Dual map and solving linear equation












1












$begingroup$



Let $f:Vto W$ be a linear transformation from a vector space $V$ to a vector space $W$. Suppose that $bin W$ satisfies $phi(b)=0$ for all $phiinker(f^*)$. Show that there exists $xin V$ such that $f(x)=b$.




I have to confess that this is part of a homework problem that was handed to me before christmas. I spent the last two weeks think on this problem and I am desperate.. I learned that if $f$ is injective, $f^*$ is surjective and I expect this to be the dual statement to this but when I try to mimic the proof, I fail. Any help would be nice..






linear-algebra vector-spaces linear-transformations dual-spaces dual-maps







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$endgroup$








  • 3




    $begingroup$
    HINT: Look up "Fredholm alternative", the finite dimensional version.
    $endgroup$
    – Giuseppe Negro
    Jan 8 at 13:45










  • $begingroup$
    You saved my day! Thank you SO much!!
    $endgroup$
    – user625722
    Jan 8 at 13:56
















1












$begingroup$



Let $f:Vto W$ be a linear transformation from a vector space $V$ to a vector space $W$. Suppose that $bin W$ satisfies $phi(b)=0$ for all $phiinker(f^*)$. Show that there exists $xin V$ such that $f(x)=b$.




I have to confess that this is part of a homework problem that was handed to me before christmas. I spent the last two weeks think on this problem and I am desperate.. I learned that if $f$ is injective, $f^*$ is surjective and I expect this to be the dual statement to this but when I try to mimic the proof, I fail. Any help would be nice..






linear-algebra vector-spaces linear-transformations dual-spaces dual-maps







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    HINT: Look up "Fredholm alternative", the finite dimensional version.
    $endgroup$
    – Giuseppe Negro
    Jan 8 at 13:45










  • $begingroup$
    You saved my day! Thank you SO much!!
    $endgroup$
    – user625722
    Jan 8 at 13:56














1












1








1





$begingroup$



Let $f:Vto W$ be a linear transformation from a vector space $V$ to a vector space $W$. Suppose that $bin W$ satisfies $phi(b)=0$ for all $phiinker(f^*)$. Show that there exists $xin V$ such that $f(x)=b$.




I have to confess that this is part of a homework problem that was handed to me before christmas. I spent the last two weeks think on this problem and I am desperate.. I learned that if $f$ is injective, $f^*$ is surjective and I expect this to be the dual statement to this but when I try to mimic the proof, I fail. Any help would be nice..






linear-algebra vector-spaces linear-transformations dual-spaces dual-maps







share|cite|improve this question











$endgroup$





Let $f:Vto W$ be a linear transformation from a vector space $V$ to a vector space $W$. Suppose that $bin W$ satisfies $phi(b)=0$ for all $phiinker(f^*)$. Show that there exists $xin V$ such that $f(x)=b$.




I have to confess that this is part of a homework problem that was handed to me before christmas. I spent the last two weeks think on this problem and I am desperate.. I learned that if $f$ is injective, $f^*$ is surjective and I expect this to be the dual statement to this but when I try to mimic the proof, I fail. Any help would be nice..






linear-algebra vector-spaces linear-transformations dual-spaces dual-maps




linear-algebra vector-spaces linear-transformations dual-spaces dual-maps






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 8 at 14:53







user593746

















asked Jan 8 at 13:42









user625722user625722

185




185








  • 3




    $begingroup$
    HINT: Look up "Fredholm alternative", the finite dimensional version.
    $endgroup$
    – Giuseppe Negro
    Jan 8 at 13:45










  • $begingroup$
    You saved my day! Thank you SO much!!
    $endgroup$
    – user625722
    Jan 8 at 13:56














  • 3




    $begingroup$
    HINT: Look up "Fredholm alternative", the finite dimensional version.
    $endgroup$
    – Giuseppe Negro
    Jan 8 at 13:45










  • $begingroup$
    You saved my day! Thank you SO much!!
    $endgroup$
    – user625722
    Jan 8 at 13:56








3




3




$begingroup$
HINT: Look up "Fredholm alternative", the finite dimensional version.
$endgroup$
– Giuseppe Negro
Jan 8 at 13:45




$begingroup$
HINT: Look up "Fredholm alternative", the finite dimensional version.
$endgroup$
– Giuseppe Negro
Jan 8 at 13:45












$begingroup$
You saved my day! Thank you SO much!!
$endgroup$
– user625722
Jan 8 at 13:56




$begingroup$
You saved my day! Thank you SO much!!
$endgroup$
– user625722
Jan 8 at 13:56










1 Answer
1






active

oldest

votes


















2












$begingroup$

This answer does not assume finite dimensionality of $V$ or $W$ (so it is a generalized version of the Fredholm alternative). That is, I claim that for any linear map $f:Vto W$,
$$bigcap_{phiinker (f^*)}kerphi=operatorname{im}f.$$



We have the following exact sequence of vector spaces:
$${0}to ker f hookrightarrow V overset{f}{longrightarrow} Wtwoheadrightarrow operatorname{coim}fto {0},$$
where $operatorname{coim}f$ is the coimage $W/operatorname{im} f$. Dualizing this exact sequence, we obtain the exact sequence
$${0}to (operatorname{coim}f)^*hookrightarrow W^*overset{f^*}{longrightarrow} V^*twoheadrightarrow (ker f)^*to{0},$$
where $(operatorname{coim}f)^*$ is considered a subspace of $W^*$ via the identification that sends ${psi} in (operatorname{coim}f)^*$ to $hatpsiin W^*$ such that
$$hatpsi(w)={psi}(w+operatorname{im}f) forall win W.$$
Similarly, $(ker f)^*$ is a quotient of $V^*$ via the restriction map sending $sigmain V^*$ to $sigma|_{ker f}in (ker f)^*$ (precisely, $(ker f)^*cong V^*/(operatorname{cokr} f)^*$, where $operatorname{cokr}f$ is the cokernel $V/ker f$). By the exactness, we get
$$ker (f^*)=(operatorname{coim}f)^*.$$
That is, if $wnotin operatorname{im}f$, then $w+operatorname{im}f$ is non-zero in $operatorname{coim}W$. Therefore, there exists ${phi}_win (operatorname{coim} W)^*$ that does not vanish at $w+operatorname{im} f$. That is, $hatphi_w(w)ne 0$. Therefore, $wnotin bigcap_{phiinker(f^*)}kerphi$. This shows that $bigcap_{phiinker(f^*)}kerphisubseteqoperatorname{im}f$. It is also obvious that $bigcap_{phiinker(f^*)}kerphisupseteqoperatorname{im}f$. The claim is now evident.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    This is a little beyond the scope of my course but a very nice answer. Thank you!
    $endgroup$
    – user625722
    Jan 8 at 14:58











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1 Answer
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1 Answer
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active

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2












$begingroup$

This answer does not assume finite dimensionality of $V$ or $W$ (so it is a generalized version of the Fredholm alternative). That is, I claim that for any linear map $f:Vto W$,
$$bigcap_{phiinker (f^*)}kerphi=operatorname{im}f.$$



We have the following exact sequence of vector spaces:
$${0}to ker f hookrightarrow V overset{f}{longrightarrow} Wtwoheadrightarrow operatorname{coim}fto {0},$$
where $operatorname{coim}f$ is the coimage $W/operatorname{im} f$. Dualizing this exact sequence, we obtain the exact sequence
$${0}to (operatorname{coim}f)^*hookrightarrow W^*overset{f^*}{longrightarrow} V^*twoheadrightarrow (ker f)^*to{0},$$
where $(operatorname{coim}f)^*$ is considered a subspace of $W^*$ via the identification that sends ${psi} in (operatorname{coim}f)^*$ to $hatpsiin W^*$ such that
$$hatpsi(w)={psi}(w+operatorname{im}f) forall win W.$$
Similarly, $(ker f)^*$ is a quotient of $V^*$ via the restriction map sending $sigmain V^*$ to $sigma|_{ker f}in (ker f)^*$ (precisely, $(ker f)^*cong V^*/(operatorname{cokr} f)^*$, where $operatorname{cokr}f$ is the cokernel $V/ker f$). By the exactness, we get
$$ker (f^*)=(operatorname{coim}f)^*.$$
That is, if $wnotin operatorname{im}f$, then $w+operatorname{im}f$ is non-zero in $operatorname{coim}W$. Therefore, there exists ${phi}_win (operatorname{coim} W)^*$ that does not vanish at $w+operatorname{im} f$. That is, $hatphi_w(w)ne 0$. Therefore, $wnotin bigcap_{phiinker(f^*)}kerphi$. This shows that $bigcap_{phiinker(f^*)}kerphisubseteqoperatorname{im}f$. It is also obvious that $bigcap_{phiinker(f^*)}kerphisupseteqoperatorname{im}f$. The claim is now evident.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    This is a little beyond the scope of my course but a very nice answer. Thank you!
    $endgroup$
    – user625722
    Jan 8 at 14:58
















2












$begingroup$

This answer does not assume finite dimensionality of $V$ or $W$ (so it is a generalized version of the Fredholm alternative). That is, I claim that for any linear map $f:Vto W$,
$$bigcap_{phiinker (f^*)}kerphi=operatorname{im}f.$$



We have the following exact sequence of vector spaces:
$${0}to ker f hookrightarrow V overset{f}{longrightarrow} Wtwoheadrightarrow operatorname{coim}fto {0},$$
where $operatorname{coim}f$ is the coimage $W/operatorname{im} f$. Dualizing this exact sequence, we obtain the exact sequence
$${0}to (operatorname{coim}f)^*hookrightarrow W^*overset{f^*}{longrightarrow} V^*twoheadrightarrow (ker f)^*to{0},$$
where $(operatorname{coim}f)^*$ is considered a subspace of $W^*$ via the identification that sends ${psi} in (operatorname{coim}f)^*$ to $hatpsiin W^*$ such that
$$hatpsi(w)={psi}(w+operatorname{im}f) forall win W.$$
Similarly, $(ker f)^*$ is a quotient of $V^*$ via the restriction map sending $sigmain V^*$ to $sigma|_{ker f}in (ker f)^*$ (precisely, $(ker f)^*cong V^*/(operatorname{cokr} f)^*$, where $operatorname{cokr}f$ is the cokernel $V/ker f$). By the exactness, we get
$$ker (f^*)=(operatorname{coim}f)^*.$$
That is, if $wnotin operatorname{im}f$, then $w+operatorname{im}f$ is non-zero in $operatorname{coim}W$. Therefore, there exists ${phi}_win (operatorname{coim} W)^*$ that does not vanish at $w+operatorname{im} f$. That is, $hatphi_w(w)ne 0$. Therefore, $wnotin bigcap_{phiinker(f^*)}kerphi$. This shows that $bigcap_{phiinker(f^*)}kerphisubseteqoperatorname{im}f$. It is also obvious that $bigcap_{phiinker(f^*)}kerphisupseteqoperatorname{im}f$. The claim is now evident.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    This is a little beyond the scope of my course but a very nice answer. Thank you!
    $endgroup$
    – user625722
    Jan 8 at 14:58














2












2








2





$begingroup$

This answer does not assume finite dimensionality of $V$ or $W$ (so it is a generalized version of the Fredholm alternative). That is, I claim that for any linear map $f:Vto W$,
$$bigcap_{phiinker (f^*)}kerphi=operatorname{im}f.$$



We have the following exact sequence of vector spaces:
$${0}to ker f hookrightarrow V overset{f}{longrightarrow} Wtwoheadrightarrow operatorname{coim}fto {0},$$
where $operatorname{coim}f$ is the coimage $W/operatorname{im} f$. Dualizing this exact sequence, we obtain the exact sequence
$${0}to (operatorname{coim}f)^*hookrightarrow W^*overset{f^*}{longrightarrow} V^*twoheadrightarrow (ker f)^*to{0},$$
where $(operatorname{coim}f)^*$ is considered a subspace of $W^*$ via the identification that sends ${psi} in (operatorname{coim}f)^*$ to $hatpsiin W^*$ such that
$$hatpsi(w)={psi}(w+operatorname{im}f) forall win W.$$
Similarly, $(ker f)^*$ is a quotient of $V^*$ via the restriction map sending $sigmain V^*$ to $sigma|_{ker f}in (ker f)^*$ (precisely, $(ker f)^*cong V^*/(operatorname{cokr} f)^*$, where $operatorname{cokr}f$ is the cokernel $V/ker f$). By the exactness, we get
$$ker (f^*)=(operatorname{coim}f)^*.$$
That is, if $wnotin operatorname{im}f$, then $w+operatorname{im}f$ is non-zero in $operatorname{coim}W$. Therefore, there exists ${phi}_win (operatorname{coim} W)^*$ that does not vanish at $w+operatorname{im} f$. That is, $hatphi_w(w)ne 0$. Therefore, $wnotin bigcap_{phiinker(f^*)}kerphi$. This shows that $bigcap_{phiinker(f^*)}kerphisubseteqoperatorname{im}f$. It is also obvious that $bigcap_{phiinker(f^*)}kerphisupseteqoperatorname{im}f$. The claim is now evident.






share|cite|improve this answer











$endgroup$



This answer does not assume finite dimensionality of $V$ or $W$ (so it is a generalized version of the Fredholm alternative). That is, I claim that for any linear map $f:Vto W$,
$$bigcap_{phiinker (f^*)}kerphi=operatorname{im}f.$$



We have the following exact sequence of vector spaces:
$${0}to ker f hookrightarrow V overset{f}{longrightarrow} Wtwoheadrightarrow operatorname{coim}fto {0},$$
where $operatorname{coim}f$ is the coimage $W/operatorname{im} f$. Dualizing this exact sequence, we obtain the exact sequence
$${0}to (operatorname{coim}f)^*hookrightarrow W^*overset{f^*}{longrightarrow} V^*twoheadrightarrow (ker f)^*to{0},$$
where $(operatorname{coim}f)^*$ is considered a subspace of $W^*$ via the identification that sends ${psi} in (operatorname{coim}f)^*$ to $hatpsiin W^*$ such that
$$hatpsi(w)={psi}(w+operatorname{im}f) forall win W.$$
Similarly, $(ker f)^*$ is a quotient of $V^*$ via the restriction map sending $sigmain V^*$ to $sigma|_{ker f}in (ker f)^*$ (precisely, $(ker f)^*cong V^*/(operatorname{cokr} f)^*$, where $operatorname{cokr}f$ is the cokernel $V/ker f$). By the exactness, we get
$$ker (f^*)=(operatorname{coim}f)^*.$$
That is, if $wnotin operatorname{im}f$, then $w+operatorname{im}f$ is non-zero in $operatorname{coim}W$. Therefore, there exists ${phi}_win (operatorname{coim} W)^*$ that does not vanish at $w+operatorname{im} f$. That is, $hatphi_w(w)ne 0$. Therefore, $wnotin bigcap_{phiinker(f^*)}kerphi$. This shows that $bigcap_{phiinker(f^*)}kerphisubseteqoperatorname{im}f$. It is also obvious that $bigcap_{phiinker(f^*)}kerphisupseteqoperatorname{im}f$. The claim is now evident.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 8 at 15:01

























answered Jan 8 at 14:47







user593746



















  • $begingroup$
    This is a little beyond the scope of my course but a very nice answer. Thank you!
    $endgroup$
    – user625722
    Jan 8 at 14:58


















  • $begingroup$
    This is a little beyond the scope of my course but a very nice answer. Thank you!
    $endgroup$
    – user625722
    Jan 8 at 14:58
















$begingroup$
This is a little beyond the scope of my course but a very nice answer. Thank you!
$endgroup$
– user625722
Jan 8 at 14:58




$begingroup$
This is a little beyond the scope of my course but a very nice answer. Thank you!
$endgroup$
– user625722
Jan 8 at 14:58


















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