Find $k$,$m$,$n$ such that $f$ is an epimorphism/monomorphism












1














I am trying to solve this task:
$f in L(X,Y)$, $dim X = n$, $dim Y = m$, $dim ker(f) = k.$

Find $k$,$m$,$n$ such that:

a) $f$ is an monomorphism

b) $f$ is an epimorphism



b) If $f$ is an epimorphism, we must be able to get any element from $Y$ by $f$. So $dim y = dim im(f)$
$$ m = n - k $$
have I right?

a) But there I am not sure about my idea - I think that it should be (intuitively) $dim X = dim im(f) $ so I get:
$$ n = n - k $$
$$ k = 0 $$
but firstly I am not sure if that is true, secondly it is only my intuitive try...






linear-algebra







share|cite|improve this question






















  • $(b)$ is correct. What is a monomorphism? I don't see monomorphism anywhere online.
    – Shubham Johri
    Jan 4 at 15:45












  • @ShubhamJohri The terms are defined in a general sense in category theory, but the exact definition one works with depends on the context. For linear maps, $f$ is a monomorphism if and only if it is injective (one-to-one), and $f$ is an epimorphism if and only if it is surjective (onto).
    – Omnomnomnom
    Jan 4 at 15:47












  • @Omnomnomnom Thanks for the definition.
    – Shubham Johri
    Jan 4 at 15:49
















1














I am trying to solve this task:
$f in L(X,Y)$, $dim X = n$, $dim Y = m$, $dim ker(f) = k.$

Find $k$,$m$,$n$ such that:

a) $f$ is an monomorphism

b) $f$ is an epimorphism



b) If $f$ is an epimorphism, we must be able to get any element from $Y$ by $f$. So $dim y = dim im(f)$
$$ m = n - k $$
have I right?

a) But there I am not sure about my idea - I think that it should be (intuitively) $dim X = dim im(f) $ so I get:
$$ n = n - k $$
$$ k = 0 $$
but firstly I am not sure if that is true, secondly it is only my intuitive try...






linear-algebra







share|cite|improve this question






















  • $(b)$ is correct. What is a monomorphism? I don't see monomorphism anywhere online.
    – Shubham Johri
    Jan 4 at 15:45












  • @ShubhamJohri The terms are defined in a general sense in category theory, but the exact definition one works with depends on the context. For linear maps, $f$ is a monomorphism if and only if it is injective (one-to-one), and $f$ is an epimorphism if and only if it is surjective (onto).
    – Omnomnomnom
    Jan 4 at 15:47












  • @Omnomnomnom Thanks for the definition.
    – Shubham Johri
    Jan 4 at 15:49














1












1








1







I am trying to solve this task:
$f in L(X,Y)$, $dim X = n$, $dim Y = m$, $dim ker(f) = k.$

Find $k$,$m$,$n$ such that:

a) $f$ is an monomorphism

b) $f$ is an epimorphism



b) If $f$ is an epimorphism, we must be able to get any element from $Y$ by $f$. So $dim y = dim im(f)$
$$ m = n - k $$
have I right?

a) But there I am not sure about my idea - I think that it should be (intuitively) $dim X = dim im(f) $ so I get:
$$ n = n - k $$
$$ k = 0 $$
but firstly I am not sure if that is true, secondly it is only my intuitive try...






linear-algebra







share|cite|improve this question













I am trying to solve this task:
$f in L(X,Y)$, $dim X = n$, $dim Y = m$, $dim ker(f) = k.$

Find $k$,$m$,$n$ such that:

a) $f$ is an monomorphism

b) $f$ is an epimorphism



b) If $f$ is an epimorphism, we must be able to get any element from $Y$ by $f$. So $dim y = dim im(f)$
$$ m = n - k $$
have I right?

a) But there I am not sure about my idea - I think that it should be (intuitively) $dim X = dim im(f) $ so I get:
$$ n = n - k $$
$$ k = 0 $$
but firstly I am not sure if that is true, secondly it is only my intuitive try...






linear-algebra




linear-algebra






share|cite|improve this question













share|cite|improve this question











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asked Jan 4 at 15:35









VirtualUserVirtualUser

54712




54712












  • $(b)$ is correct. What is a monomorphism? I don't see monomorphism anywhere online.
    – Shubham Johri
    Jan 4 at 15:45












  • @ShubhamJohri The terms are defined in a general sense in category theory, but the exact definition one works with depends on the context. For linear maps, $f$ is a monomorphism if and only if it is injective (one-to-one), and $f$ is an epimorphism if and only if it is surjective (onto).
    – Omnomnomnom
    Jan 4 at 15:47












  • @Omnomnomnom Thanks for the definition.
    – Shubham Johri
    Jan 4 at 15:49


















  • $(b)$ is correct. What is a monomorphism? I don't see monomorphism anywhere online.
    – Shubham Johri
    Jan 4 at 15:45












  • @ShubhamJohri The terms are defined in a general sense in category theory, but the exact definition one works with depends on the context. For linear maps, $f$ is a monomorphism if and only if it is injective (one-to-one), and $f$ is an epimorphism if and only if it is surjective (onto).
    – Omnomnomnom
    Jan 4 at 15:47












  • @Omnomnomnom Thanks for the definition.
    – Shubham Johri
    Jan 4 at 15:49
















$(b)$ is correct. What is a monomorphism? I don't see monomorphism anywhere online.
– Shubham Johri
Jan 4 at 15:45






$(b)$ is correct. What is a monomorphism? I don't see monomorphism anywhere online.
– Shubham Johri
Jan 4 at 15:45














@ShubhamJohri The terms are defined in a general sense in category theory, but the exact definition one works with depends on the context. For linear maps, $f$ is a monomorphism if and only if it is injective (one-to-one), and $f$ is an epimorphism if and only if it is surjective (onto).
– Omnomnomnom
Jan 4 at 15:47






@ShubhamJohri The terms are defined in a general sense in category theory, but the exact definition one works with depends on the context. For linear maps, $f$ is a monomorphism if and only if it is injective (one-to-one), and $f$ is an epimorphism if and only if it is surjective (onto).
– Omnomnomnom
Jan 4 at 15:47














@Omnomnomnom Thanks for the definition.
– Shubham Johri
Jan 4 at 15:49




@Omnomnomnom Thanks for the definition.
– Shubham Johri
Jan 4 at 15:49










1 Answer
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Your answers are correct.



The rank nullity theorem states that for linear maps $f: X to Y$, $dim ker(f) + dim operatorname{im}(f) = dim(X)$. So for this context, we have $k+ dim operatorname{im}(f) = n$.



For a: whatever your definition of a monomorphism, we should be able to conclude that $f$ is a monomorphism if and only if $ker(f) = {0}$, which is to say that $k=0$. The rank-nullity theorem confirms your intuition.



For b: whatever your definition of an epimorphism, we should be able to conclude that $f$ is an epimorphism if and only if $operatorname{im}(f) = Y$. So, $dim operatorname{im}(f) = dim Y$, and by the rank-nullity theorem $dim operatorname{im}(f) = n-k$.






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    Your answers are correct.



    The rank nullity theorem states that for linear maps $f: X to Y$, $dim ker(f) + dim operatorname{im}(f) = dim(X)$. So for this context, we have $k+ dim operatorname{im}(f) = n$.



    For a: whatever your definition of a monomorphism, we should be able to conclude that $f$ is a monomorphism if and only if $ker(f) = {0}$, which is to say that $k=0$. The rank-nullity theorem confirms your intuition.



    For b: whatever your definition of an epimorphism, we should be able to conclude that $f$ is an epimorphism if and only if $operatorname{im}(f) = Y$. So, $dim operatorname{im}(f) = dim Y$, and by the rank-nullity theorem $dim operatorname{im}(f) = n-k$.






    share|cite|improve this answer


























      3














      Your answers are correct.



      The rank nullity theorem states that for linear maps $f: X to Y$, $dim ker(f) + dim operatorname{im}(f) = dim(X)$. So for this context, we have $k+ dim operatorname{im}(f) = n$.



      For a: whatever your definition of a monomorphism, we should be able to conclude that $f$ is a monomorphism if and only if $ker(f) = {0}$, which is to say that $k=0$. The rank-nullity theorem confirms your intuition.



      For b: whatever your definition of an epimorphism, we should be able to conclude that $f$ is an epimorphism if and only if $operatorname{im}(f) = Y$. So, $dim operatorname{im}(f) = dim Y$, and by the rank-nullity theorem $dim operatorname{im}(f) = n-k$.






      share|cite|improve this answer
























        3












        3








        3






        Your answers are correct.



        The rank nullity theorem states that for linear maps $f: X to Y$, $dim ker(f) + dim operatorname{im}(f) = dim(X)$. So for this context, we have $k+ dim operatorname{im}(f) = n$.



        For a: whatever your definition of a monomorphism, we should be able to conclude that $f$ is a monomorphism if and only if $ker(f) = {0}$, which is to say that $k=0$. The rank-nullity theorem confirms your intuition.



        For b: whatever your definition of an epimorphism, we should be able to conclude that $f$ is an epimorphism if and only if $operatorname{im}(f) = Y$. So, $dim operatorname{im}(f) = dim Y$, and by the rank-nullity theorem $dim operatorname{im}(f) = n-k$.






        share|cite|improve this answer












        Your answers are correct.



        The rank nullity theorem states that for linear maps $f: X to Y$, $dim ker(f) + dim operatorname{im}(f) = dim(X)$. So for this context, we have $k+ dim operatorname{im}(f) = n$.



        For a: whatever your definition of a monomorphism, we should be able to conclude that $f$ is a monomorphism if and only if $ker(f) = {0}$, which is to say that $k=0$. The rank-nullity theorem confirms your intuition.



        For b: whatever your definition of an epimorphism, we should be able to conclude that $f$ is an epimorphism if and only if $operatorname{im}(f) = Y$. So, $dim operatorname{im}(f) = dim Y$, and by the rank-nullity theorem $dim operatorname{im}(f) = n-k$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 4 at 15:44









        OmnomnomnomOmnomnomnom

        127k788176




        127k788176






























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