Find $k$,$m$,$n$ such that $f$ is an epimorphism/monomorphism
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
asked Jan 4 at 15:35
VirtualUserVirtualUser
54712
add a comment |
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
asked Jan 4 at 15:35
VirtualUserVirtualUser
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
add a comment |
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
asked Jan 4 at 15:35
VirtualUserVirtualUser
54712
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
asked Jan 4 at 15:35
VirtualUserVirtualUser
54712
asked Jan 4 at 15:35
VirtualUserVirtualUser
54712
asked Jan 4 at 15:35
VirtualUserVirtualUser
54712
asked Jan 4 at 15:35
VirtualUserVirtualUser
54712
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
add a comment |
$(b)$ 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
add a comment |
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 $dim operatorname{im}(f) = n-k$.
answered Jan 4 at 15:44
OmnomnomnomOmnomnomnom
127k788176
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 $dim operatorname{im}(f) = n-k$.
answered Jan 4 at 15:44
OmnomnomnomOmnomnomnom
127k788176
add a comment |
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 $dim operatorname{im}(f) = n-k$.
answered Jan 4 at 15:44
OmnomnomnomOmnomnomnom
127k788176
add a comment |
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 $dim operatorname{im}(f) = n-k$.
answered Jan 4 at 15:44
OmnomnomnomOmnomnomnom
127k788176
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 $dim operatorname{im}(f) = n-k$.
answered Jan 4 at 15:44
OmnomnomnomOmnomnomnom
127k788176
answered Jan 4 at 15:44
OmnomnomnomOmnomnomnom
127k788176
answered Jan 4 at 15:44
OmnomnomnomOmnomnomnom
127k788176
answered Jan 4 at 15:44
OmnomnomnomOmnomnomnom
127k788176
127k788176
add a comment |
add a comment |
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$(b)$ 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