Mixing
If $X = Y + operatorname{im}(F)$, prove that $dim (X) le 2 cdot dim (Y)$
$begingroup$
I am considering this task
$X$ is linear space, $dim X < infty$, $Y subset X$ is linear
subspace and $X = Y + operatorname{im}(F)$ for some linear
transformation $F in L(Y,X)$. Prove that $dim X le 2 cdot dim Y$
and there is equality iff $F$ is monomorphism and $Y cap
> operatorname{im}(F) = {0}$.
my try
I think that the best option is use this formula:
$$ dim (operatorname{im}(F)) + dim Y = dim (Y + operatorname{im}(F)) + dim (Y cap operatorname{im}(F)) $$
but we know that
$$ X = Y + operatorname{im}(F) $$
so
$$ dim (operatorname{im}(F)) + dim Y = dim X + dim (Y cap operatorname{im}(F)) $$
Moreover
$$ dim (Y cap operatorname{im}(F)) le dim (operatorname{im}(F))$$
so
$$ dim Y le dim X $$
but unfortunately it is trivial because it comes from
$$ X = Y + operatorname{im}(F) $$
I am trying on different ways but in each case I finish at something like that.
This task has been added to site a long time ago but I want to use this formula (only if it's possible) and I want to show my way of thinking there.
linear-algebra dimension-theory
asked Jan 6 at 9:59
VirtualUserVirtualUser
68912
$endgroup$
add a comment |
$begingroup$
I am considering this task
$X$ is linear space, $dim X < infty$, $Y subset X$ is linear
subspace and $X = Y + operatorname{im}(F)$ for some linear
transformation $F in L(Y,X)$. Prove that $dim X le 2 cdot dim Y$
and there is equality iff $F$ is monomorphism and $Y cap
> operatorname{im}(F) = {0}$.
my try
I think that the best option is use this formula:
$$ dim (operatorname{im}(F)) + dim Y = dim (Y + operatorname{im}(F)) + dim (Y cap operatorname{im}(F)) $$
but we know that
$$ X = Y + operatorname{im}(F) $$
so
$$ dim (operatorname{im}(F)) + dim Y = dim X + dim (Y cap operatorname{im}(F)) $$
Moreover
$$ dim (Y cap operatorname{im}(F)) le dim (operatorname{im}(F))$$
so
$$ dim Y le dim X $$
but unfortunately it is trivial because it comes from
$$ X = Y + operatorname{im}(F) $$
I am trying on different ways but in each case I finish at something like that.
This task has been added to site a long time ago but I want to use this formula (only if it's possible) and I want to show my way of thinking there.
linear-algebra dimension-theory
asked Jan 6 at 9:59
VirtualUserVirtualUser
68912
$endgroup$
add a comment |
$begingroup$
I am considering this task
$X$ is linear space, $dim X < infty$, $Y subset X$ is linear
subspace and $X = Y + operatorname{im}(F)$ for some linear
transformation $F in L(Y,X)$. Prove that $dim X le 2 cdot dim Y$
and there is equality iff $F$ is monomorphism and $Y cap
> operatorname{im}(F) = {0}$.
my try
I think that the best option is use this formula:
$$ dim (operatorname{im}(F)) + dim Y = dim (Y + operatorname{im}(F)) + dim (Y cap operatorname{im}(F)) $$
but we know that
$$ X = Y + operatorname{im}(F) $$
so
$$ dim (operatorname{im}(F)) + dim Y = dim X + dim (Y cap operatorname{im}(F)) $$
Moreover
$$ dim (Y cap operatorname{im}(F)) le dim (operatorname{im}(F))$$
so
$$ dim Y le dim X $$
but unfortunately it is trivial because it comes from
$$ X = Y + operatorname{im}(F) $$
I am trying on different ways but in each case I finish at something like that.
This task has been added to site a long time ago but I want to use this formula (only if it's possible) and I want to show my way of thinking there.
linear-algebra dimension-theory
asked Jan 6 at 9:59
VirtualUserVirtualUser
68912
$endgroup$
I am considering this task
$X$ is linear space, $dim X < infty$, $Y subset X$ is linear
subspace and $X = Y + operatorname{im}(F)$ for some linear
transformation $F in L(Y,X)$. Prove that $dim X le 2 cdot dim Y$
and there is equality iff $F$ is monomorphism and $Y cap
> operatorname{im}(F) = {0}$.
my try
I think that the best option is use this formula:
$$ dim (operatorname{im}(F)) + dim Y = dim (Y + operatorname{im}(F)) + dim (Y cap operatorname{im}(F)) $$
but we know that
$$ X = Y + operatorname{im}(F) $$
so
$$ dim (operatorname{im}(F)) + dim Y = dim X + dim (Y cap operatorname{im}(F)) $$
Moreover
$$ dim (Y cap operatorname{im}(F)) le dim (operatorname{im}(F))$$
so
$$ dim Y le dim X $$
but unfortunately it is trivial because it comes from
$$ X = Y + operatorname{im}(F) $$
I am trying on different ways but in each case I finish at something like that.
This task has been added to site a long time ago but I want to use this formula (only if it's possible) and I want to show my way of thinking there.
linear-algebra dimension-theory
linear-algebra dimension-theory
asked Jan 6 at 9:59
VirtualUserVirtualUser
68912
asked Jan 6 at 9:59
VirtualUserVirtualUser
68912
edited Jan 6 at 23:15
VirtualUser
asked Jan 6 at 9:59
VirtualUserVirtualUser
68912
asked Jan 6 at 9:59
VirtualUserVirtualUser
68912
asked Jan 6 at 9:59
VirtualUserVirtualUser
68912
68912
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint: remember that, given a linear transformation $F:Yrightarrow X$ between finite dimensional linear spaces, then you have the rank-nullity theorem:
$$text{dim}(Y) = text{dim}(ker(F)) +text{dim}(text{im}(F))$$
which yields two consequences:
$text{dim(im}(F))leq text{dim}(Y)$;
$text{dim(im}(F)) = text{dim}(Y)$ if and only if $text{dim}(ker(F))=0$, i.e. if and only if $F$ is a monomorphism.
Item 1 helps you with proving $text{dim}(X)leq 2cdot text{dim}(Y)$ while item 2 is needed for the second part of your question.
answered Jan 6 at 15:31
F.BattistoniF.Battistoni
515
$endgroup$
add a comment |
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
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3063669%2fif-x-y-operatornameimf-prove-that-dim-x-le-2-cdot-dim-y%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
$begingroup$
Hint: remember that, given a linear transformation $F:Yrightarrow X$ between finite dimensional linear spaces, then you have the rank-nullity theorem:
$$text{dim}(Y) = text{dim}(ker(F)) +text{dim}(text{im}(F))$$
which yields two consequences:
$text{dim(im}(F))leq text{dim}(Y)$;
$text{dim(im}(F)) = text{dim}(Y)$ if and only if $text{dim}(ker(F))=0$, i.e. if and only if $F$ is a monomorphism.
Item 1 helps you with proving $text{dim}(X)leq 2cdot text{dim}(Y)$ while item 2 is needed for the second part of your question.
answered Jan 6 at 15:31
F.BattistoniF.Battistoni
515
$endgroup$
add a comment |
$begingroup$
Hint: remember that, given a linear transformation $F:Yrightarrow X$ between finite dimensional linear spaces, then you have the rank-nullity theorem:
$$text{dim}(Y) = text{dim}(ker(F)) +text{dim}(text{im}(F))$$
which yields two consequences:
$text{dim(im}(F))leq text{dim}(Y)$;
$text{dim(im}(F)) = text{dim}(Y)$ if and only if $text{dim}(ker(F))=0$, i.e. if and only if $F$ is a monomorphism.
Item 1 helps you with proving $text{dim}(X)leq 2cdot text{dim}(Y)$ while item 2 is needed for the second part of your question.
answered Jan 6 at 15:31
F.BattistoniF.Battistoni
515
$endgroup$
add a comment |
$begingroup$
Hint: remember that, given a linear transformation $F:Yrightarrow X$ between finite dimensional linear spaces, then you have the rank-nullity theorem:
$$text{dim}(Y) = text{dim}(ker(F)) +text{dim}(text{im}(F))$$
which yields two consequences:
$text{dim(im}(F))leq text{dim}(Y)$;
$text{dim(im}(F)) = text{dim}(Y)$ if and only if $text{dim}(ker(F))=0$, i.e. if and only if $F$ is a monomorphism.
Item 1 helps you with proving $text{dim}(X)leq 2cdot text{dim}(Y)$ while item 2 is needed for the second part of your question.
answered Jan 6 at 15:31
F.BattistoniF.Battistoni
515
$endgroup$
Hint: remember that, given a linear transformation $F:Yrightarrow X$ between finite dimensional linear spaces, then you have the rank-nullity theorem:
$$text{dim}(Y) = text{dim}(ker(F)) +text{dim}(text{im}(F))$$
which yields two consequences:
$text{dim(im}(F))leq text{dim}(Y)$;
$text{dim(im}(F)) = text{dim}(Y)$ if and only if $text{dim}(ker(F))=0$, i.e. if and only if $F$ is a monomorphism.
Item 1 helps you with proving $text{dim}(X)leq 2cdot text{dim}(Y)$ while item 2 is needed for the second part of your question.
answered Jan 6 at 15:31
F.BattistoniF.Battistoni
515
answered Jan 6 at 15:31
F.BattistoniF.Battistoni
515
answered Jan 6 at 15:31
F.BattistoniF.Battistoni
515
answered Jan 6 at 15:31
F.BattistoniF.Battistoni
515
515
add a comment |
add a comment |
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.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3063669%2fif-x-y-operatornameimf-prove-that-dim-x-le-2-cdot-dim-y%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
