Suppose $V, $and $W$ are both finite dimensional . Prove that there exits an injective linear map from $V$ to...
$begingroup$
Suppose $V$ and $W$ are both finite dimensional. Prove that there exits an injective linear map from $V$ to $W$ if and only if $dim V leq dim W$.
Proof: Suppose $V$ and $W$ are both finite dimensional. Then we have $dim V = dim(text{null} (T)) + dim (text{range} (T))$. Then we have $dim V= dim text{range} (T)leq dim W$.
So there exists an injective map, so that $dim V leq dim W$.
Conversely, suppose $dim Vleq dim W$. Then let $dim V = n, dim W = m$. Define a basis for $V$ by $v_1,ldots,v_n $ for $V$ and $w_1,ldots,w_m$ for $W$.
Can someone please help me? I dont really know how to show there is an injective linear map .
linear-algebra
$endgroup$
add a comment |
$begingroup$
Suppose $V$ and $W$ are both finite dimensional. Prove that there exits an injective linear map from $V$ to $W$ if and only if $dim V leq dim W$.
Proof: Suppose $V$ and $W$ are both finite dimensional. Then we have $dim V = dim(text{null} (T)) + dim (text{range} (T))$. Then we have $dim V= dim text{range} (T)leq dim W$.
So there exists an injective map, so that $dim V leq dim W$.
Conversely, suppose $dim Vleq dim W$. Then let $dim V = n, dim W = m$. Define a basis for $V$ by $v_1,ldots,v_n $ for $V$ and $w_1,ldots,w_m$ for $W$.
Can someone please help me? I dont really know how to show there is an injective linear map .
linear-algebra
$endgroup$
1
$begingroup$
You can just show that there exists one, for example the one which maps $v_i$ to $w_i$.
$endgroup$
– anderstood
Sep 28 '16 at 21:47
$begingroup$
Note: For "only if" you could show that the image of a basis of $V$ is linearly independent in $W$ using the injective property. (You haven't, incidentally, said what $T$ is).
$endgroup$
– Mark Bennet
Sep 28 '16 at 21:52
$begingroup$
$T$ can be defined as $T: V→W$
$endgroup$
– user5956
Sep 28 '16 at 21:53
add a comment |
$begingroup$
Suppose $V$ and $W$ are both finite dimensional. Prove that there exits an injective linear map from $V$ to $W$ if and only if $dim V leq dim W$.
Proof: Suppose $V$ and $W$ are both finite dimensional. Then we have $dim V = dim(text{null} (T)) + dim (text{range} (T))$. Then we have $dim V= dim text{range} (T)leq dim W$.
So there exists an injective map, so that $dim V leq dim W$.
Conversely, suppose $dim Vleq dim W$. Then let $dim V = n, dim W = m$. Define a basis for $V$ by $v_1,ldots,v_n $ for $V$ and $w_1,ldots,w_m$ for $W$.
Can someone please help me? I dont really know how to show there is an injective linear map .
linear-algebra
$endgroup$
Suppose $V$ and $W$ are both finite dimensional. Prove that there exits an injective linear map from $V$ to $W$ if and only if $dim V leq dim W$.
Proof: Suppose $V$ and $W$ are both finite dimensional. Then we have $dim V = dim(text{null} (T)) + dim (text{range} (T))$. Then we have $dim V= dim text{range} (T)leq dim W$.
So there exists an injective map, so that $dim V leq dim W$.
Conversely, suppose $dim Vleq dim W$. Then let $dim V = n, dim W = m$. Define a basis for $V$ by $v_1,ldots,v_n $ for $V$ and $w_1,ldots,w_m$ for $W$.
Can someone please help me? I dont really know how to show there is an injective linear map .
linear-algebra
linear-algebra
edited Sep 28 '16 at 22:28
Lonidard
2,93311021
2,93311021
asked Sep 28 '16 at 21:44
user5956user5956
82
82
1
$begingroup$
You can just show that there exists one, for example the one which maps $v_i$ to $w_i$.
$endgroup$
– anderstood
Sep 28 '16 at 21:47
$begingroup$
Note: For "only if" you could show that the image of a basis of $V$ is linearly independent in $W$ using the injective property. (You haven't, incidentally, said what $T$ is).
$endgroup$
– Mark Bennet
Sep 28 '16 at 21:52
$begingroup$
$T$ can be defined as $T: V→W$
$endgroup$
– user5956
Sep 28 '16 at 21:53
add a comment |
1
$begingroup$
You can just show that there exists one, for example the one which maps $v_i$ to $w_i$.
$endgroup$
– anderstood
Sep 28 '16 at 21:47
$begingroup$
Note: For "only if" you could show that the image of a basis of $V$ is linearly independent in $W$ using the injective property. (You haven't, incidentally, said what $T$ is).
$endgroup$
– Mark Bennet
Sep 28 '16 at 21:52
$begingroup$
$T$ can be defined as $T: V→W$
$endgroup$
– user5956
Sep 28 '16 at 21:53
1
1
$begingroup$
You can just show that there exists one, for example the one which maps $v_i$ to $w_i$.
$endgroup$
– anderstood
Sep 28 '16 at 21:47
$begingroup$
You can just show that there exists one, for example the one which maps $v_i$ to $w_i$.
$endgroup$
– anderstood
Sep 28 '16 at 21:47
$begingroup$
Note: For "only if" you could show that the image of a basis of $V$ is linearly independent in $W$ using the injective property. (You haven't, incidentally, said what $T$ is).
$endgroup$
– Mark Bennet
Sep 28 '16 at 21:52
$begingroup$
Note: For "only if" you could show that the image of a basis of $V$ is linearly independent in $W$ using the injective property. (You haven't, incidentally, said what $T$ is).
$endgroup$
– Mark Bennet
Sep 28 '16 at 21:52
$begingroup$
$T$ can be defined as $T: V→W$
$endgroup$
– user5956
Sep 28 '16 at 21:53
$begingroup$
$T$ can be defined as $T: V→W$
$endgroup$
– user5956
Sep 28 '16 at 21:53
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint: define $T$ by $v_ilongmapsto w_i$, $i=1,dots n$.
$endgroup$
$begingroup$
So if I define T as you say we have $T: V → W$ by $Tv_1 = w_1,...,Tv_n = w_n$. and $n leq m.$ So $T(v_1,...,v_n) = a_1v_1 + ....+ a_nv_n$ and each $a_1v_1 + ...+ a_nv_n = 0 $ if and only if $(v_1,...,v_n) = (0,....,0)$ since $(v_1,...,v_n)$ is linearly independent , we have $null T = {{0,...,0}}$ so $T $ is one to one. Can someone please verify this?
$endgroup$
– user5956
Sep 28 '16 at 21:51
$begingroup$
HI Martin , can you please give me some feedback? thanks
$endgroup$
– user5956
Sep 28 '16 at 23:27
$begingroup$
@user5956, right.
$endgroup$
– Martín-Blas Pérez Pinilla
Sep 29 '16 at 7:40
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%2f1945646%2fsuppose-v-and-w-are-both-finite-dimensional-prove-that-there-exits-an-inj%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: define $T$ by $v_ilongmapsto w_i$, $i=1,dots n$.
$endgroup$
$begingroup$
So if I define T as you say we have $T: V → W$ by $Tv_1 = w_1,...,Tv_n = w_n$. and $n leq m.$ So $T(v_1,...,v_n) = a_1v_1 + ....+ a_nv_n$ and each $a_1v_1 + ...+ a_nv_n = 0 $ if and only if $(v_1,...,v_n) = (0,....,0)$ since $(v_1,...,v_n)$ is linearly independent , we have $null T = {{0,...,0}}$ so $T $ is one to one. Can someone please verify this?
$endgroup$
– user5956
Sep 28 '16 at 21:51
$begingroup$
HI Martin , can you please give me some feedback? thanks
$endgroup$
– user5956
Sep 28 '16 at 23:27
$begingroup$
@user5956, right.
$endgroup$
– Martín-Blas Pérez Pinilla
Sep 29 '16 at 7:40
add a comment |
$begingroup$
Hint: define $T$ by $v_ilongmapsto w_i$, $i=1,dots n$.
$endgroup$
$begingroup$
So if I define T as you say we have $T: V → W$ by $Tv_1 = w_1,...,Tv_n = w_n$. and $n leq m.$ So $T(v_1,...,v_n) = a_1v_1 + ....+ a_nv_n$ and each $a_1v_1 + ...+ a_nv_n = 0 $ if and only if $(v_1,...,v_n) = (0,....,0)$ since $(v_1,...,v_n)$ is linearly independent , we have $null T = {{0,...,0}}$ so $T $ is one to one. Can someone please verify this?
$endgroup$
– user5956
Sep 28 '16 at 21:51
$begingroup$
HI Martin , can you please give me some feedback? thanks
$endgroup$
– user5956
Sep 28 '16 at 23:27
$begingroup$
@user5956, right.
$endgroup$
– Martín-Blas Pérez Pinilla
Sep 29 '16 at 7:40
add a comment |
$begingroup$
Hint: define $T$ by $v_ilongmapsto w_i$, $i=1,dots n$.
$endgroup$
Hint: define $T$ by $v_ilongmapsto w_i$, $i=1,dots n$.
answered Sep 28 '16 at 21:47
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.2k42871
34.2k42871
$begingroup$
So if I define T as you say we have $T: V → W$ by $Tv_1 = w_1,...,Tv_n = w_n$. and $n leq m.$ So $T(v_1,...,v_n) = a_1v_1 + ....+ a_nv_n$ and each $a_1v_1 + ...+ a_nv_n = 0 $ if and only if $(v_1,...,v_n) = (0,....,0)$ since $(v_1,...,v_n)$ is linearly independent , we have $null T = {{0,...,0}}$ so $T $ is one to one. Can someone please verify this?
$endgroup$
– user5956
Sep 28 '16 at 21:51
$begingroup$
HI Martin , can you please give me some feedback? thanks
$endgroup$
– user5956
Sep 28 '16 at 23:27
$begingroup$
@user5956, right.
$endgroup$
– Martín-Blas Pérez Pinilla
Sep 29 '16 at 7:40
add a comment |
$begingroup$
So if I define T as you say we have $T: V → W$ by $Tv_1 = w_1,...,Tv_n = w_n$. and $n leq m.$ So $T(v_1,...,v_n) = a_1v_1 + ....+ a_nv_n$ and each $a_1v_1 + ...+ a_nv_n = 0 $ if and only if $(v_1,...,v_n) = (0,....,0)$ since $(v_1,...,v_n)$ is linearly independent , we have $null T = {{0,...,0}}$ so $T $ is one to one. Can someone please verify this?
$endgroup$
– user5956
Sep 28 '16 at 21:51
$begingroup$
HI Martin , can you please give me some feedback? thanks
$endgroup$
– user5956
Sep 28 '16 at 23:27
$begingroup$
@user5956, right.
$endgroup$
– Martín-Blas Pérez Pinilla
Sep 29 '16 at 7:40
$begingroup$
So if I define T as you say we have $T: V → W$ by $Tv_1 = w_1,...,Tv_n = w_n$. and $n leq m.$ So $T(v_1,...,v_n) = a_1v_1 + ....+ a_nv_n$ and each $a_1v_1 + ...+ a_nv_n = 0 $ if and only if $(v_1,...,v_n) = (0,....,0)$ since $(v_1,...,v_n)$ is linearly independent , we have $null T = {{0,...,0}}$ so $T $ is one to one. Can someone please verify this?
$endgroup$
– user5956
Sep 28 '16 at 21:51
$begingroup$
So if I define T as you say we have $T: V → W$ by $Tv_1 = w_1,...,Tv_n = w_n$. and $n leq m.$ So $T(v_1,...,v_n) = a_1v_1 + ....+ a_nv_n$ and each $a_1v_1 + ...+ a_nv_n = 0 $ if and only if $(v_1,...,v_n) = (0,....,0)$ since $(v_1,...,v_n)$ is linearly independent , we have $null T = {{0,...,0}}$ so $T $ is one to one. Can someone please verify this?
$endgroup$
– user5956
Sep 28 '16 at 21:51
$begingroup$
HI Martin , can you please give me some feedback? thanks
$endgroup$
– user5956
Sep 28 '16 at 23:27
$begingroup$
HI Martin , can you please give me some feedback? thanks
$endgroup$
– user5956
Sep 28 '16 at 23:27
$begingroup$
@user5956, right.
$endgroup$
– Martín-Blas Pérez Pinilla
Sep 29 '16 at 7:40
$begingroup$
@user5956, right.
$endgroup$
– Martín-Blas Pérez Pinilla
Sep 29 '16 at 7:40
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%2f1945646%2fsuppose-v-and-w-are-both-finite-dimensional-prove-that-there-exits-an-inj%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
$begingroup$
You can just show that there exists one, for example the one which maps $v_i$ to $w_i$.
$endgroup$
– anderstood
Sep 28 '16 at 21:47
$begingroup$
Note: For "only if" you could show that the image of a basis of $V$ is linearly independent in $W$ using the injective property. (You haven't, incidentally, said what $T$ is).
$endgroup$
– Mark Bennet
Sep 28 '16 at 21:52
$begingroup$
$T$ can be defined as $T: V→W$
$endgroup$
– user5956
Sep 28 '16 at 21:53