If rank$(T)le$rank$(T^3)$, then intersection of range and null space of $T$ is zero












0












$begingroup$


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







share|cite|improve this question









$endgroup$








  • 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
















0












$begingroup$


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







share|cite|improve this question









$endgroup$








  • 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














0












0








0


2



$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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














  • 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










1 Answer
1






active

oldest

votes


















2












$begingroup$

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 $$0=ker(Tmid_{textrm{im}(T)}) = ker(T)cap textrm{im}{T}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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













Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3033844%2fif-rankt-lerankt3-then-intersection-of-range-and-null-space-of-t-is%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

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 $$0=ker(Tmid_{textrm{im}(T)}) = ker(T)cap textrm{im}{T}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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


















2












$begingroup$

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 $$0=ker(Tmid_{textrm{im}(T)}) = ker(T)cap textrm{im}{T}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















2












2








2





$begingroup$

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 $$0=ker(Tmid_{textrm{im}(T)}) = ker(T)cap textrm{im}{T}$$






share|cite|improve this answer









$endgroup$



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 $$0=ker(Tmid_{textrm{im}(T)}) = ker(T)cap textrm{im}{T}$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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




















  • $begingroup$
    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




















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3033844%2fif-rankt-lerankt3-then-intersection-of-range-and-null-space-of-t-is%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

mysqli_query(): Empty query in /home/lucindabrummitt/public_html/blog/wp-includes/wp-db.php on line 1924

Image
0 I just uploaded a new plugin to my wordpress site and I got this warning: mysqli_query(): Empty query in /home/lucindabrummitt/public_html/blog/wp-includes/wp-db.php on line 1924 Usually if I have a problem, I can find answers on google but this one seems pretty particular and I'm not finding any info on it except for this forum. I'm not terribly techie but if someone can tell me how to read this and how to change it I can follow directions pretty well. Thanks All! PS If I don't have the proper tags here could someone add the appropriate tags? php mysql wordpress share | improve this question asked Nov 16 '18 at 20:41 ...
Read more

How to change which sound is reproduced for terminal bell?

Image
3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question ...
Read more

Can I use Tabulator js library in my java Spring + Thymeleaf project?

Image
0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"...
Read more