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Show that $Γ$ is isomorphic to $L(V/W,V')$

0

$begingroup$

Question: let $V$, $V'$ be vector spaces over field $K$ and $W$ be subspace of $V$ then show that $Gamma = Tin L(V,V')vertforall win W: T(w) = 0$ is subspace of $L(V,V')$. Further show that this subspace is isomorphic to $L(V/W,V')$ and if $V$, $V'$ are finite dimensional then $dimGamma = (dim V - dim W)dim V'$.

Note: $L(V,V')$ is vector space of all linear transformations from $V$ to $V'$.

My attempt: I had finished the first part. I had shown that $Γ$ is subspace of $L(V,V')$. Further I can see, $W$ is contained in $ker T$ for all $Tin L(V,V')$. I know, I need to use isomorphism theorems. But I didn't able to go further.
Please help me.

linear-algebra linear-transformations vector-space-isomorphism

share|cite|improve this question

edited Apr 8 at 13:41

Martín-Blas Pérez Pinilla

35.5k42972

asked Apr 8 at 13:25

Akash PatalwanshiAkash Patalwanshi

1,0171820

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I assume you mean the subspace is isomorphic to $L(V/W,V')$ and not $L(V,V')$.
  $endgroup$
  – Floris Claassens
  Apr 8 at 13:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @FlorisClaassens sir, yes.
  $endgroup$
  – Akash Patalwanshi
  Apr 8 at 13:52
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, but there was a typo which now has been edited out.
  $endgroup$
  – Floris Claassens
  Apr 8 at 13:52
  

  
  
  
  

add a comment | 

0

$begingroup$

Question: let $V$, $V'$ be vector spaces over field $K$ and $W$ be subspace of $V$ then show that $Gamma = Tin L(V,V')vertforall win W: T(w) = 0$ is subspace of $L(V,V')$. Further show that this subspace is isomorphic to $L(V/W,V')$ and if $V$, $V'$ are finite dimensional then $dimGamma = (dim V - dim W)dim V'$.

Note: $L(V,V')$ is vector space of all linear transformations from $V$ to $V'$.

My attempt: I had finished the first part. I had shown that $Γ$ is subspace of $L(V,V')$. Further I can see, $W$ is contained in $ker T$ for all $Tin L(V,V')$. I know, I need to use isomorphism theorems. But I didn't able to go further.
Please help me.

linear-algebra linear-transformations vector-space-isomorphism

share|cite|improve this question

edited Apr 8 at 13:41

Martín-Blas Pérez Pinilla

35.5k42972

asked Apr 8 at 13:25

Akash PatalwanshiAkash Patalwanshi

1,0171820

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I assume you mean the subspace is isomorphic to $L(V/W,V')$ and not $L(V,V')$.
  $endgroup$
  – Floris Claassens
  Apr 8 at 13:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @FlorisClaassens sir, yes.
  $endgroup$
  – Akash Patalwanshi
  Apr 8 at 13:52
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, but there was a typo which now has been edited out.
  $endgroup$
  – Floris Claassens
  Apr 8 at 13:52
  

  
  
  
  

add a comment | 

$begingroup$

Question: let $V$, $V'$ be vector spaces over field $K$ and $W$ be subspace of $V$ then show that $Gamma = Tin L(V,V')vertforall win W: T(w) = 0$ is subspace of $L(V,V')$. Further show that this subspace is isomorphic to $L(V/W,V')$ and if $V$, $V'$ are finite dimensional then $dimGamma = (dim V - dim W)dim V'$.

Note: $L(V,V')$ is vector space of all linear transformations from $V$ to $V'$.

My attempt: I had finished the first part. I had shown that $Γ$ is subspace of $L(V,V')$. Further I can see, $W$ is contained in $ker T$ for all $Tin L(V,V')$. I know, I need to use isomorphism theorems. But I didn't able to go further.
Please help me.

linear-algebra linear-transformations vector-space-isomorphism

share|cite|improve this question

edited Apr 8 at 13:41

Martín-Blas Pérez Pinilla

35.5k42972

asked Apr 8 at 13:25

Akash PatalwanshiAkash Patalwanshi

1,0171820

$endgroup$

share|cite|improve this question

edited Apr 8 at 13:41

Martín-Blas Pérez Pinilla

35.5k42972

asked Apr 8 at 13:25

Akash PatalwanshiAkash Patalwanshi

1,0171820

share|cite|improve this question

edited Apr 8 at 13:41

Martín-Blas Pérez Pinilla

35.5k42972

asked Apr 8 at 13:25

Akash PatalwanshiAkash Patalwanshi

1,0171820

edited Apr 8 at 13:41

Martín-Blas Pérez Pinilla

35.5k42972

edited Apr 8 at 13:41

Martín-Blas Pérez Pinilla

35.5k42972

asked Apr 8 at 13:25

Akash PatalwanshiAkash Patalwanshi

1,0171820

asked Apr 8 at 13:25

Akash PatalwanshiAkash Patalwanshi

1,0171820

1 Answer
1

active

oldest

votes

2

$begingroup$

Hint: check the (obviously? linear) function
$$f:Gammalongrightarrow L(V/W,V′)$$
$$Tlongmapsto f(T)$$
$$f(T):V/Wlongrightarrow V′$$
$$v + Wlongmapsto T(v).$$
Why is well-defined? Why is isomorphism?...

Edit: about the well-definedness of $f(T)$: if $Tin Gamma$, by definition,
$$forall win W: T(w) = 0.$$
This means that for all $vin V, win W$:
$T(v + w) = T(v)...$

Edit: about the injectivity of $F$ ($[] =$ class of equivalence):

$$f(T) = 0iffforall [v]in V/W: T(v) = f(T) = 0iff T = 0.$$
(why the last $iff$?)

About the surjectivity: write $V$ as direct sum of $W$ with another subspace (call it $E$). Now taking $Sin L(V/W,V')$, we must define a $bar SinGamma$ with $f(bar S) = S$. This means that for $vin E: bar S(v) =cdots$

share|cite|improve this answer

edited Apr 10 at 7:33

answered Apr 8 at 13:36

Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla

35.5k42972

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sir, how $f$ is defined? What is definition of$f$? and please elaborate.
  $endgroup$
  – Akash Patalwanshi
  Apr 8 at 13:50
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @AkashPatalwanshi, read again. For each $TinGamma$, $f(T)$ is the linear function (check!) from $V/W$ in $V'$ defined by $f(T)(v + W) = T(v)$. Start checking that is well-defined (independent of the representative).
  $endgroup$
  – Martín-Blas Pérez Pinilla
  Apr 8 at 16:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sir thanks for adding more details. Sir yes now I can see $f$ is well defined. Further to check injectivity of $f$, I think we just need to check $kerf=o$ where $o$ is zero element of $L(V/W,V')$. Since $Kerf=Tin Gamma :f(T)=o$ and hence $kerf=Tin Gamma: f(T)(v+W)=o(v+W)$. $kerf=Tin Gamma: T(v)=0_V'$ then how to proceed further? and if we prove injectivity then too we need to prove $f$ is surjective and that will more difficult for me because I didn't able to see how for every $T$ in $L(V/W,V')$ there exists element in $Gamma$. Please help.
  $endgroup$
  – Akash Patalwanshi
  Apr 9 at 12:40
  
  

  
  
  
  

add a comment | 

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2

$begingroup$

Hint: check the (obviously? linear) function
$$f:Gammalongrightarrow L(V/W,V′)$$
$$Tlongmapsto f(T)$$
$$f(T):V/Wlongrightarrow V′$$
$$v + Wlongmapsto T(v).$$
Why is well-defined? Why is isomorphism?...

Edit: about the well-definedness of $f(T)$: if $Tin Gamma$, by definition,
$$forall win W: T(w) = 0.$$
This means that for all $vin V, win W$:
$T(v + w) = T(v)...$

Edit: about the injectivity of $F$ ($[] =$ class of equivalence):

$$f(T) = 0iffforall [v]in V/W: T(v) = f(T) = 0iff T = 0.$$
(why the last $iff$?)

About the surjectivity: write $V$ as direct sum of $W$ with another subspace (call it $E$). Now taking $Sin L(V/W,V')$, we must define a $bar SinGamma$ with $f(bar S) = S$. This means that for $vin E: bar S(v) =cdots$

share|cite|improve this answer

edited Apr 10 at 7:33

answered Apr 8 at 13:36

Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla

35.5k42972

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sir, how $f$ is defined? What is definition of$f$? and please elaborate.
  $endgroup$
  – Akash Patalwanshi
  Apr 8 at 13:50
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @AkashPatalwanshi, read again. For each $TinGamma$, $f(T)$ is the linear function (check!) from $V/W$ in $V'$ defined by $f(T)(v + W) = T(v)$. Start checking that is well-defined (independent of the representative).
  $endgroup$
  – Martín-Blas Pérez Pinilla
  Apr 8 at 16:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sir thanks for adding more details. Sir yes now I can see $f$ is well defined. Further to check injectivity of $f$, I think we just need to check $kerf=o$ where $o$ is zero element of $L(V/W,V')$. Since $Kerf=Tin Gamma :f(T)=o$ and hence $kerf=Tin Gamma: f(T)(v+W)=o(v+W)$. $kerf=Tin Gamma: T(v)=0_V'$ then how to proceed further? and if we prove injectivity then too we need to prove $f$ is surjective and that will more difficult for me because I didn't able to see how for every $T$ in $L(V/W,V')$ there exists element in $Gamma$. Please help.
  $endgroup$
  – Akash Patalwanshi
  Apr 9 at 12:40
  
  

  
  
  
  

add a comment | 

2

$begingroup$

Hint: check the (obviously? linear) function
$$f:Gammalongrightarrow L(V/W,V′)$$
$$Tlongmapsto f(T)$$
$$f(T):V/Wlongrightarrow V′$$
$$v + Wlongmapsto T(v).$$
Why is well-defined? Why is isomorphism?...

Edit: about the well-definedness of $f(T)$: if $Tin Gamma$, by definition,
$$forall win W: T(w) = 0.$$
This means that for all $vin V, win W$:
$T(v + w) = T(v)...$

Edit: about the injectivity of $F$ ($[] =$ class of equivalence):

$$f(T) = 0iffforall [v]in V/W: T(v) = f(T) = 0iff T = 0.$$
(why the last $iff$?)

About the surjectivity: write $V$ as direct sum of $W$ with another subspace (call it $E$). Now taking $Sin L(V/W,V')$, we must define a $bar SinGamma$ with $f(bar S) = S$. This means that for $vin E: bar S(v) =cdots$

share|cite|improve this answer

edited Apr 10 at 7:33

answered Apr 8 at 13:36

Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla

35.5k42972

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sir, how $f$ is defined? What is definition of$f$? and please elaborate.
  $endgroup$
  – Akash Patalwanshi
  Apr 8 at 13:50
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @AkashPatalwanshi, read again. For each $TinGamma$, $f(T)$ is the linear function (check!) from $V/W$ in $V'$ defined by $f(T)(v + W) = T(v)$. Start checking that is well-defined (independent of the representative).
  $endgroup$
  – Martín-Blas Pérez Pinilla
  Apr 8 at 16:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sir thanks for adding more details. Sir yes now I can see $f$ is well defined. Further to check injectivity of $f$, I think we just need to check $kerf=o$ where $o$ is zero element of $L(V/W,V')$. Since $Kerf=Tin Gamma :f(T)=o$ and hence $kerf=Tin Gamma: f(T)(v+W)=o(v+W)$. $kerf=Tin Gamma: T(v)=0_V'$ then how to proceed further? and if we prove injectivity then too we need to prove $f$ is surjective and that will more difficult for me because I didn't able to see how for every $T$ in $L(V/W,V')$ there exists element in $Gamma$. Please help.
  $endgroup$
  – Akash Patalwanshi
  Apr 9 at 12:40
  
  

  
  
  
  

add a comment | 

$begingroup$

Hint: check the (obviously? linear) function
$$f:Gammalongrightarrow L(V/W,V′)$$
$$Tlongmapsto f(T)$$
$$f(T):V/Wlongrightarrow V′$$
$$v + Wlongmapsto T(v).$$
Why is well-defined? Why is isomorphism?...

Edit: about the well-definedness of $f(T)$: if $Tin Gamma$, by definition,
$$forall win W: T(w) = 0.$$
This means that for all $vin V, win W$:
$T(v + w) = T(v)...$

Edit: about the injectivity of $F$ ($[] =$ class of equivalence):

$$f(T) = 0iffforall [v]in V/W: T(v) = f(T) = 0iff T = 0.$$
(why the last $iff$?)

About the surjectivity: write $V$ as direct sum of $W$ with another subspace (call it $E$). Now taking $Sin L(V/W,V')$, we must define a $bar SinGamma$ with $f(bar S) = S$. This means that for $vin E: bar S(v) =cdots$

share|cite|improve this answer

edited Apr 10 at 7:33

answered Apr 8 at 13:36

Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla

35.5k42972

$endgroup$

share|cite|improve this answer

edited Apr 10 at 7:33

answered Apr 8 at 13:36

Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla

35.5k42972

answered Apr 8 at 13:36

Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla

35.5k42972

answered Apr 8 at 13:36

Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla

35.5k42972