The dual basis spans the vector space of linear transformations.
$begingroup$
Proof attempt:
Let $T in L(V,Bbb{F})$ where $L(V,Bbb{F})$ is the space of linear functionals of $V$ and where $Bbb{F}$ is some field. Let $mathscr{F}={Phi_1,cdots Phi_n}$ be the dual basis of, $B_v={v_1,cdots, v_n}$, the basis of $V$. Note $T(v)=T(sum_{i=1}^{n}c_i v_i)=sum_{i=1}^{n}c_i T(v_i)$ such that $forall i, T(v_i)inBbb{F}$. Since $T(v_i)inBbb{F}$ implies $exists lambda_iinBbb{F}, T(v_i)=lambda_i=lambda_icdot 1=lambda_icdotPhi(v_i)$. Then $T(v)=sum_{i=1}^{n}(c_ilambda_i)Phi(v_i)$ and $T=sum_{i=1}^n(c_ilambda_i)Phi_i$, as desired.
linear-algebra proof-verification linear-transformations dual-spaces
asked Dec 24 '18 at 3:56
TheLast CipherTheLast Cipher
757715
$endgroup$
add a comment |
$begingroup$
Proof attempt:
Let $T in L(V,Bbb{F})$ where $L(V,Bbb{F})$ is the space of linear functionals of $V$ and where $Bbb{F}$ is some field. Let $mathscr{F}={Phi_1,cdots Phi_n}$ be the dual basis of, $B_v={v_1,cdots, v_n}$, the basis of $V$. Note $T(v)=T(sum_{i=1}^{n}c_i v_i)=sum_{i=1}^{n}c_i T(v_i)$ such that $forall i, T(v_i)inBbb{F}$. Since $T(v_i)inBbb{F}$ implies $exists lambda_iinBbb{F}, T(v_i)=lambda_i=lambda_icdot 1=lambda_icdotPhi(v_i)$. Then $T(v)=sum_{i=1}^{n}(c_ilambda_i)Phi(v_i)$ and $T=sum_{i=1}^n(c_ilambda_i)Phi_i$, as desired.
linear-algebra proof-verification linear-transformations dual-spaces
asked Dec 24 '18 at 3:56
TheLast CipherTheLast Cipher
757715
$endgroup$
add a comment |
$begingroup$
Proof attempt:
Let $T in L(V,Bbb{F})$ where $L(V,Bbb{F})$ is the space of linear functionals of $V$ and where $Bbb{F}$ is some field. Let $mathscr{F}={Phi_1,cdots Phi_n}$ be the dual basis of, $B_v={v_1,cdots, v_n}$, the basis of $V$. Note $T(v)=T(sum_{i=1}^{n}c_i v_i)=sum_{i=1}^{n}c_i T(v_i)$ such that $forall i, T(v_i)inBbb{F}$. Since $T(v_i)inBbb{F}$ implies $exists lambda_iinBbb{F}, T(v_i)=lambda_i=lambda_icdot 1=lambda_icdotPhi(v_i)$. Then $T(v)=sum_{i=1}^{n}(c_ilambda_i)Phi(v_i)$ and $T=sum_{i=1}^n(c_ilambda_i)Phi_i$, as desired.
linear-algebra proof-verification linear-transformations dual-spaces
asked Dec 24 '18 at 3:56
TheLast CipherTheLast Cipher
757715
$endgroup$
Proof attempt:
Let $T in L(V,Bbb{F})$ where $L(V,Bbb{F})$ is the space of linear functionals of $V$ and where $Bbb{F}$ is some field. Let $mathscr{F}={Phi_1,cdots Phi_n}$ be the dual basis of, $B_v={v_1,cdots, v_n}$, the basis of $V$. Note $T(v)=T(sum_{i=1}^{n}c_i v_i)=sum_{i=1}^{n}c_i T(v_i)$ such that $forall i, T(v_i)inBbb{F}$. Since $T(v_i)inBbb{F}$ implies $exists lambda_iinBbb{F}, T(v_i)=lambda_i=lambda_icdot 1=lambda_icdotPhi(v_i)$. Then $T(v)=sum_{i=1}^{n}(c_ilambda_i)Phi(v_i)$ and $T=sum_{i=1}^n(c_ilambda_i)Phi_i$, as desired.
linear-algebra proof-verification linear-transformations dual-spaces
linear-algebra proof-verification linear-transformations dual-spaces
asked Dec 24 '18 at 3:56
TheLast CipherTheLast Cipher
757715
asked Dec 24 '18 at 3:56
TheLast CipherTheLast Cipher
757715
asked Dec 24 '18 at 3:56
TheLast CipherTheLast Cipher
757715
asked Dec 24 '18 at 3:56
TheLast CipherTheLast Cipher
757715
asked Dec 24 '18 at 3:56
TheLast CipherTheLast Cipher
757715
757715
add a comment |
add a comment |
1 Answer
1
active
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votes
$begingroup$
Since you are asking for criticism, a couple of (nit picky) remarks:
- The name 'dual basis' suggest that this set already spans the space of linear functionals. Perhaps it would be better to give a definition of ${Phi_i}_i$ and just assert that it spans $L(V,mathbb{F})$.
- In the last two sentences, it should be $Phi_i(v_i)$.
Other than that, it seems fine.
Another approach, although ad-hoc, is to just claim that $T = sum_{i = 1}^n T(v_i)Phi_i$, since given $v = sum_ic_iv_i$ we have that
$$
begin{align}
T(v) &= sum_ic_iT(v_i) = sum_ic_iT(v_i) cdot 1 = sum_ic_iT(v_i) Phi_i(v_i)\
&= sum_iT(v_i) cdot Phi(c_iv_i) = sum_{i} T(v_i)Phi_i(v).
end{align}
$$
with the latter equality given by the fact that
$$
Phi_i(v) = sum_jc_jPhi_i(v_j) = c_iPhi_i(v_i) = Phi_i(c_iv_i).
$$
answered Dec 24 '18 at 4:21
Guido A.Guido A.
8,3351730
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Since you are asking for criticism, a couple of (nit picky) remarks:
- The name 'dual basis' suggest that this set already spans the space of linear functionals. Perhaps it would be better to give a definition of ${Phi_i}_i$ and just assert that it spans $L(V,mathbb{F})$.
- In the last two sentences, it should be $Phi_i(v_i)$.
Other than that, it seems fine.
Another approach, although ad-hoc, is to just claim that $T = sum_{i = 1}^n T(v_i)Phi_i$, since given $v = sum_ic_iv_i$ we have that
$$
begin{align}
T(v) &= sum_ic_iT(v_i) = sum_ic_iT(v_i) cdot 1 = sum_ic_iT(v_i) Phi_i(v_i)\
&= sum_iT(v_i) cdot Phi(c_iv_i) = sum_{i} T(v_i)Phi_i(v).
end{align}
$$
with the latter equality given by the fact that
$$
Phi_i(v) = sum_jc_jPhi_i(v_j) = c_iPhi_i(v_i) = Phi_i(c_iv_i).
$$
answered Dec 24 '18 at 4:21
Guido A.Guido A.
8,3351730
$endgroup$
add a comment |
$begingroup$
Since you are asking for criticism, a couple of (nit picky) remarks:
- The name 'dual basis' suggest that this set already spans the space of linear functionals. Perhaps it would be better to give a definition of ${Phi_i}_i$ and just assert that it spans $L(V,mathbb{F})$.
- In the last two sentences, it should be $Phi_i(v_i)$.
Other than that, it seems fine.
Another approach, although ad-hoc, is to just claim that $T = sum_{i = 1}^n T(v_i)Phi_i$, since given $v = sum_ic_iv_i$ we have that
$$
begin{align}
T(v) &= sum_ic_iT(v_i) = sum_ic_iT(v_i) cdot 1 = sum_ic_iT(v_i) Phi_i(v_i)\
&= sum_iT(v_i) cdot Phi(c_iv_i) = sum_{i} T(v_i)Phi_i(v).
end{align}
$$
with the latter equality given by the fact that
$$
Phi_i(v) = sum_jc_jPhi_i(v_j) = c_iPhi_i(v_i) = Phi_i(c_iv_i).
$$
answered Dec 24 '18 at 4:21
Guido A.Guido A.
8,3351730
$endgroup$
add a comment |
$begingroup$
Since you are asking for criticism, a couple of (nit picky) remarks:
- The name 'dual basis' suggest that this set already spans the space of linear functionals. Perhaps it would be better to give a definition of ${Phi_i}_i$ and just assert that it spans $L(V,mathbb{F})$.
- In the last two sentences, it should be $Phi_i(v_i)$.
Other than that, it seems fine.
Another approach, although ad-hoc, is to just claim that $T = sum_{i = 1}^n T(v_i)Phi_i$, since given $v = sum_ic_iv_i$ we have that
$$
begin{align}
T(v) &= sum_ic_iT(v_i) = sum_ic_iT(v_i) cdot 1 = sum_ic_iT(v_i) Phi_i(v_i)\
&= sum_iT(v_i) cdot Phi(c_iv_i) = sum_{i} T(v_i)Phi_i(v).
end{align}
$$
with the latter equality given by the fact that
$$
Phi_i(v) = sum_jc_jPhi_i(v_j) = c_iPhi_i(v_i) = Phi_i(c_iv_i).
$$
answered Dec 24 '18 at 4:21
Guido A.Guido A.
8,3351730
$endgroup$
Since you are asking for criticism, a couple of (nit picky) remarks:
- The name 'dual basis' suggest that this set already spans the space of linear functionals. Perhaps it would be better to give a definition of ${Phi_i}_i$ and just assert that it spans $L(V,mathbb{F})$.
- In the last two sentences, it should be $Phi_i(v_i)$.
Other than that, it seems fine.
Another approach, although ad-hoc, is to just claim that $T = sum_{i = 1}^n T(v_i)Phi_i$, since given $v = sum_ic_iv_i$ we have that
$$
begin{align}
T(v) &= sum_ic_iT(v_i) = sum_ic_iT(v_i) cdot 1 = sum_ic_iT(v_i) Phi_i(v_i)\
&= sum_iT(v_i) cdot Phi(c_iv_i) = sum_{i} T(v_i)Phi_i(v).
end{align}
$$
with the latter equality given by the fact that
$$
Phi_i(v) = sum_jc_jPhi_i(v_j) = c_iPhi_i(v_i) = Phi_i(c_iv_i).
$$
answered Dec 24 '18 at 4:21
Guido A.Guido A.
8,3351730
answered Dec 24 '18 at 4:21
Guido A.Guido A.
8,3351730
answered Dec 24 '18 at 4:21
Guido A.Guido A.
8,3351730
answered Dec 24 '18 at 4:21
Guido A.Guido A.
8,3351730
8,3351730
add a comment |
add a comment |
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