Prove that if $phi$ is invertible operator of V space, then $phi$ and $phi^{-1}$ has the same invariant...












0












$begingroup$


Prove that if $phi$ is invertible operator of V space, then $phi$ and $phi^{-1}$ has the same invariant subspaces. I know how to prove this statement for finite V. (Wish it is right) : Let $W$ be invariant subspace for $phi$, then $phi(W)subset W$. If W is finite and $phi$ is invertible $Rightarrow$ $phimid_W : W rightarrow W$ is bijection, that's why for every $vin W$ there exist $v'in W$ such that $phi(v) = v' Rightarrow phi^{-1}(phi(v)) = phi^{-1}(v')Rightarrow v = phi^{-1}(v') Rightarrow phi^{-1}(W)subset W$.



But I don't know how to deal with infinite V. Can I do the same or it is completely different solution? Can you help me?






operator-theory invariant-subspace







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  • $begingroup$
    Are you assuming that $V$ is finite-dimensional?
    $endgroup$
    – José Carlos Santos
    Dec 4 '18 at 11:54










  • $begingroup$
    Actually, it can be finite or infinite, that is why I want to know how solve it for two cases. (If I did not answered you question, please, let me know)
    $endgroup$
    – ErlGrey
    Dec 4 '18 at 11:59
















0












$begingroup$


Prove that if $phi$ is invertible operator of V space, then $phi$ and $phi^{-1}$ has the same invariant subspaces. I know how to prove this statement for finite V. (Wish it is right) : Let $W$ be invariant subspace for $phi$, then $phi(W)subset W$. If W is finite and $phi$ is invertible $Rightarrow$ $phimid_W : W rightarrow W$ is bijection, that's why for every $vin W$ there exist $v'in W$ such that $phi(v) = v' Rightarrow phi^{-1}(phi(v)) = phi^{-1}(v')Rightarrow v = phi^{-1}(v') Rightarrow phi^{-1}(W)subset W$.



But I don't know how to deal with infinite V. Can I do the same or it is completely different solution? Can you help me?






operator-theory invariant-subspace







share|cite|improve this question









$endgroup$












  • $begingroup$
    Are you assuming that $V$ is finite-dimensional?
    $endgroup$
    – José Carlos Santos
    Dec 4 '18 at 11:54










  • $begingroup$
    Actually, it can be finite or infinite, that is why I want to know how solve it for two cases. (If I did not answered you question, please, let me know)
    $endgroup$
    – ErlGrey
    Dec 4 '18 at 11:59














0












0








0





$begingroup$


Prove that if $phi$ is invertible operator of V space, then $phi$ and $phi^{-1}$ has the same invariant subspaces. I know how to prove this statement for finite V. (Wish it is right) : Let $W$ be invariant subspace for $phi$, then $phi(W)subset W$. If W is finite and $phi$ is invertible $Rightarrow$ $phimid_W : W rightarrow W$ is bijection, that's why for every $vin W$ there exist $v'in W$ such that $phi(v) = v' Rightarrow phi^{-1}(phi(v)) = phi^{-1}(v')Rightarrow v = phi^{-1}(v') Rightarrow phi^{-1}(W)subset W$.



But I don't know how to deal with infinite V. Can I do the same or it is completely different solution? Can you help me?






operator-theory invariant-subspace







share|cite|improve this question









$endgroup$




Prove that if $phi$ is invertible operator of V space, then $phi$ and $phi^{-1}$ has the same invariant subspaces. I know how to prove this statement for finite V. (Wish it is right) : Let $W$ be invariant subspace for $phi$, then $phi(W)subset W$. If W is finite and $phi$ is invertible $Rightarrow$ $phimid_W : W rightarrow W$ is bijection, that's why for every $vin W$ there exist $v'in W$ such that $phi(v) = v' Rightarrow phi^{-1}(phi(v)) = phi^{-1}(v')Rightarrow v = phi^{-1}(v') Rightarrow phi^{-1}(W)subset W$.



But I don't know how to deal with infinite V. Can I do the same or it is completely different solution? Can you help me?






operator-theory invariant-subspace




operator-theory invariant-subspace






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 4 '18 at 11:52









ErlGreyErlGrey

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32












  • $begingroup$
    Are you assuming that $V$ is finite-dimensional?
    $endgroup$
    – José Carlos Santos
    Dec 4 '18 at 11:54










  • $begingroup$
    Actually, it can be finite or infinite, that is why I want to know how solve it for two cases. (If I did not answered you question, please, let me know)
    $endgroup$
    – ErlGrey
    Dec 4 '18 at 11:59


















  • $begingroup$
    Are you assuming that $V$ is finite-dimensional?
    $endgroup$
    – José Carlos Santos
    Dec 4 '18 at 11:54










  • $begingroup$
    Actually, it can be finite or infinite, that is why I want to know how solve it for two cases. (If I did not answered you question, please, let me know)
    $endgroup$
    – ErlGrey
    Dec 4 '18 at 11:59
















$begingroup$
Are you assuming that $V$ is finite-dimensional?
$endgroup$
– José Carlos Santos
Dec 4 '18 at 11:54




$begingroup$
Are you assuming that $V$ is finite-dimensional?
$endgroup$
– José Carlos Santos
Dec 4 '18 at 11:54












$begingroup$
Actually, it can be finite or infinite, that is why I want to know how solve it for two cases. (If I did not answered you question, please, let me know)
$endgroup$
– ErlGrey
Dec 4 '18 at 11:59




$begingroup$
Actually, it can be finite or infinite, that is why I want to know how solve it for two cases. (If I did not answered you question, please, let me know)
$endgroup$
– ErlGrey
Dec 4 '18 at 11:59










1 Answer
1






active

oldest

votes


















1












$begingroup$

I suppose that the assertion that you wish to prove is this:




If $V$ is a vector space, $phicolon Vlongrightarrow V$ is an invertible linear map and $Wsubset V$ is a subspace such that $phi(W)subset W$, then $phi^{-1}(W)subset W$.




If that's so, then you cannot prove it, since it is false. Let $V$ be the real vector space of all functions from $mathbb R$ into itself, let $bigl(phi(f)bigr)(x)=f(x-1)$ and let $W={fcolonmathbb{R}longrightarrowmathbb{R},|,x<0implies f(x)=0}$. Then $phi(W)subset W$. However, if you define$$begin{array}{rccc}fcolon&mathbb R&longrightarrow&mathbb R\&x&mapsto&begin{cases}0&text{ if }x<0\1&text{ otherwise,}end{cases}end{array}$$then $fin W$, but $phi^{-1}(f)notin W$, since $bigl(phi^{-1}(f)bigr)left(-frac12right)=1$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    oh, I see. Thank you so much for counterexample. am I right that $(phi^{-1}(f))(-frac{1}{2}) = 1$, because $(phi^{-1}(f))(x) = f(x+1)$? And is it true statement for finite V and is my solution right then?
    $endgroup$
    – ErlGrey
    Dec 4 '18 at 12:27










  • $begingroup$
    Your computation is correct. But I am quite confused with your term “finite”. Do you mean finite set? Or finite-dimensional space?
    $endgroup$
    – José Carlos Santos
    Dec 4 '18 at 12:29












  • $begingroup$
    I meant finite-dimensional, because if I am not mistaken if V is finite dimensional we can say that $phi|_W : W rightarrow W$ is bijection
    $endgroup$
    – ErlGrey
    Dec 4 '18 at 12:35












  • $begingroup$
    Then, yes, you are absolutely right.
    $endgroup$
    – José Carlos Santos
    Dec 4 '18 at 12:55











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

I suppose that the assertion that you wish to prove is this:




If $V$ is a vector space, $phicolon Vlongrightarrow V$ is an invertible linear map and $Wsubset V$ is a subspace such that $phi(W)subset W$, then $phi^{-1}(W)subset W$.




If that's so, then you cannot prove it, since it is false. Let $V$ be the real vector space of all functions from $mathbb R$ into itself, let $bigl(phi(f)bigr)(x)=f(x-1)$ and let $W={fcolonmathbb{R}longrightarrowmathbb{R},|,x<0implies f(x)=0}$. Then $phi(W)subset W$. However, if you define$$begin{array}{rccc}fcolon&mathbb R&longrightarrow&mathbb R\&x&mapsto&begin{cases}0&text{ if }x<0\1&text{ otherwise,}end{cases}end{array}$$then $fin W$, but $phi^{-1}(f)notin W$, since $bigl(phi^{-1}(f)bigr)left(-frac12right)=1$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    oh, I see. Thank you so much for counterexample. am I right that $(phi^{-1}(f))(-frac{1}{2}) = 1$, because $(phi^{-1}(f))(x) = f(x+1)$? And is it true statement for finite V and is my solution right then?
    $endgroup$
    – ErlGrey
    Dec 4 '18 at 12:27










  • $begingroup$
    Your computation is correct. But I am quite confused with your term “finite”. Do you mean finite set? Or finite-dimensional space?
    $endgroup$
    – José Carlos Santos
    Dec 4 '18 at 12:29












  • $begingroup$
    I meant finite-dimensional, because if I am not mistaken if V is finite dimensional we can say that $phi|_W : W rightarrow W$ is bijection
    $endgroup$
    – ErlGrey
    Dec 4 '18 at 12:35












  • $begingroup$
    Then, yes, you are absolutely right.
    $endgroup$
    – José Carlos Santos
    Dec 4 '18 at 12:55
















1












$begingroup$

I suppose that the assertion that you wish to prove is this:




If $V$ is a vector space, $phicolon Vlongrightarrow V$ is an invertible linear map and $Wsubset V$ is a subspace such that $phi(W)subset W$, then $phi^{-1}(W)subset W$.




If that's so, then you cannot prove it, since it is false. Let $V$ be the real vector space of all functions from $mathbb R$ into itself, let $bigl(phi(f)bigr)(x)=f(x-1)$ and let $W={fcolonmathbb{R}longrightarrowmathbb{R},|,x<0implies f(x)=0}$. Then $phi(W)subset W$. However, if you define$$begin{array}{rccc}fcolon&mathbb R&longrightarrow&mathbb R\&x&mapsto&begin{cases}0&text{ if }x<0\1&text{ otherwise,}end{cases}end{array}$$then $fin W$, but $phi^{-1}(f)notin W$, since $bigl(phi^{-1}(f)bigr)left(-frac12right)=1$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    oh, I see. Thank you so much for counterexample. am I right that $(phi^{-1}(f))(-frac{1}{2}) = 1$, because $(phi^{-1}(f))(x) = f(x+1)$? And is it true statement for finite V and is my solution right then?
    $endgroup$
    – ErlGrey
    Dec 4 '18 at 12:27










  • $begingroup$
    Your computation is correct. But I am quite confused with your term “finite”. Do you mean finite set? Or finite-dimensional space?
    $endgroup$
    – José Carlos Santos
    Dec 4 '18 at 12:29












  • $begingroup$
    I meant finite-dimensional, because if I am not mistaken if V is finite dimensional we can say that $phi|_W : W rightarrow W$ is bijection
    $endgroup$
    – ErlGrey
    Dec 4 '18 at 12:35












  • $begingroup$
    Then, yes, you are absolutely right.
    $endgroup$
    – José Carlos Santos
    Dec 4 '18 at 12:55














1












1








1





$begingroup$

I suppose that the assertion that you wish to prove is this:




If $V$ is a vector space, $phicolon Vlongrightarrow V$ is an invertible linear map and $Wsubset V$ is a subspace such that $phi(W)subset W$, then $phi^{-1}(W)subset W$.




If that's so, then you cannot prove it, since it is false. Let $V$ be the real vector space of all functions from $mathbb R$ into itself, let $bigl(phi(f)bigr)(x)=f(x-1)$ and let $W={fcolonmathbb{R}longrightarrowmathbb{R},|,x<0implies f(x)=0}$. Then $phi(W)subset W$. However, if you define$$begin{array}{rccc}fcolon&mathbb R&longrightarrow&mathbb R\&x&mapsto&begin{cases}0&text{ if }x<0\1&text{ otherwise,}end{cases}end{array}$$then $fin W$, but $phi^{-1}(f)notin W$, since $bigl(phi^{-1}(f)bigr)left(-frac12right)=1$.






share|cite|improve this answer









$endgroup$



I suppose that the assertion that you wish to prove is this:




If $V$ is a vector space, $phicolon Vlongrightarrow V$ is an invertible linear map and $Wsubset V$ is a subspace such that $phi(W)subset W$, then $phi^{-1}(W)subset W$.




If that's so, then you cannot prove it, since it is false. Let $V$ be the real vector space of all functions from $mathbb R$ into itself, let $bigl(phi(f)bigr)(x)=f(x-1)$ and let $W={fcolonmathbb{R}longrightarrowmathbb{R},|,x<0implies f(x)=0}$. Then $phi(W)subset W$. However, if you define$$begin{array}{rccc}fcolon&mathbb R&longrightarrow&mathbb R\&x&mapsto&begin{cases}0&text{ if }x<0\1&text{ otherwise,}end{cases}end{array}$$then $fin W$, but $phi^{-1}(f)notin W$, since $bigl(phi^{-1}(f)bigr)left(-frac12right)=1$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 4 '18 at 12:16









José Carlos SantosJosé Carlos Santos

158k22126228




158k22126228












  • $begingroup$
    oh, I see. Thank you so much for counterexample. am I right that $(phi^{-1}(f))(-frac{1}{2}) = 1$, because $(phi^{-1}(f))(x) = f(x+1)$? And is it true statement for finite V and is my solution right then?
    $endgroup$
    – ErlGrey
    Dec 4 '18 at 12:27










  • $begingroup$
    Your computation is correct. But I am quite confused with your term “finite”. Do you mean finite set? Or finite-dimensional space?
    $endgroup$
    – José Carlos Santos
    Dec 4 '18 at 12:29












  • $begingroup$
    I meant finite-dimensional, because if I am not mistaken if V is finite dimensional we can say that $phi|_W : W rightarrow W$ is bijection
    $endgroup$
    – ErlGrey
    Dec 4 '18 at 12:35












  • $begingroup$
    Then, yes, you are absolutely right.
    $endgroup$
    – José Carlos Santos
    Dec 4 '18 at 12:55


















  • $begingroup$
    oh, I see. Thank you so much for counterexample. am I right that $(phi^{-1}(f))(-frac{1}{2}) = 1$, because $(phi^{-1}(f))(x) = f(x+1)$? And is it true statement for finite V and is my solution right then?
    $endgroup$
    – ErlGrey
    Dec 4 '18 at 12:27










  • $begingroup$
    Your computation is correct. But I am quite confused with your term “finite”. Do you mean finite set? Or finite-dimensional space?
    $endgroup$
    – José Carlos Santos
    Dec 4 '18 at 12:29












  • $begingroup$
    I meant finite-dimensional, because if I am not mistaken if V is finite dimensional we can say that $phi|_W : W rightarrow W$ is bijection
    $endgroup$
    – ErlGrey
    Dec 4 '18 at 12:35












  • $begingroup$
    Then, yes, you are absolutely right.
    $endgroup$
    – José Carlos Santos
    Dec 4 '18 at 12:55
















$begingroup$
oh, I see. Thank you so much for counterexample. am I right that $(phi^{-1}(f))(-frac{1}{2}) = 1$, because $(phi^{-1}(f))(x) = f(x+1)$? And is it true statement for finite V and is my solution right then?
$endgroup$
– ErlGrey
Dec 4 '18 at 12:27




$begingroup$
oh, I see. Thank you so much for counterexample. am I right that $(phi^{-1}(f))(-frac{1}{2}) = 1$, because $(phi^{-1}(f))(x) = f(x+1)$? And is it true statement for finite V and is my solution right then?
$endgroup$
– ErlGrey
Dec 4 '18 at 12:27












$begingroup$
Your computation is correct. But I am quite confused with your term “finite”. Do you mean finite set? Or finite-dimensional space?
$endgroup$
– José Carlos Santos
Dec 4 '18 at 12:29






$begingroup$
Your computation is correct. But I am quite confused with your term “finite”. Do you mean finite set? Or finite-dimensional space?
$endgroup$
– José Carlos Santos
Dec 4 '18 at 12:29














$begingroup$
I meant finite-dimensional, because if I am not mistaken if V is finite dimensional we can say that $phi|_W : W rightarrow W$ is bijection
$endgroup$
– ErlGrey
Dec 4 '18 at 12:35






$begingroup$
I meant finite-dimensional, because if I am not mistaken if V is finite dimensional we can say that $phi|_W : W rightarrow W$ is bijection
$endgroup$
– ErlGrey
Dec 4 '18 at 12:35














$begingroup$
Then, yes, you are absolutely right.
$endgroup$
– José Carlos Santos
Dec 4 '18 at 12:55




$begingroup$
Then, yes, you are absolutely right.
$endgroup$
– José Carlos Santos
Dec 4 '18 at 12:55


















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