What is the intuition behind fundamental theorem of linear maps
$begingroup$
I am self-studying Linear Algebra Done Right. I follow the proof in the book on Fundamental Theorem of Linear Maps.
$$text{dim }V = text{dim null }T + text{dim range } T$$
My understanding is that, the basis of null $T$ with length $m $ can be extended to a basis of $V$ with length $m+v$. And then we can show that dim range $T = n$.
But I don't see the intuition/ picture of this. Why the dimension of $V$ is the sum of dimension of its null space and the range of $T$?
linear-algebra linear-transformations
edited Jan 24 at 12:04
José Carlos Santos
165k22132235
asked Jan 24 at 11:47
JOHN JOHN
4119
$endgroup$
add a comment |
$begingroup$
I am self-studying Linear Algebra Done Right. I follow the proof in the book on Fundamental Theorem of Linear Maps.
$$text{dim }V = text{dim null }T + text{dim range } T$$
My understanding is that, the basis of null $T$ with length $m $ can be extended to a basis of $V$ with length $m+v$. And then we can show that dim range $T = n$.
But I don't see the intuition/ picture of this. Why the dimension of $V$ is the sum of dimension of its null space and the range of $T$?
linear-algebra linear-transformations
edited Jan 24 at 12:04
José Carlos Santos
165k22132235
asked Jan 24 at 11:47
JOHN JOHN
4119
$endgroup$
$begingroup$
Consider a map from $mathbb{R}^2$ to $mathbb{R}^1$ given by sending $(x,y)$ to $x$. This is a linear map, and you can see that the null space and the range each have dimension one. The theorem is that this kind of picture is general.
$endgroup$
– Rylee Lyman
Jan 24 at 12:10
add a comment |
$begingroup$
I am self-studying Linear Algebra Done Right. I follow the proof in the book on Fundamental Theorem of Linear Maps.
$$text{dim }V = text{dim null }T + text{dim range } T$$
My understanding is that, the basis of null $T$ with length $m $ can be extended to a basis of $V$ with length $m+v$. And then we can show that dim range $T = n$.
But I don't see the intuition/ picture of this. Why the dimension of $V$ is the sum of dimension of its null space and the range of $T$?
linear-algebra linear-transformations
edited Jan 24 at 12:04
José Carlos Santos
165k22132235
asked Jan 24 at 11:47
JOHN JOHN
4119
$endgroup$
I am self-studying Linear Algebra Done Right. I follow the proof in the book on Fundamental Theorem of Linear Maps.
$$text{dim }V = text{dim null }T + text{dim range } T$$
My understanding is that, the basis of null $T$ with length $m $ can be extended to a basis of $V$ with length $m+v$. And then we can show that dim range $T = n$.
But I don't see the intuition/ picture of this. Why the dimension of $V$ is the sum of dimension of its null space and the range of $T$?
linear-algebra linear-transformations
linear-algebra linear-transformations
edited Jan 24 at 12:04
José Carlos Santos
165k22132235
asked Jan 24 at 11:47
JOHN JOHN
4119
edited Jan 24 at 12:04
José Carlos Santos
165k22132235
asked Jan 24 at 11:47
JOHN JOHN
4119
edited Jan 24 at 12:04
José Carlos Santos
165k22132235
edited Jan 24 at 12:04
José Carlos Santos
165k22132235
edited Jan 24 at 12:04
José Carlos Santos
165k22132235
165k22132235
asked Jan 24 at 11:47
JOHN JOHN
4119
asked Jan 24 at 11:47
JOHN JOHN
4119
asked Jan 24 at 11:47
JOHN JOHN
4119
4119
$begingroup$
Consider a map from $mathbb{R}^2$ to $mathbb{R}^1$ given by sending $(x,y)$ to $x$. This is a linear map, and you can see that the null space and the range each have dimension one. The theorem is that this kind of picture is general.
$endgroup$
– Rylee Lyman
Jan 24 at 12:10
add a comment |
$begingroup$
Consider a map from $mathbb{R}^2$ to $mathbb{R}^1$ given by sending $(x,y)$ to $x$. This is a linear map, and you can see that the null space and the range each have dimension one. The theorem is that this kind of picture is general.
$endgroup$
– Rylee Lyman
Jan 24 at 12:10
$begingroup$
Consider a map from $mathbb{R}^2$ to $mathbb{R}^1$ given by sending $(x,y)$ to $x$. This is a linear map, and you can see that the null space and the range each have dimension one. The theorem is that this kind of picture is general.
$endgroup$
– Rylee Lyman
Jan 24 at 12:10
$begingroup$
Consider a map from $mathbb{R}^2$ to $mathbb{R}^1$ given by sending $(x,y)$ to $x$. This is a linear map, and you can see that the null space and the range each have dimension one. The theorem is that this kind of picture is general.
$endgroup$
– Rylee Lyman
Jan 24 at 12:10
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Take a basis ${n_1,ldots,n_k}$ of $operatorname{null}T$. Let ${v_1,ldots,v_l}subset V$ be such that ${n_1,ldots,n_k}cup{v_1,ldots,v_l}$ is a basis of $V$. Then $bigl{T(v_1)ldots,T(v_l)bigr}$ is a basis of $operatorname{range}T$ and therefore$$dimoperatorname{range}T=l=k+l-k=dim V-dimoperatorname{null}T.$$
answered Jan 24 at 12:03
José Carlos SantosJosé Carlos Santos
165k22132235
$endgroup$
add a comment |
$begingroup$
What seems like the "intuition" to me involves concepts you may not have studied yet.
Say $T:Vto W$ and $Z$ is the nullspace of $T$. Now it's clear that $T$ "induces" a map $$tilde T:V/Zto W$$ in a "natural" way. Since we factored out the nullspace of $T$ it's clear that $tilde T$ is injective, so it's clear that the dimension of the range of $tilde T$ is equal to $dim(V/Z)=dim(V)-dim(Z)$. And it's clear that the range of $tilde T$ is the same as the range of $T$.
answered Jan 24 at 14:09
David C. UllrichDavid C. Ullrich
61.2k43994
$endgroup$
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%2f3085783%2fwhat-is-the-intuition-behind-fundamental-theorem-of-linear-maps%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Take a basis ${n_1,ldots,n_k}$ of $operatorname{null}T$. Let ${v_1,ldots,v_l}subset V$ be such that ${n_1,ldots,n_k}cup{v_1,ldots,v_l}$ is a basis of $V$. Then $bigl{T(v_1)ldots,T(v_l)bigr}$ is a basis of $operatorname{range}T$ and therefore$$dimoperatorname{range}T=l=k+l-k=dim V-dimoperatorname{null}T.$$
answered Jan 24 at 12:03
José Carlos SantosJosé Carlos Santos
165k22132235
$endgroup$
add a comment |
$begingroup$
Take a basis ${n_1,ldots,n_k}$ of $operatorname{null}T$. Let ${v_1,ldots,v_l}subset V$ be such that ${n_1,ldots,n_k}cup{v_1,ldots,v_l}$ is a basis of $V$. Then $bigl{T(v_1)ldots,T(v_l)bigr}$ is a basis of $operatorname{range}T$ and therefore$$dimoperatorname{range}T=l=k+l-k=dim V-dimoperatorname{null}T.$$
answered Jan 24 at 12:03
José Carlos SantosJosé Carlos Santos
165k22132235
$endgroup$
add a comment |
$begingroup$
Take a basis ${n_1,ldots,n_k}$ of $operatorname{null}T$. Let ${v_1,ldots,v_l}subset V$ be such that ${n_1,ldots,n_k}cup{v_1,ldots,v_l}$ is a basis of $V$. Then $bigl{T(v_1)ldots,T(v_l)bigr}$ is a basis of $operatorname{range}T$ and therefore$$dimoperatorname{range}T=l=k+l-k=dim V-dimoperatorname{null}T.$$
answered Jan 24 at 12:03
José Carlos SantosJosé Carlos Santos
165k22132235
$endgroup$
Take a basis ${n_1,ldots,n_k}$ of $operatorname{null}T$. Let ${v_1,ldots,v_l}subset V$ be such that ${n_1,ldots,n_k}cup{v_1,ldots,v_l}$ is a basis of $V$. Then $bigl{T(v_1)ldots,T(v_l)bigr}$ is a basis of $operatorname{range}T$ and therefore$$dimoperatorname{range}T=l=k+l-k=dim V-dimoperatorname{null}T.$$
answered Jan 24 at 12:03
José Carlos SantosJosé Carlos Santos
165k22132235
answered Jan 24 at 12:03
José Carlos SantosJosé Carlos Santos
165k22132235
answered Jan 24 at 12:03
José Carlos SantosJosé Carlos Santos
165k22132235
answered Jan 24 at 12:03
José Carlos SantosJosé Carlos Santos
165k22132235
165k22132235
add a comment |
add a comment |
$begingroup$
What seems like the "intuition" to me involves concepts you may not have studied yet.
Say $T:Vto W$ and $Z$ is the nullspace of $T$. Now it's clear that $T$ "induces" a map $$tilde T:V/Zto W$$ in a "natural" way. Since we factored out the nullspace of $T$ it's clear that $tilde T$ is injective, so it's clear that the dimension of the range of $tilde T$ is equal to $dim(V/Z)=dim(V)-dim(Z)$. And it's clear that the range of $tilde T$ is the same as the range of $T$.
answered Jan 24 at 14:09
David C. UllrichDavid C. Ullrich
61.2k43994
$endgroup$
add a comment |
$begingroup$
What seems like the "intuition" to me involves concepts you may not have studied yet.
Say $T:Vto W$ and $Z$ is the nullspace of $T$. Now it's clear that $T$ "induces" a map $$tilde T:V/Zto W$$ in a "natural" way. Since we factored out the nullspace of $T$ it's clear that $tilde T$ is injective, so it's clear that the dimension of the range of $tilde T$ is equal to $dim(V/Z)=dim(V)-dim(Z)$. And it's clear that the range of $tilde T$ is the same as the range of $T$.
answered Jan 24 at 14:09
David C. UllrichDavid C. Ullrich
61.2k43994
$endgroup$
add a comment |
$begingroup$
What seems like the "intuition" to me involves concepts you may not have studied yet.
Say $T:Vto W$ and $Z$ is the nullspace of $T$. Now it's clear that $T$ "induces" a map $$tilde T:V/Zto W$$ in a "natural" way. Since we factored out the nullspace of $T$ it's clear that $tilde T$ is injective, so it's clear that the dimension of the range of $tilde T$ is equal to $dim(V/Z)=dim(V)-dim(Z)$. And it's clear that the range of $tilde T$ is the same as the range of $T$.
answered Jan 24 at 14:09
David C. UllrichDavid C. Ullrich
61.2k43994
$endgroup$
What seems like the "intuition" to me involves concepts you may not have studied yet.
Say $T:Vto W$ and $Z$ is the nullspace of $T$. Now it's clear that $T$ "induces" a map $$tilde T:V/Zto W$$ in a "natural" way. Since we factored out the nullspace of $T$ it's clear that $tilde T$ is injective, so it's clear that the dimension of the range of $tilde T$ is equal to $dim(V/Z)=dim(V)-dim(Z)$. And it's clear that the range of $tilde T$ is the same as the range of $T$.
answered Jan 24 at 14:09
David C. UllrichDavid C. Ullrich
61.2k43994
answered Jan 24 at 14:09
David C. UllrichDavid C. Ullrich
61.2k43994
answered Jan 24 at 14:09
David C. UllrichDavid C. Ullrich
61.2k43994
answered Jan 24 at 14:09
David C. UllrichDavid C. Ullrich
61.2k43994
61.2k43994
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%2f3085783%2fwhat-is-the-intuition-behind-fundamental-theorem-of-linear-maps%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
$begingroup$
Consider a map from $mathbb{R}^2$ to $mathbb{R}^1$ given by sending $(x,y)$ to $x$. This is a linear map, and you can see that the null space and the range each have dimension one. The theorem is that this kind of picture is general.
$endgroup$
– Rylee Lyman
Jan 24 at 12:10