Showing that image of a certain linear map is either trivial or a straight line












2














From S.L Linear Algebra:




Let $A$ be a non-zero vector in $R^2$. Let $F: mathbb{R}^2
rightarrow W$
be a linear map such that $F(A)=O$. Show that the image
of $F$ is either a straight line or ${0}$.






I've taken following theorems from the book to try and construct the answer (proofs of theorems are omitted):




Theorem 3.2. Let $V$ be a vector space. Let $L: V rightarrow W$ be a linear map of $V$ into another space $W$. Let $n$ be the
dimension of $V$, $q$ the dimension of the kernel of $L$, and $s$ the
dimension of the image of $L$. Then $n = q + s$. In other words,



$$dim V= dim operatorname{Ker} L + dimoperatorname{Im } L$$






Answer that I have constructed assumes all possibilities of dimension that kernel might have (due to cardinality, and Theorem 3.2, $dimoperatorname{Ker}F in {0, 1, 2}$ of a linear map $F$).



Possibility 1) $dimoperatorname{Ker} F = 2$



If the dimension of kernel is $2$, that is, image is zero dimensional according to Theorem 3.2 ($dim operatorname{Im} F = dimmathbb{R}^{2} - dimoperatorname{Ker}F = 2 - 2 = 0$), then considering that kernel is a subspace, $mathbb{R}^2 = operatorname{Ker}F$, and therefore $F$ is a zero map having the image of ${0}$.



Possibility 2) $dimoperatorname{Ker}F = 1$



If kernel is $1$-dimensional, then so is the image according to Theorem 3.2 and thus we have a straight line as the image of $F$, considering that we have one-dimensional image under linear map $F$.



Possibility 3) $dimoperatorname{Ker}F ={0} $ (Presumably impossible)



Kernel can't be zero-dimensional, since it is spanned by two dimensional vectors that are not zero vectors.



Conclusion:



Hence the image of $F$ is either a straight line or ${0}$.





Is this answer sufficient and true? My explanation of Possibility 2) concerns me the most, it might not be specific enough.



Thank you!










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  • 1




    $ker F$ can't be $0$ dimensional because $0neq Ainker F$.
    – Federico
    Nov 23 at 19:16










  • @Federico That is what I tried to imply in the possibility 3, when I mentioned that kernel not being "spanned" by zero vectors makes it non trivial and therefore $> 0$ dimensional.
    – ShellRox
    Nov 23 at 22:15
















2














From S.L Linear Algebra:




Let $A$ be a non-zero vector in $R^2$. Let $F: mathbb{R}^2
rightarrow W$
be a linear map such that $F(A)=O$. Show that the image
of $F$ is either a straight line or ${0}$.






I've taken following theorems from the book to try and construct the answer (proofs of theorems are omitted):




Theorem 3.2. Let $V$ be a vector space. Let $L: V rightarrow W$ be a linear map of $V$ into another space $W$. Let $n$ be the
dimension of $V$, $q$ the dimension of the kernel of $L$, and $s$ the
dimension of the image of $L$. Then $n = q + s$. In other words,



$$dim V= dim operatorname{Ker} L + dimoperatorname{Im } L$$






Answer that I have constructed assumes all possibilities of dimension that kernel might have (due to cardinality, and Theorem 3.2, $dimoperatorname{Ker}F in {0, 1, 2}$ of a linear map $F$).



Possibility 1) $dimoperatorname{Ker} F = 2$



If the dimension of kernel is $2$, that is, image is zero dimensional according to Theorem 3.2 ($dim operatorname{Im} F = dimmathbb{R}^{2} - dimoperatorname{Ker}F = 2 - 2 = 0$), then considering that kernel is a subspace, $mathbb{R}^2 = operatorname{Ker}F$, and therefore $F$ is a zero map having the image of ${0}$.



Possibility 2) $dimoperatorname{Ker}F = 1$



If kernel is $1$-dimensional, then so is the image according to Theorem 3.2 and thus we have a straight line as the image of $F$, considering that we have one-dimensional image under linear map $F$.



Possibility 3) $dimoperatorname{Ker}F ={0} $ (Presumably impossible)



Kernel can't be zero-dimensional, since it is spanned by two dimensional vectors that are not zero vectors.



Conclusion:



Hence the image of $F$ is either a straight line or ${0}$.





Is this answer sufficient and true? My explanation of Possibility 2) concerns me the most, it might not be specific enough.



Thank you!










share|cite|improve this question




















  • 1




    $ker F$ can't be $0$ dimensional because $0neq Ainker F$.
    – Federico
    Nov 23 at 19:16










  • @Federico That is what I tried to imply in the possibility 3, when I mentioned that kernel not being "spanned" by zero vectors makes it non trivial and therefore $> 0$ dimensional.
    – ShellRox
    Nov 23 at 22:15














2












2








2







From S.L Linear Algebra:




Let $A$ be a non-zero vector in $R^2$. Let $F: mathbb{R}^2
rightarrow W$
be a linear map such that $F(A)=O$. Show that the image
of $F$ is either a straight line or ${0}$.






I've taken following theorems from the book to try and construct the answer (proofs of theorems are omitted):




Theorem 3.2. Let $V$ be a vector space. Let $L: V rightarrow W$ be a linear map of $V$ into another space $W$. Let $n$ be the
dimension of $V$, $q$ the dimension of the kernel of $L$, and $s$ the
dimension of the image of $L$. Then $n = q + s$. In other words,



$$dim V= dim operatorname{Ker} L + dimoperatorname{Im } L$$






Answer that I have constructed assumes all possibilities of dimension that kernel might have (due to cardinality, and Theorem 3.2, $dimoperatorname{Ker}F in {0, 1, 2}$ of a linear map $F$).



Possibility 1) $dimoperatorname{Ker} F = 2$



If the dimension of kernel is $2$, that is, image is zero dimensional according to Theorem 3.2 ($dim operatorname{Im} F = dimmathbb{R}^{2} - dimoperatorname{Ker}F = 2 - 2 = 0$), then considering that kernel is a subspace, $mathbb{R}^2 = operatorname{Ker}F$, and therefore $F$ is a zero map having the image of ${0}$.



Possibility 2) $dimoperatorname{Ker}F = 1$



If kernel is $1$-dimensional, then so is the image according to Theorem 3.2 and thus we have a straight line as the image of $F$, considering that we have one-dimensional image under linear map $F$.



Possibility 3) $dimoperatorname{Ker}F ={0} $ (Presumably impossible)



Kernel can't be zero-dimensional, since it is spanned by two dimensional vectors that are not zero vectors.



Conclusion:



Hence the image of $F$ is either a straight line or ${0}$.





Is this answer sufficient and true? My explanation of Possibility 2) concerns me the most, it might not be specific enough.



Thank you!










share|cite|improve this question















From S.L Linear Algebra:




Let $A$ be a non-zero vector in $R^2$. Let $F: mathbb{R}^2
rightarrow W$
be a linear map such that $F(A)=O$. Show that the image
of $F$ is either a straight line or ${0}$.






I've taken following theorems from the book to try and construct the answer (proofs of theorems are omitted):




Theorem 3.2. Let $V$ be a vector space. Let $L: V rightarrow W$ be a linear map of $V$ into another space $W$. Let $n$ be the
dimension of $V$, $q$ the dimension of the kernel of $L$, and $s$ the
dimension of the image of $L$. Then $n = q + s$. In other words,



$$dim V= dim operatorname{Ker} L + dimoperatorname{Im } L$$






Answer that I have constructed assumes all possibilities of dimension that kernel might have (due to cardinality, and Theorem 3.2, $dimoperatorname{Ker}F in {0, 1, 2}$ of a linear map $F$).



Possibility 1) $dimoperatorname{Ker} F = 2$



If the dimension of kernel is $2$, that is, image is zero dimensional according to Theorem 3.2 ($dim operatorname{Im} F = dimmathbb{R}^{2} - dimoperatorname{Ker}F = 2 - 2 = 0$), then considering that kernel is a subspace, $mathbb{R}^2 = operatorname{Ker}F$, and therefore $F$ is a zero map having the image of ${0}$.



Possibility 2) $dimoperatorname{Ker}F = 1$



If kernel is $1$-dimensional, then so is the image according to Theorem 3.2 and thus we have a straight line as the image of $F$, considering that we have one-dimensional image under linear map $F$.



Possibility 3) $dimoperatorname{Ker}F ={0} $ (Presumably impossible)



Kernel can't be zero-dimensional, since it is spanned by two dimensional vectors that are not zero vectors.



Conclusion:



Hence the image of $F$ is either a straight line or ${0}$.





Is this answer sufficient and true? My explanation of Possibility 2) concerns me the most, it might not be specific enough.



Thank you!







linear-algebra functions linear-transformations






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edited Nov 23 at 19:19









Bernard

118k638111




118k638111










asked Nov 23 at 19:12









ShellRox

25328




25328








  • 1




    $ker F$ can't be $0$ dimensional because $0neq Ainker F$.
    – Federico
    Nov 23 at 19:16










  • @Federico That is what I tried to imply in the possibility 3, when I mentioned that kernel not being "spanned" by zero vectors makes it non trivial and therefore $> 0$ dimensional.
    – ShellRox
    Nov 23 at 22:15














  • 1




    $ker F$ can't be $0$ dimensional because $0neq Ainker F$.
    – Federico
    Nov 23 at 19:16










  • @Federico That is what I tried to imply in the possibility 3, when I mentioned that kernel not being "spanned" by zero vectors makes it non trivial and therefore $> 0$ dimensional.
    – ShellRox
    Nov 23 at 22:15








1




1




$ker F$ can't be $0$ dimensional because $0neq Ainker F$.
– Federico
Nov 23 at 19:16




$ker F$ can't be $0$ dimensional because $0neq Ainker F$.
– Federico
Nov 23 at 19:16












@Federico That is what I tried to imply in the possibility 3, when I mentioned that kernel not being "spanned" by zero vectors makes it non trivial and therefore $> 0$ dimensional.
– ShellRox
Nov 23 at 22:15




@Federico That is what I tried to imply in the possibility 3, when I mentioned that kernel not being "spanned" by zero vectors makes it non trivial and therefore $> 0$ dimensional.
– ShellRox
Nov 23 at 22:15










2 Answers
2






active

oldest

votes


















1














Your approach is correct!



P1) $dim(Im F)=0 implies Im(F)={0}$, because the image of a linear function is a subspace and hence $0$ is in it, and can't have anything else because its dimension is zero. So $F(x)=0 forall x$



P2) we have $dim(Ker F)=1$, applying the theorem you get $dim(Im T)=1$ and you can use the fact that two vector spaces are isomorphic (they are "the same space") if their dimension are equal, hence you can say that $Im(T)cong mathbb{R}$ which is a very nice way to justify that "$Im(T)$ is a straight line".



P3) can't be the case that $dim(Ker T)=0$ because this would implie $Ker(T)={0}$, but we know that $Anot=0$ and $Ain Ker(T)$



Your answer is good too! But it seems like it need to be more "direct" in a way... but the question isn't too direct either... I assumed that "being a straight line" is the same that "have dimension one"... but justifying that dimension one implies being isomorphic to the reals is also a good argument (because they are often called THE line).






share|cite|improve this answer

















  • 1




    Thank you! So since two subspaces can be associated with bijective mapping (injective and surjective) they are isomorphic in this case. I have one last question, on my possibility 3 explanation, was I correct when I said that kernel not being "spanned" by only zero vectors makes it non trivial and higher dimensional?
    – ShellRox
    Nov 23 at 22:11






  • 1




    If I got what you said... yes! By definition if a subspace is "spanned" by some vectors, say $W=Span(v_1,cdots,v_n)$, then $dim (W)$ is the smallest number of non-zero vectors that we can get from $v_1,cdots, v_n$ and the Span still be $W$. Knowing that some $v_i$ is not zero you won't have all $W$ if you delet all $v_n$
    – Robson
    Nov 23 at 22:21








  • 1




    Now I get it! Thank you!
    – ShellRox
    Nov 23 at 23:48



















1














You're trying to use the Rank-Nullity Theorem.
$$ r + n = text{dim of the domain}$$.



You know 2 $ge$ n $ge$ 1, since A $in$ Ket(T). Thus 0 $le$ r $le$ 1.






share|cite|improve this answer





















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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

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    active

    oldest

    votes









    1














    Your approach is correct!



    P1) $dim(Im F)=0 implies Im(F)={0}$, because the image of a linear function is a subspace and hence $0$ is in it, and can't have anything else because its dimension is zero. So $F(x)=0 forall x$



    P2) we have $dim(Ker F)=1$, applying the theorem you get $dim(Im T)=1$ and you can use the fact that two vector spaces are isomorphic (they are "the same space") if their dimension are equal, hence you can say that $Im(T)cong mathbb{R}$ which is a very nice way to justify that "$Im(T)$ is a straight line".



    P3) can't be the case that $dim(Ker T)=0$ because this would implie $Ker(T)={0}$, but we know that $Anot=0$ and $Ain Ker(T)$



    Your answer is good too! But it seems like it need to be more "direct" in a way... but the question isn't too direct either... I assumed that "being a straight line" is the same that "have dimension one"... but justifying that dimension one implies being isomorphic to the reals is also a good argument (because they are often called THE line).






    share|cite|improve this answer

















    • 1




      Thank you! So since two subspaces can be associated with bijective mapping (injective and surjective) they are isomorphic in this case. I have one last question, on my possibility 3 explanation, was I correct when I said that kernel not being "spanned" by only zero vectors makes it non trivial and higher dimensional?
      – ShellRox
      Nov 23 at 22:11






    • 1




      If I got what you said... yes! By definition if a subspace is "spanned" by some vectors, say $W=Span(v_1,cdots,v_n)$, then $dim (W)$ is the smallest number of non-zero vectors that we can get from $v_1,cdots, v_n$ and the Span still be $W$. Knowing that some $v_i$ is not zero you won't have all $W$ if you delet all $v_n$
      – Robson
      Nov 23 at 22:21








    • 1




      Now I get it! Thank you!
      – ShellRox
      Nov 23 at 23:48
















    1














    Your approach is correct!



    P1) $dim(Im F)=0 implies Im(F)={0}$, because the image of a linear function is a subspace and hence $0$ is in it, and can't have anything else because its dimension is zero. So $F(x)=0 forall x$



    P2) we have $dim(Ker F)=1$, applying the theorem you get $dim(Im T)=1$ and you can use the fact that two vector spaces are isomorphic (they are "the same space") if their dimension are equal, hence you can say that $Im(T)cong mathbb{R}$ which is a very nice way to justify that "$Im(T)$ is a straight line".



    P3) can't be the case that $dim(Ker T)=0$ because this would implie $Ker(T)={0}$, but we know that $Anot=0$ and $Ain Ker(T)$



    Your answer is good too! But it seems like it need to be more "direct" in a way... but the question isn't too direct either... I assumed that "being a straight line" is the same that "have dimension one"... but justifying that dimension one implies being isomorphic to the reals is also a good argument (because they are often called THE line).






    share|cite|improve this answer

















    • 1




      Thank you! So since two subspaces can be associated with bijective mapping (injective and surjective) they are isomorphic in this case. I have one last question, on my possibility 3 explanation, was I correct when I said that kernel not being "spanned" by only zero vectors makes it non trivial and higher dimensional?
      – ShellRox
      Nov 23 at 22:11






    • 1




      If I got what you said... yes! By definition if a subspace is "spanned" by some vectors, say $W=Span(v_1,cdots,v_n)$, then $dim (W)$ is the smallest number of non-zero vectors that we can get from $v_1,cdots, v_n$ and the Span still be $W$. Knowing that some $v_i$ is not zero you won't have all $W$ if you delet all $v_n$
      – Robson
      Nov 23 at 22:21








    • 1




      Now I get it! Thank you!
      – ShellRox
      Nov 23 at 23:48














    1












    1








    1






    Your approach is correct!



    P1) $dim(Im F)=0 implies Im(F)={0}$, because the image of a linear function is a subspace and hence $0$ is in it, and can't have anything else because its dimension is zero. So $F(x)=0 forall x$



    P2) we have $dim(Ker F)=1$, applying the theorem you get $dim(Im T)=1$ and you can use the fact that two vector spaces are isomorphic (they are "the same space") if their dimension are equal, hence you can say that $Im(T)cong mathbb{R}$ which is a very nice way to justify that "$Im(T)$ is a straight line".



    P3) can't be the case that $dim(Ker T)=0$ because this would implie $Ker(T)={0}$, but we know that $Anot=0$ and $Ain Ker(T)$



    Your answer is good too! But it seems like it need to be more "direct" in a way... but the question isn't too direct either... I assumed that "being a straight line" is the same that "have dimension one"... but justifying that dimension one implies being isomorphic to the reals is also a good argument (because they are often called THE line).






    share|cite|improve this answer












    Your approach is correct!



    P1) $dim(Im F)=0 implies Im(F)={0}$, because the image of a linear function is a subspace and hence $0$ is in it, and can't have anything else because its dimension is zero. So $F(x)=0 forall x$



    P2) we have $dim(Ker F)=1$, applying the theorem you get $dim(Im T)=1$ and you can use the fact that two vector spaces are isomorphic (they are "the same space") if their dimension are equal, hence you can say that $Im(T)cong mathbb{R}$ which is a very nice way to justify that "$Im(T)$ is a straight line".



    P3) can't be the case that $dim(Ker T)=0$ because this would implie $Ker(T)={0}$, but we know that $Anot=0$ and $Ain Ker(T)$



    Your answer is good too! But it seems like it need to be more "direct" in a way... but the question isn't too direct either... I assumed that "being a straight line" is the same that "have dimension one"... but justifying that dimension one implies being isomorphic to the reals is also a good argument (because they are often called THE line).







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 23 at 22:04









    Robson

    771221




    771221








    • 1




      Thank you! So since two subspaces can be associated with bijective mapping (injective and surjective) they are isomorphic in this case. I have one last question, on my possibility 3 explanation, was I correct when I said that kernel not being "spanned" by only zero vectors makes it non trivial and higher dimensional?
      – ShellRox
      Nov 23 at 22:11






    • 1




      If I got what you said... yes! By definition if a subspace is "spanned" by some vectors, say $W=Span(v_1,cdots,v_n)$, then $dim (W)$ is the smallest number of non-zero vectors that we can get from $v_1,cdots, v_n$ and the Span still be $W$. Knowing that some $v_i$ is not zero you won't have all $W$ if you delet all $v_n$
      – Robson
      Nov 23 at 22:21








    • 1




      Now I get it! Thank you!
      – ShellRox
      Nov 23 at 23:48














    • 1




      Thank you! So since two subspaces can be associated with bijective mapping (injective and surjective) they are isomorphic in this case. I have one last question, on my possibility 3 explanation, was I correct when I said that kernel not being "spanned" by only zero vectors makes it non trivial and higher dimensional?
      – ShellRox
      Nov 23 at 22:11






    • 1




      If I got what you said... yes! By definition if a subspace is "spanned" by some vectors, say $W=Span(v_1,cdots,v_n)$, then $dim (W)$ is the smallest number of non-zero vectors that we can get from $v_1,cdots, v_n$ and the Span still be $W$. Knowing that some $v_i$ is not zero you won't have all $W$ if you delet all $v_n$
      – Robson
      Nov 23 at 22:21








    • 1




      Now I get it! Thank you!
      – ShellRox
      Nov 23 at 23:48








    1




    1




    Thank you! So since two subspaces can be associated with bijective mapping (injective and surjective) they are isomorphic in this case. I have one last question, on my possibility 3 explanation, was I correct when I said that kernel not being "spanned" by only zero vectors makes it non trivial and higher dimensional?
    – ShellRox
    Nov 23 at 22:11




    Thank you! So since two subspaces can be associated with bijective mapping (injective and surjective) they are isomorphic in this case. I have one last question, on my possibility 3 explanation, was I correct when I said that kernel not being "spanned" by only zero vectors makes it non trivial and higher dimensional?
    – ShellRox
    Nov 23 at 22:11




    1




    1




    If I got what you said... yes! By definition if a subspace is "spanned" by some vectors, say $W=Span(v_1,cdots,v_n)$, then $dim (W)$ is the smallest number of non-zero vectors that we can get from $v_1,cdots, v_n$ and the Span still be $W$. Knowing that some $v_i$ is not zero you won't have all $W$ if you delet all $v_n$
    – Robson
    Nov 23 at 22:21






    If I got what you said... yes! By definition if a subspace is "spanned" by some vectors, say $W=Span(v_1,cdots,v_n)$, then $dim (W)$ is the smallest number of non-zero vectors that we can get from $v_1,cdots, v_n$ and the Span still be $W$. Knowing that some $v_i$ is not zero you won't have all $W$ if you delet all $v_n$
    – Robson
    Nov 23 at 22:21






    1




    1




    Now I get it! Thank you!
    – ShellRox
    Nov 23 at 23:48




    Now I get it! Thank you!
    – ShellRox
    Nov 23 at 23:48











    1














    You're trying to use the Rank-Nullity Theorem.
    $$ r + n = text{dim of the domain}$$.



    You know 2 $ge$ n $ge$ 1, since A $in$ Ket(T). Thus 0 $le$ r $le$ 1.






    share|cite|improve this answer


























      1














      You're trying to use the Rank-Nullity Theorem.
      $$ r + n = text{dim of the domain}$$.



      You know 2 $ge$ n $ge$ 1, since A $in$ Ket(T). Thus 0 $le$ r $le$ 1.






      share|cite|improve this answer
























        1












        1








        1






        You're trying to use the Rank-Nullity Theorem.
        $$ r + n = text{dim of the domain}$$.



        You know 2 $ge$ n $ge$ 1, since A $in$ Ket(T). Thus 0 $le$ r $le$ 1.






        share|cite|improve this answer












        You're trying to use the Rank-Nullity Theorem.
        $$ r + n = text{dim of the domain}$$.



        You know 2 $ge$ n $ge$ 1, since A $in$ Ket(T). Thus 0 $le$ r $le$ 1.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 23 at 20:19









        Joel Pereira

        65119




        65119






























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