Mixing
Existence of a subspace of the domain of a linear map with specific properties
$begingroup$
Let $V$, $W$ be two finite-dimensional vector spaces over $mathbb{K}$ and $T in L(V,W)$.
Does a subspace $U$ of $V$ with the following properties exist:
$$U cap mathrm{Ker}(T) = {0}$$ and $$ mathrm{Im}(T) = { T(u);|;uin U }$$
My solution:
Let $B_0 = { v_1, dots, v_m } $ be a basis of $mathrm{Ker}(T)$.
Because every basis of a subspace can be expanded to a basis of its vector space, $B_0$ can be expanded to a basis $B$ of $V$:
$$ B = {v_1, dots, v_m, v_{m+1}, dots, v_{n}} $$
Then let $ U = mathrm{span}{v_{m+1}, dots, v_n }$ be the span of the extended vectors.
U now fulfills the first property because the intersection of the span of two linearly independent sets only contains the zero vector.
For the second property I don't know how to get to the property, so far I have:
I now know that $ V = U oplus mathrm{Ker}(T)$ and from
$$ mathrm{dim};V = mathrm{dim}(mathrm{Ker}(T)) + mathrm{dim};U$$ and
$$mathrm{dim}; V = mathrm{dim}(mathrm{Ker}(T)) + mathrm{dim}(mathrm{Im}(T))$$
that $mathrm{dim};U = mathrm{dim}(mathrm{Im}(T))$.
And I also know that $T$ restricted to $U$ is injective.
Do these observation help me solve the problem or could someone provide a hint? Thank you
linear-algebra abstract-algebra vector-spaces linear-transformations
asked Jan 17 at 17:35
strelsolstrelsol
245
$endgroup$
add a comment |
$begingroup$
Let $V$, $W$ be two finite-dimensional vector spaces over $mathbb{K}$ and $T in L(V,W)$.
Does a subspace $U$ of $V$ with the following properties exist:
$$U cap mathrm{Ker}(T) = {0}$$ and $$ mathrm{Im}(T) = { T(u);|;uin U }$$
My solution:
Let $B_0 = { v_1, dots, v_m } $ be a basis of $mathrm{Ker}(T)$.
Because every basis of a subspace can be expanded to a basis of its vector space, $B_0$ can be expanded to a basis $B$ of $V$:
$$ B = {v_1, dots, v_m, v_{m+1}, dots, v_{n}} $$
Then let $ U = mathrm{span}{v_{m+1}, dots, v_n }$ be the span of the extended vectors.
U now fulfills the first property because the intersection of the span of two linearly independent sets only contains the zero vector.
For the second property I don't know how to get to the property, so far I have:
I now know that $ V = U oplus mathrm{Ker}(T)$ and from
$$ mathrm{dim};V = mathrm{dim}(mathrm{Ker}(T)) + mathrm{dim};U$$ and
$$mathrm{dim}; V = mathrm{dim}(mathrm{Ker}(T)) + mathrm{dim}(mathrm{Im}(T))$$
that $mathrm{dim};U = mathrm{dim}(mathrm{Im}(T))$.
And I also know that $T$ restricted to $U$ is injective.
Do these observation help me solve the problem or could someone provide a hint? Thank you
linear-algebra abstract-algebra vector-spaces linear-transformations
asked Jan 17 at 17:35
strelsolstrelsol
245
$endgroup$
1
$begingroup$
For $z = T(v) in Im(T)$, write (uniquely as you showed) $v = u + a$ with $u in U$ and $a in ker(T)$. So you get $z = T(v) = T(u)$.
$endgroup$
– C. Zhihao
Jan 17 at 17:41
$begingroup$
Thank you for your help!
$endgroup$
– strelsol
Jan 17 at 17:59
add a comment |
$begingroup$
Let $V$, $W$ be two finite-dimensional vector spaces over $mathbb{K}$ and $T in L(V,W)$.
Does a subspace $U$ of $V$ with the following properties exist:
$$U cap mathrm{Ker}(T) = {0}$$ and $$ mathrm{Im}(T) = { T(u);|;uin U }$$
My solution:
Let $B_0 = { v_1, dots, v_m } $ be a basis of $mathrm{Ker}(T)$.
Because every basis of a subspace can be expanded to a basis of its vector space, $B_0$ can be expanded to a basis $B$ of $V$:
$$ B = {v_1, dots, v_m, v_{m+1}, dots, v_{n}} $$
Then let $ U = mathrm{span}{v_{m+1}, dots, v_n }$ be the span of the extended vectors.
U now fulfills the first property because the intersection of the span of two linearly independent sets only contains the zero vector.
For the second property I don't know how to get to the property, so far I have:
I now know that $ V = U oplus mathrm{Ker}(T)$ and from
$$ mathrm{dim};V = mathrm{dim}(mathrm{Ker}(T)) + mathrm{dim};U$$ and
$$mathrm{dim}; V = mathrm{dim}(mathrm{Ker}(T)) + mathrm{dim}(mathrm{Im}(T))$$
that $mathrm{dim};U = mathrm{dim}(mathrm{Im}(T))$.
And I also know that $T$ restricted to $U$ is injective.
Do these observation help me solve the problem or could someone provide a hint? Thank you
linear-algebra abstract-algebra vector-spaces linear-transformations
asked Jan 17 at 17:35
strelsolstrelsol
245
$endgroup$
Let $V$, $W$ be two finite-dimensional vector spaces over $mathbb{K}$ and $T in L(V,W)$.
Does a subspace $U$ of $V$ with the following properties exist:
$$U cap mathrm{Ker}(T) = {0}$$ and $$ mathrm{Im}(T) = { T(u);|;uin U }$$
My solution:
Let $B_0 = { v_1, dots, v_m } $ be a basis of $mathrm{Ker}(T)$.
Because every basis of a subspace can be expanded to a basis of its vector space, $B_0$ can be expanded to a basis $B$ of $V$:
$$ B = {v_1, dots, v_m, v_{m+1}, dots, v_{n}} $$
Then let $ U = mathrm{span}{v_{m+1}, dots, v_n }$ be the span of the extended vectors.
U now fulfills the first property because the intersection of the span of two linearly independent sets only contains the zero vector.
For the second property I don't know how to get to the property, so far I have:
I now know that $ V = U oplus mathrm{Ker}(T)$ and from
$$ mathrm{dim};V = mathrm{dim}(mathrm{Ker}(T)) + mathrm{dim};U$$ and
$$mathrm{dim}; V = mathrm{dim}(mathrm{Ker}(T)) + mathrm{dim}(mathrm{Im}(T))$$
that $mathrm{dim};U = mathrm{dim}(mathrm{Im}(T))$.
And I also know that $T$ restricted to $U$ is injective.
Do these observation help me solve the problem or could someone provide a hint? Thank you
linear-algebra abstract-algebra vector-spaces linear-transformations
linear-algebra abstract-algebra vector-spaces linear-transformations
asked Jan 17 at 17:35
strelsolstrelsol
245
asked Jan 17 at 17:35
strelsolstrelsol
245
edited Jan 17 at 19:19
user635162
asked Jan 17 at 17:35
strelsolstrelsol
245
asked Jan 17 at 17:35
strelsolstrelsol
245
asked Jan 17 at 17:35
strelsolstrelsol
245
245
1
$begingroup$
For $z = T(v) in Im(T)$, write (uniquely as you showed) $v = u + a$ with $u in U$ and $a in ker(T)$. So you get $z = T(v) = T(u)$.
$endgroup$
– C. Zhihao
Jan 17 at 17:41
$begingroup$
Thank you for your help!
$endgroup$
– strelsol
Jan 17 at 17:59
add a comment |
1
$begingroup$
For $z = T(v) in Im(T)$, write (uniquely as you showed) $v = u + a$ with $u in U$ and $a in ker(T)$. So you get $z = T(v) = T(u)$.
$endgroup$
– C. Zhihao
Jan 17 at 17:41
$begingroup$
Thank you for your help!
$endgroup$
– strelsol
Jan 17 at 17:59
1
1
$begingroup$
For $z = T(v) in Im(T)$, write (uniquely as you showed) $v = u + a$ with $u in U$ and $a in ker(T)$. So you get $z = T(v) = T(u)$.
$endgroup$
– C. Zhihao
Jan 17 at 17:41
$begingroup$
For $z = T(v) in Im(T)$, write (uniquely as you showed) $v = u + a$ with $u in U$ and $a in ker(T)$. So you get $z = T(v) = T(u)$.
$endgroup$
– C. Zhihao
Jan 17 at 17:41
$begingroup$
Thank you for your help!
$endgroup$
– strelsol
Jan 17 at 17:59
$begingroup$
Thank you for your help!
$endgroup$
– strelsol
Jan 17 at 17:59
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Try this:
Since $V=Uoplus ker(T)$, any $vin V$ can be written uniquely as $v=u+w$ with $uin U$ and $win ker(T)$. Now compute
$$T(v)=T(u+w)=T(u)+T(w)=T(u)+0=T(u).$$
It now follows that $T(V)=T(U)$ as required.
answered Jan 17 at 19:26
David HillDavid Hill
9,1461619
$endgroup$
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%2f3077270%2fexistence-of-a-subspace-of-the-domain-of-a-linear-map-with-specific-properties%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
$begingroup$
Try this:
Since $V=Uoplus ker(T)$, any $vin V$ can be written uniquely as $v=u+w$ with $uin U$ and $win ker(T)$. Now compute
$$T(v)=T(u+w)=T(u)+T(w)=T(u)+0=T(u).$$
It now follows that $T(V)=T(U)$ as required.
answered Jan 17 at 19:26
David HillDavid Hill
9,1461619
$endgroup$
add a comment |
$begingroup$
Try this:
Since $V=Uoplus ker(T)$, any $vin V$ can be written uniquely as $v=u+w$ with $uin U$ and $win ker(T)$. Now compute
$$T(v)=T(u+w)=T(u)+T(w)=T(u)+0=T(u).$$
It now follows that $T(V)=T(U)$ as required.
answered Jan 17 at 19:26
David HillDavid Hill
9,1461619
$endgroup$
add a comment |
$begingroup$
Try this:
Since $V=Uoplus ker(T)$, any $vin V$ can be written uniquely as $v=u+w$ with $uin U$ and $win ker(T)$. Now compute
$$T(v)=T(u+w)=T(u)+T(w)=T(u)+0=T(u).$$
It now follows that $T(V)=T(U)$ as required.
answered Jan 17 at 19:26
David HillDavid Hill
9,1461619
$endgroup$
Try this:
Since $V=Uoplus ker(T)$, any $vin V$ can be written uniquely as $v=u+w$ with $uin U$ and $win ker(T)$. Now compute
$$T(v)=T(u+w)=T(u)+T(w)=T(u)+0=T(u).$$
It now follows that $T(V)=T(U)$ as required.
answered Jan 17 at 19:26
David HillDavid Hill
9,1461619
answered Jan 17 at 19:26
David HillDavid Hill
9,1461619
answered Jan 17 at 19:26
David HillDavid Hill
9,1461619
answered Jan 17 at 19:26
David HillDavid Hill
9,1461619
9,1461619
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%2f3077270%2fexistence-of-a-subspace-of-the-domain-of-a-linear-map-with-specific-properties%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
1
$begingroup$
For $z = T(v) in Im(T)$, write (uniquely as you showed) $v = u + a$ with $u in U$ and $a in ker(T)$. So you get $z = T(v) = T(u)$.
$endgroup$
– C. Zhihao
Jan 17 at 17:41
$begingroup$
Thank you for your help!
$endgroup$
– strelsol
Jan 17 at 17:59