everything, and I'll be back
– Stu
Nov 16 at 18:28
@Batominovski you are right, and that is the reason why I simply wrote positive definite right at the beginning. Also a more general result would also hold only using duality bracket $<. ,.>_{Etimes E^*}$ and $<. ,.>_{Ftimes F^*}$ (without introducing scalar products $<. ,.>_E$ and $<. ,.>_F$
– Picaud Vincent
Nov 16 at 18:30
Shorter proof (sorry it is nearly a complete rewrite, but the idea is the same)
– Picaud Vincent
Nov 16 at 21:39
add a comment |
1
I would like to note that the equality $ker (s^top)=big(text{im}(s)big)^perp$ is true for any nondegenerate symmetric bilinear form $langle_,_rangle_E$. However, the last equality $big(text{im}(s)big)^perpcaptext{im}(s)={0_E}$ must rely on some strong conditions on $langle_,_rangle_E$ such as positive-definiteness.
– Batominovski
Nov 16 at 18:27
thanks a lot I need some time to understand everything, and I'll be back
– Stu
Nov 16 at 18:28
@Batominovski you are right, and that is the reason why I simply wrote positive definite right at the beginning. Also a more general result would also hold only using duality bracket $<. ,.>_{Etimes E^*}$ and $<. ,.>_{Ftimes F^*}$ (without introducing scalar products $<. ,.>_E$ and $<. ,.>_F$
– Picaud Vincent
Nov 16 at 18:30
Shorter proof (sorry it is nearly a complete rewrite, but the idea is the same)
– Picaud Vincent
Nov 16 at 21:39
1
1
I would like to note that the equality $ker (s^top)=big(text{im}(s)big)^perp$ is true for any nondegenerate symmetric bilinear form $langle_,_rangle_E$. However, the last equality $big(text{im}(s)big)^perpcaptext{im}(s)={0_E}$ must rely on some strong conditions on $langle_,_rangle_E$ such as positive-definiteness.
– Batominovski
Nov 16 at 18:27
I would like to note that the equality $ker (s^top)=big(text{im}(s)big)^perp$ is true for any nondegenerate symmetric bilinear form $langle_,_rangle_E$. However, the last equality $big(text{im}(s)big)^perpcaptext{im}(s)={0_E}$ must rely on some strong conditions on $langle_,_rangle_E$ such as positive-definiteness.
– Batominovski
Nov 16 at 18:27
thanks a lot I need some time to understand everything, and I'll be back
– Stu
Nov 16 at 18:28
thanks a lot I need some time to understand everything, and I'll be back
– Stu
Nov 16 at 18:28
@Batominovski you are right, and that is the reason why I simply wrote positive definite right at the beginning. Also a more general result would also hold only using duality bracket $<. ,.>_{Etimes E^*}$ and $<. ,.>_{Ftimes F^*}$ (without introducing scalar products $<. ,.>_E$ and $<. ,.>_F$
– Picaud Vincent
Nov 16 at 18:30
@Batominovski you are right, and that is the reason why I simply wrote positive definite right at the beginning. Also a more general result would also hold only using duality bracket $<. ,.>_{Etimes E^*}$ and $<. ,.>_{Ftimes F^*}$ (without introducing scalar products $<. ,.>_E$ and $<. ,.>_F$
– Picaud Vincent
Nov 16 at 18:30
Shorter proof (sorry it is nearly a complete rewrite, but the idea is the same)
– Picaud Vincent
Nov 16 at 21:39
Shorter proof (sorry it is nearly a complete rewrite, but the idea is the same)
– Picaud Vincent
Nov 16 at 21:39
add a comment |
up vote
1
down vote
In order that the transpose is a map $Fto E$ you need some assumptions: both vector spaces need to be equipped with a nondegenerate bilinear form, so that we can define isomorphisms $Eto E^*$ and $Fto F^*$ (the dual spaces), assuming finite dimensionality.
In particular, if the base field is $mathbb{R}$, the two space might be inner product spaces.
If $langle{cdot},{cdot}rangle_E$ is the form on $E$, then for each $vin E$, the map
$$
e_vcolon Eto K,qquad e_v(x)=langle v,xrangle_E
$$
is an element of $E^*$. If $vne0$, then nondegeneracy implies there exists $xin E$ with $langle v,xrangle_Ene0$, so $ecolon Eto E^*$, $vmapsto e_v$ is an isomorphism (it is clear from bilinearity that $e_v$ is linear, for every $vin E$, and $e$ is linear as well).
Then the transpose ${}^{t!}s$ is the dual map composed with these isomorphisms:
$$
{}^{t!}s=e^{-1}circ s^*circ e
$$
(I denote by $e$ both maps $Eto E^*$ and $Fto F^*$, no confusion should arise).
If $w=s(v)$ and ${}^{t!}s(w)=0$, then also $s^*circ e_w=0$, which means $e_wcirc s=0$, that is,
$$
e_w(s(x))=0quadtext{for all $xin V$}
$$
hence
$$
langle w,s(x)rangle_F=0
$$
In particular, $langle s(v),s(v)rangle_F=0$, so by nondegeneracy, $s(v)=0$ and $w=0$.
answered Nov 16 at 23:16
egreg
173k1383198
add a comment |
up vote
1
down vote
In order that the transpose is a map $Fto E$ you need some assumptions: both vector spaces need to be equipped with a nondegenerate bilinear form, so that we can define isomorphisms $Eto E^*$ and $Fto F^*$ (the dual spaces), assuming finite dimensionality.
In particular, if the base field is $mathbb{R}$, the two space might be inner product spaces.
If $langle{cdot},{cdot}rangle_E$ is the form on $E$, then for each $vin E$, the map
$$
e_vcolon Eto K,qquad e_v(x)=langle v,xrangle_E
$$
is an element of $E^*$. If $vne0$, then nondegeneracy implies there exists $xin E$ with $langle v,xrangle_Ene0$, so $ecolon Eto E^*$, $vmapsto e_v$ is an isomorphism (it is clear from bilinearity that $e_v$ is linear, for every $vin E$, and $e$ is linear as well).
Then the transpose ${}^{t!}s$ is the dual map composed with these isomorphisms:
$$
{}^{t!}s=e^{-1}circ s^*circ e
$$
(I denote by $e$ both maps $Eto E^*$ and $Fto F^*$, no confusion should arise).
If $w=s(v)$ and ${}^{t!}s(w)=0$, then also $s^*circ e_w=0$, which means $e_wcirc s=0$, that is,
$$
e_w(s(x))=0quadtext{for all $xin V$}
$$
hence
$$
langle w,s(x)rangle_F=0
$$
In particular, $langle s(v),s(v)rangle_F=0$, so by nondegeneracy, $s(v)=0$ and $w=0$.
answered Nov 16 at 23:16
egreg
173k1383198
add a comment |
up vote
1
down vote
up vote
1
down vote
In order that the transpose is a map $Fto E$ you need some assumptions: both vector spaces need to be equipped with a nondegenerate bilinear form, so that we can define isomorphisms $Eto E^*$ and $Fto F^*$ (the dual spaces), assuming finite dimensionality.
In particular, if the base field is $mathbb{R}$, the two space might be inner product spaces.
If $langle{cdot},{cdot}rangle_E$ is the form on $E$, then for each $vin E$, the map
$$
e_vcolon Eto K,qquad e_v(x)=langle v,xrangle_E
$$
is an element of $E^*$. If $vne0$, then nondegeneracy implies there exists $xin E$ with $langle v,xrangle_Ene0$, so $ecolon Eto E^*$, $vmapsto e_v$ is an isomorphism (it is clear from bilinearity that $e_v$ is linear, for every $vin E$, and $e$ is linear as well).
Then the transpose ${}^{t!}s$ is the dual map composed with these isomorphisms:
$$
{}^{t!}s=e^{-1}circ s^*circ e
$$
(I denote by $e$ both maps $Eto E^*$ and $Fto F^*$, no confusion should arise).
If $w=s(v)$ and ${}^{t!}s(w)=0$, then also $s^*circ e_w=0$, which means $e_wcirc s=0$, that is,
$$
e_w(s(x))=0quadtext{for all $xin V$}
$$
hence
$$
langle w,s(x)rangle_F=0
$$
In particular, $langle s(v),s(v)rangle_F=0$, so by nondegeneracy, $s(v)=0$ and $w=0$.
answered Nov 16 at 23:16
egreg
173k1383198
In order that the transpose is a map $Fto E$ you need some assumptions: both vector spaces need to be equipped with a nondegenerate bilinear form, so that we can define isomorphisms $Eto E^*$ and $Fto F^*$ (the dual spaces), assuming finite dimensionality.
In particular, if the base field is $mathbb{R}$, the two space might be inner product spaces.
If $langle{cdot},{cdot}rangle_E$ is the form on $E$, then for each $vin E$, the map
$$
e_vcolon Eto K,qquad e_v(x)=langle v,xrangle_E
$$
is an element of $E^*$. If $vne0$, then nondegeneracy implies there exists $xin E$ with $langle v,xrangle_Ene0$, so $ecolon Eto E^*$, $vmapsto e_v$ is an isomorphism (it is clear from bilinearity that $e_v$ is linear, for every $vin E$, and $e$ is linear as well).
Then the transpose ${}^{t!}s$ is the dual map composed with these isomorphisms:
$$
{}^{t!}s=e^{-1}circ s^*circ e
$$
(I denote by $e$ both maps $Eto E^*$ and $Fto F^*$, no confusion should arise).
If $w=s(v)$ and ${}^{t!}s(w)=0$, then also $s^*circ e_w=0$, which means $e_wcirc s=0$, that is,
$$
e_w(s(x))=0quadtext{for all $xin V$}
$$
hence
$$
langle w,s(x)rangle_F=0
$$
In particular, $langle s(v),s(v)rangle_F=0$, so by nondegeneracy, $s(v)=0$ and $w=0$.
answered Nov 16 at 23:16
egreg
173k1383198
answered Nov 16 at 23:16
egreg
173k1383198
answered Nov 16 at 23:16
egreg
173k1383198
answered Nov 16 at 23:16
egreg
173k1383198
173k1383198
add a comment |
add a comment |
up vote
0
down vote
If you represent a linear map by a matrix then the image of the map is the sub-space spanned of the columns of the matrix. And the kernel of the map is the sub-space that is perpendicular to the rows of the matrix (because the inner product of each row with any vector in the kernel must be zero).
So the image of $s$ is the sub-space that is spanned by the columns of the matrix representing $s$ (relative to specific bases in $E$ and $F$).
And the kernel of $^ts$ is the sub-space that is perpendicular to the sub-space spanned by the rows of the matrix representing $^ts$.
But the columns of the first matrix are the rows of the second matrix. The result follows.
answered Nov 16 at 16:48
gandalf61
7,177523
add a comment |
up vote
0
down vote
If you represent a linear map by a matrix then the image of the map is the sub-space spanned of the columns of the matrix. And the kernel of the map is the sub-space that is perpendicular to the rows of the matrix (because the inner product of each row with any vector in the kernel must be zero).
So the image of $s$ is the sub-space that is spanned by the columns of the matrix representing $s$ (relative to specific bases in $E$ and $F$).
And the kernel of $^ts$ is the sub-space that is perpendicular to the sub-space spanned by the rows of the matrix representing $^ts$.
But the columns of the first matrix are the rows of the second matrix. The result follows.
answered Nov 16 at 16:48
gandalf61
7,177523
add a comment |
up vote
0
down vote
up vote
0
down vote
If you represent a linear map by a matrix then the image of the map is the sub-space spanned of the columns of the matrix. And the kernel of the map is the sub-space that is perpendicular to the rows of the matrix (because the inner product of each row with any vector in the kernel must be zero).
So the image of $s$ is the sub-space that is spanned by the columns of the matrix representing $s$ (relative to specific bases in $E$ and $F$).
And the kernel of $^ts$ is the sub-space that is perpendicular to the sub-space spanned by the rows of the matrix representing $^ts$.
But the columns of the first matrix are the rows of the second matrix. The result follows.
answered Nov 16 at 16:48
gandalf61
7,177523
If you represent a linear map by a matrix then the image of the map is the sub-space spanned of the columns of the matrix. And the kernel of the map is the sub-space that is perpendicular to the rows of the matrix (because the inner product of each row with any vector in the kernel must be zero).
So the image of $s$ is the sub-space that is spanned by the columns of the matrix representing $s$ (relative to specific bases in $E$ and $F$).
And the kernel of $^ts$ is the sub-space that is perpendicular to the sub-space spanned by the rows of the matrix representing $^ts$.
But the columns of the first matrix are the rows of the second matrix. The result follows.
answered Nov 16 at 16:48
gandalf61
7,177523
answered Nov 16 at 16:48
gandalf61
7,177523
answered Nov 16 at 16:48
gandalf61
7,177523
answered Nov 16 at 16:48
gandalf61
7,177523
7,177523
add a comment |
add a comment |
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I think the problem is not clear. What is the base field here? Presumably, you have nondegenerate bilinear forms on $E$ and $F$. However, not all bilinear forms work. If the base field is of a positive characteristic, then this is false. Even when the characteristic of the base field is $0$, you can come up with a nondegenerate symmetric bilinear form that contradicts the claim.
– Batominovski
Nov 16 at 17:03
1
So, I would like the OP to confirm whether: (1) the base field is $mathbb{R}$, (2) $E$ and $F$ are finite-dimensional vector spaces, and (3) the vector spaces are equipped with positive-definite symmetric bilinear forms.
– Batominovski
Nov 16 at 17:05
I added more informations, but I didn't want to use the inner product at first...but it seems I must do.
– Stu
Nov 16 at 17:23