If rank$(T)le$rank$(T^3)$, then intersection of range and null space of $T$ is zero
$begingroup$
If $T:Vto V$ be linear trannsformation on a vector space $V$ with rank$(T)le$ rank$(T^3)$. Then, what can be said about the null space and range of $T$. Typically, do they have empty intersection, or, is one the subset of another, or does there exist a subspace $W$ of $V$which equals their intersection.
I know that range of $T^3$ is a subspace of $T$ and null space of $T$ is a subspace of $T^3$. But, am unable to proceed further. Any hints. Thanks beforehand.
linear-algebra linear-transformations
asked Dec 10 '18 at 12:22
vidyarthividyarthi
3,0611833
$endgroup$
add a comment |
$begingroup$
If $T:Vto V$ be linear trannsformation on a vector space $V$ with rank$(T)le$ rank$(T^3)$. Then, what can be said about the null space and range of $T$. Typically, do they have empty intersection, or, is one the subset of another, or does there exist a subspace $W$ of $V$which equals their intersection.
I know that range of $T^3$ is a subspace of $T$ and null space of $T$ is a subspace of $T^3$. But, am unable to proceed further. Any hints. Thanks beforehand.
linear-algebra linear-transformations
asked Dec 10 '18 at 12:22
vidyarthividyarthi
3,0611833
$endgroup$
1
$begingroup$
If the linear transformation has finite dimensional range, then this hypothesis says that $Im(T) = Im(T^3)$.
$endgroup$
– ncmathsadist
Dec 10 '18 at 12:33
add a comment |
$begingroup$
If $T:Vto V$ be linear trannsformation on a vector space $V$ with rank$(T)le$ rank$(T^3)$. Then, what can be said about the null space and range of $T$. Typically, do they have empty intersection, or, is one the subset of another, or does there exist a subspace $W$ of $V$which equals their intersection.
I know that range of $T^3$ is a subspace of $T$ and null space of $T$ is a subspace of $T^3$. But, am unable to proceed further. Any hints. Thanks beforehand.
linear-algebra linear-transformations
asked Dec 10 '18 at 12:22
vidyarthividyarthi
3,0611833
$endgroup$
If $T:Vto V$ be linear trannsformation on a vector space $V$ with rank$(T)le$ rank$(T^3)$. Then, what can be said about the null space and range of $T$. Typically, do they have empty intersection, or, is one the subset of another, or does there exist a subspace $W$ of $V$which equals their intersection.
I know that range of $T^3$ is a subspace of $T$ and null space of $T$ is a subspace of $T^3$. But, am unable to proceed further. Any hints. Thanks beforehand.
linear-algebra linear-transformations
linear-algebra linear-transformations
asked Dec 10 '18 at 12:22
vidyarthividyarthi
3,0611833
asked Dec 10 '18 at 12:22
vidyarthividyarthi
3,0611833
asked Dec 10 '18 at 12:22
vidyarthividyarthi
3,0611833
asked Dec 10 '18 at 12:22
vidyarthividyarthi
3,0611833
asked Dec 10 '18 at 12:22
vidyarthividyarthi
3,0611833
3,0611833
1
$begingroup$
If the linear transformation has finite dimensional range, then this hypothesis says that $Im(T) = Im(T^3)$.
$endgroup$
– ncmathsadist
Dec 10 '18 at 12:33
add a comment |
1
$begingroup$
If the linear transformation has finite dimensional range, then this hypothesis says that $Im(T) = Im(T^3)$.
$endgroup$
– ncmathsadist
Dec 10 '18 at 12:33
1
1
$begingroup$
If the linear transformation has finite dimensional range, then this hypothesis says that $Im(T) = Im(T^3)$.
$endgroup$
– ncmathsadist
Dec 10 '18 at 12:33
$begingroup$
If the linear transformation has finite dimensional range, then this hypothesis says that $Im(T) = Im(T^3)$.
$endgroup$
– ncmathsadist
Dec 10 '18 at 12:33
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
well first of all, lets consider $T$ as a linear map (this makes stuff in my opinion often a lot clearer). We know that $rank(T)=dim textrm{im}(T)$, But we know that $text{im}(T^3)=textrm{im}(Tmid_{textrm{im}(T^2)})$, in particular $textrm{im}T^3subset textrm{im} T$ but that just means in dimensions that $dim textrm{im}(T) ge dim textrm{im}(T^3) $ in particular we get with your inequality $$dim textrm{im}(T) = dim textrm{im}(T^3)$$
But this just means that $Tmid_{textrm{im}(T)}$ is an isomorphism. So $Tmid_{textrm{im}(T)}$ has kernel zero. But this immediately gives $$0=ker(Tmid_{textrm{im}(T)}) = ker(T)cap textrm{im}{T}$$
answered Dec 10 '18 at 12:33
EnkiduEnkidu
1,44419
$endgroup$
$begingroup$
how does ranks being equal mean that the transformation is an isomorphism?
$endgroup$
– vidyarthi
Dec 10 '18 at 13:22
1
$begingroup$
first of all, it is a linear map, second, it is not a rank, it is the dimension (as we are talking about spaces and not maps). Well, if the dimensions are equal, then the cardinality of your basis are the same. In particular (since the image of a basis is a generating set of the image) any image of a basis is linearly independent. but that just means that it is injective. now we get that the image is a subspace of another vector space with same finite dimension, hence they have to coincide, which gives surjectivity.
$endgroup$
– Enkidu
Dec 10 '18 at 13:26
$begingroup$
good explanation! so, if at any time, we have rank$(T)=rank(T^n)$, does that mean that the map is an isomorphism?
$endgroup$
– vidyarthi
Dec 10 '18 at 13:33
$begingroup$
No! it means that it is an isomorphism if you restrict it to the image. a good example to keep in mind is a projection onto a subspace. But in general, maps like this will be the projection onto a subspace followed by a isomorphism on the subspace. a good exapmle is: $begin{pmatrix}0&0&0\0 & 0 & 1 \ 0 & 1 & 0end{pmatrix}$
$endgroup$
– Enkidu
Dec 10 '18 at 13:36
add a comment |
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1 Answer
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1 Answer
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$begingroup$
well first of all, lets consider $T$ as a linear map (this makes stuff in my opinion often a lot clearer). We know that $rank(T)=dim textrm{im}(T)$, But we know that $text{im}(T^3)=textrm{im}(Tmid_{textrm{im}(T^2)})$, in particular $textrm{im}T^3subset textrm{im} T$ but that just means in dimensions that $dim textrm{im}(T) ge dim textrm{im}(T^3) $ in particular we get with your inequality $$dim textrm{im}(T) = dim textrm{im}(T^3)$$
But this just means that $Tmid_{textrm{im}(T)}$ is an isomorphism. So $Tmid_{textrm{im}(T)}$ has kernel zero. But this immediately gives $$0=ker(Tmid_{textrm{im}(T)}) = ker(T)cap textrm{im}{T}$$
answered Dec 10 '18 at 12:33
EnkiduEnkidu
1,44419
$endgroup$
$begingroup$
how does ranks being equal mean that the transformation is an isomorphism?
$endgroup$
– vidyarthi
Dec 10 '18 at 13:22
1
$begingroup$
first of all, it is a linear map, second, it is not a rank, it is the dimension (as we are talking about spaces and not maps). Well, if the dimensions are equal, then the cardinality of your basis are the same. In particular (since the image of a basis is a generating set of the image) any image of a basis is linearly independent. but that just means that it is injective. now we get that the image is a subspace of another vector space with same finite dimension, hence they have to coincide, which gives surjectivity.
$endgroup$
– Enkidu
Dec 10 '18 at 13:26
$begingroup$
good explanation! so, if at any time, we have rank$(T)=rank(T^n)$, does that mean that the map is an isomorphism?
$endgroup$
– vidyarthi
Dec 10 '18 at 13:33
$begingroup$
No! it means that it is an isomorphism if you restrict it to the image. a good example to keep in mind is a projection onto a subspace. But in general, maps like this will be the projection onto a subspace followed by a isomorphism on the subspace. a good exapmle is: $begin{pmatrix}0&0&0\0 & 0 & 1 \ 0 & 1 & 0end{pmatrix}$
$endgroup$
– Enkidu
Dec 10 '18 at 13:36
add a comment |
$begingroup$
well first of all, lets consider $T$ as a linear map (this makes stuff in my opinion often a lot clearer). We know that $rank(T)=dim textrm{im}(T)$, But we know that $text{im}(T^3)=textrm{im}(Tmid_{textrm{im}(T^2)})$, in particular $textrm{im}T^3subset textrm{im} T$ but that just means in dimensions that $dim textrm{im}(T) ge dim textrm{im}(T^3) $ in particular we get with your inequality $$dim textrm{im}(T) = dim textrm{im}(T^3)$$
But this just means that $Tmid_{textrm{im}(T)}$ is an isomorphism. So $Tmid_{textrm{im}(T)}$ has kernel zero. But this immediately gives $$0=ker(Tmid_{textrm{im}(T)}) = ker(T)cap textrm{im}{T}$$
answered Dec 10 '18 at 12:33
EnkiduEnkidu
1,44419
$endgroup$
$begingroup$
how does ranks being equal mean that the transformation is an isomorphism?
$endgroup$
– vidyarthi
Dec 10 '18 at 13:22
1
$begingroup$
first of all, it is a linear map, second, it is not a rank, it is the dimension (as we are talking about spaces and not maps). Well, if the dimensions are equal, then the cardinality of your basis are the same. In particular (since the image of a basis is a generating set of the image) any image of a basis is linearly independent. but that just means that it is injective. now we get that the image is a subspace of another vector space with same finite dimension, hence they have to coincide, which gives surjectivity.
$endgroup$
– Enkidu
Dec 10 '18 at 13:26
$begingroup$
good explanation! so, if at any time, we have rank$(T)=rank(T^n)$, does that mean that the map is an isomorphism?
$endgroup$
– vidyarthi
Dec 10 '18 at 13:33
$begingroup$
No! it means that it is an isomorphism if you restrict it to the image. a good example to keep in mind is a projection onto a subspace. But in general, maps like this will be the projection onto a subspace followed by a isomorphism on the subspace. a good exapmle is: $begin{pmatrix}0&0&0\0 & 0 & 1 \ 0 & 1 & 0end{pmatrix}$
$endgroup$
– Enkidu
Dec 10 '18 at 13:36
add a comment |
$begingroup$
well first of all, lets consider $T$ as a linear map (this makes stuff in my opinion often a lot clearer). We know that $rank(T)=dim textrm{im}(T)$, But we know that $text{im}(T^3)=textrm{im}(Tmid_{textrm{im}(T^2)})$, in particular $textrm{im}T^3subset textrm{im} T$ but that just means in dimensions that $dim textrm{im}(T) ge dim textrm{im}(T^3) $ in particular we get with your inequality $$dim textrm{im}(T) = dim textrm{im}(T^3)$$
But this just means that $Tmid_{textrm{im}(T)}$ is an isomorphism. So $Tmid_{textrm{im}(T)}$ has kernel zero. But this immediately gives $$0=ker(Tmid_{textrm{im}(T)}) = ker(T)cap textrm{im}{T}$$
answered Dec 10 '18 at 12:33
EnkiduEnkidu
1,44419
$endgroup$
well first of all, lets consider $T$ as a linear map (this makes stuff in my opinion often a lot clearer). We know that $rank(T)=dim textrm{im}(T)$, But we know that $text{im}(T^3)=textrm{im}(Tmid_{textrm{im}(T^2)})$, in particular $textrm{im}T^3subset textrm{im} T$ but that just means in dimensions that $dim textrm{im}(T) ge dim textrm{im}(T^3) $ in particular we get with your inequality $$dim textrm{im}(T) = dim textrm{im}(T^3)$$
But this just means that $Tmid_{textrm{im}(T)}$ is an isomorphism. So $Tmid_{textrm{im}(T)}$ has kernel zero. But this immediately gives $$0=ker(Tmid_{textrm{im}(T)}) = ker(T)cap textrm{im}{T}$$
answered Dec 10 '18 at 12:33
EnkiduEnkidu
1,44419
answered Dec 10 '18 at 12:33
EnkiduEnkidu
1,44419
answered Dec 10 '18 at 12:33
EnkiduEnkidu
1,44419
answered Dec 10 '18 at 12:33
EnkiduEnkidu
1,44419
1,44419
$begingroup$
how does ranks being equal mean that the transformation is an isomorphism?
$endgroup$
– vidyarthi
Dec 10 '18 at 13:22
1
$begingroup$
first of all, it is a linear map, second, it is not a rank, it is the dimension (as we are talking about spaces and not maps). Well, if the dimensions are equal, then the cardinality of your basis are the same. In particular (since the image of a basis is a generating set of the image) any image of a basis is linearly independent. but that just means that it is injective. now we get that the image is a subspace of another vector space with same finite dimension, hence they have to coincide, which gives surjectivity.
$endgroup$
– Enkidu
Dec 10 '18 at 13:26
$begingroup$
good explanation! so, if at any time, we have rank$(T)=rank(T^n)$, does that mean that the map is an isomorphism?
$endgroup$
– vidyarthi
Dec 10 '18 at 13:33
$begingroup$
No! it means that it is an isomorphism if you restrict it to the image. a good example to keep in mind is a projection onto a subspace. But in general, maps like this will be the projection onto a subspace followed by a isomorphism on the subspace. a good exapmle is: $begin{pmatrix}0&0&0\0 & 0 & 1 \ 0 & 1 & 0end{pmatrix}$
$endgroup$
– Enkidu
Dec 10 '18 at 13:36
add a comment |
$begingroup$
how does ranks being equal mean that the transformation is an isomorphism?
$endgroup$
– vidyarthi
Dec 10 '18 at 13:22
1
$begingroup$
first of all, it is a linear map, second, it is not a rank, it is the dimension (as we are talking about spaces and not maps). Well, if the dimensions are equal, then the cardinality of your basis are the same. In particular (since the image of a basis is a generating set of the image) any image of a basis is linearly independent. but that just means that it is injective. now we get that the image is a subspace of another vector space with same finite dimension, hence they have to coincide, which gives surjectivity.
$endgroup$
– Enkidu
Dec 10 '18 at 13:26
$begingroup$
good explanation! so, if at any time, we have rank$(T)=rank(T^n)$, does that mean that the map is an isomorphism?
$endgroup$
– vidyarthi
Dec 10 '18 at 13:33
$begingroup$
No! it means that it is an isomorphism if you restrict it to the image. a good example to keep in mind is a projection onto a subspace. But in general, maps like this will be the projection onto a subspace followed by a isomorphism on the subspace. a good exapmle is: $begin{pmatrix}0&0&0\0 & 0 & 1 \ 0 & 1 & 0end{pmatrix}$
$endgroup$
– Enkidu
Dec 10 '18 at 13:36
$begingroup$
how does ranks being equal mean that the transformation is an isomorphism?
$endgroup$
– vidyarthi
Dec 10 '18 at 13:22
$begingroup$
how does ranks being equal mean that the transformation is an isomorphism?
$endgroup$
– vidyarthi
Dec 10 '18 at 13:22
1
1
$begingroup$
first of all, it is a linear map, second, it is not a rank, it is the dimension (as we are talking about spaces and not maps). Well, if the dimensions are equal, then the cardinality of your basis are the same. In particular (since the image of a basis is a generating set of the image) any image of a basis is linearly independent. but that just means that it is injective. now we get that the image is a subspace of another vector space with same finite dimension, hence they have to coincide, which gives surjectivity.
$endgroup$
– Enkidu
Dec 10 '18 at 13:26
$begingroup$
first of all, it is a linear map, second, it is not a rank, it is the dimension (as we are talking about spaces and not maps). Well, if the dimensions are equal, then the cardinality of your basis are the same. In particular (since the image of a basis is a generating set of the image) any image of a basis is linearly independent. but that just means that it is injective. now we get that the image is a subspace of another vector space with same finite dimension, hence they have to coincide, which gives surjectivity.
$endgroup$
– Enkidu
Dec 10 '18 at 13:26
$begingroup$
good explanation! so, if at any time, we have rank$(T)=rank(T^n)$, does that mean that the map is an isomorphism?
$endgroup$
– vidyarthi
Dec 10 '18 at 13:33
$begingroup$
good explanation! so, if at any time, we have rank$(T)=rank(T^n)$, does that mean that the map is an isomorphism?
$endgroup$
– vidyarthi
Dec 10 '18 at 13:33
$begingroup$
No! it means that it is an isomorphism if you restrict it to the image. a good example to keep in mind is a projection onto a subspace. But in general, maps like this will be the projection onto a subspace followed by a isomorphism on the subspace. a good exapmle is: $begin{pmatrix}0&0&0\0 & 0 & 1 \ 0 & 1 & 0end{pmatrix}$
$endgroup$
– Enkidu
Dec 10 '18 at 13:36
$begingroup$
No! it means that it is an isomorphism if you restrict it to the image. a good example to keep in mind is a projection onto a subspace. But in general, maps like this will be the projection onto a subspace followed by a isomorphism on the subspace. a good exapmle is: $begin{pmatrix}0&0&0\0 & 0 & 1 \ 0 & 1 & 0end{pmatrix}$
$endgroup$
– Enkidu
Dec 10 '18 at 13:36
add a comment |
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$begingroup$
If the linear transformation has finite dimensional range, then this hypothesis says that $Im(T) = Im(T^3)$.
$endgroup$
– ncmathsadist
Dec 10 '18 at 12:33