Find a basis for the image of T, where T is a linear transformation. How do you find an image of just a...
$begingroup$
Doesn't it have to be "an image of V under T, where V is a subset" or "an image of v under T, where v is a vector"? What is the image of JUST a transformation?
And if you were given the matrix representation of that transformation, how might you figure out the "image of the transformation" is just from the matrix/the RREF of the matrix?
matrices linear-transformations
asked Dec 16 '18 at 0:09
James RonaldJames Ronald
1257
$endgroup$
add a comment |
$begingroup$
Doesn't it have to be "an image of V under T, where V is a subset" or "an image of v under T, where v is a vector"? What is the image of JUST a transformation?
And if you were given the matrix representation of that transformation, how might you figure out the "image of the transformation" is just from the matrix/the RREF of the matrix?
matrices linear-transformations
asked Dec 16 '18 at 0:09
James RonaldJames Ronald
1257
$endgroup$
$begingroup$
$V$ is tacitly assumed to be the entire space on which $T$ is defined.
$endgroup$
– amd
Dec 16 '18 at 0:11
add a comment |
$begingroup$
Doesn't it have to be "an image of V under T, where V is a subset" or "an image of v under T, where v is a vector"? What is the image of JUST a transformation?
And if you were given the matrix representation of that transformation, how might you figure out the "image of the transformation" is just from the matrix/the RREF of the matrix?
matrices linear-transformations
asked Dec 16 '18 at 0:09
James RonaldJames Ronald
1257
$endgroup$
Doesn't it have to be "an image of V under T, where V is a subset" or "an image of v under T, where v is a vector"? What is the image of JUST a transformation?
And if you were given the matrix representation of that transformation, how might you figure out the "image of the transformation" is just from the matrix/the RREF of the matrix?
matrices linear-transformations
matrices linear-transformations
asked Dec 16 '18 at 0:09
James RonaldJames Ronald
1257
asked Dec 16 '18 at 0:09
James RonaldJames Ronald
1257
asked Dec 16 '18 at 0:09
James RonaldJames Ronald
1257
asked Dec 16 '18 at 0:09
James RonaldJames Ronald
1257
asked Dec 16 '18 at 0:09
James RonaldJames Ronald
1257
1257
$begingroup$
$V$ is tacitly assumed to be the entire space on which $T$ is defined.
$endgroup$
– amd
Dec 16 '18 at 0:11
add a comment |
$begingroup$
$V$ is tacitly assumed to be the entire space on which $T$ is defined.
$endgroup$
– amd
Dec 16 '18 at 0:11
$begingroup$
$V$ is tacitly assumed to be the entire space on which $T$ is defined.
$endgroup$
– amd
Dec 16 '18 at 0:11
$begingroup$
$V$ is tacitly assumed to be the entire space on which $T$ is defined.
$endgroup$
– amd
Dec 16 '18 at 0:11
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The image of a linear transformation $T : V to W$ (or indeed any function) is a synonym for the range of the transformation. It's the set
$$operatorname{Im} T = operatorname{Range} T = { Tv : v in V }.$$
Given the standard matrix $mathcal{M}(T)$ of $T$ (when $T$ maps between Euclidean spaces), the range is the columnspace of $mathcal{M}(T)$, meaning the span of the columns of $mathcal{M}(T)$. This cannot be deduced purely from the reduced row echelon form of the matrix, as elementary row operations do not preserve the columnspace.
However, if you take note of which columns in the RREF contain pivots (leading $1$s), the corresponding columns of the standard matrix will form a basis for the columnspace of $mathcal{M}(T)$, and hence the image of $T$. Alternatively, row reducing $mathcal{M}(T)^top$ will also yield a basis for the image.
answered Dec 16 '18 at 0:33
Theo BenditTheo Bendit
18.4k12152
$endgroup$
add a comment |
$begingroup$
The image of a linear transformation is actually the span of the columns of the matrix rel a basis (see this answer of mine).
Thus, you just need a basis for the column space. To do that, you can read off which columns form a basis for the column space of the RREF. The corresponding columns in the original matrix are a basis (for the original column space, and hence the image).
answered Dec 16 '18 at 0:28
Chris CusterChris Custer
13.2k3827
$endgroup$
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%2f3042091%2ffind-a-basis-for-the-image-of-t-where-t-is-a-linear-transformation-how-do-you%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
$begingroup$
The image of a linear transformation $T : V to W$ (or indeed any function) is a synonym for the range of the transformation. It's the set
$$operatorname{Im} T = operatorname{Range} T = { Tv : v in V }.$$
Given the standard matrix $mathcal{M}(T)$ of $T$ (when $T$ maps between Euclidean spaces), the range is the columnspace of $mathcal{M}(T)$, meaning the span of the columns of $mathcal{M}(T)$. This cannot be deduced purely from the reduced row echelon form of the matrix, as elementary row operations do not preserve the columnspace.
However, if you take note of which columns in the RREF contain pivots (leading $1$s), the corresponding columns of the standard matrix will form a basis for the columnspace of $mathcal{M}(T)$, and hence the image of $T$. Alternatively, row reducing $mathcal{M}(T)^top$ will also yield a basis for the image.
answered Dec 16 '18 at 0:33
Theo BenditTheo Bendit
18.4k12152
$endgroup$
add a comment |
$begingroup$
The image of a linear transformation $T : V to W$ (or indeed any function) is a synonym for the range of the transformation. It's the set
$$operatorname{Im} T = operatorname{Range} T = { Tv : v in V }.$$
Given the standard matrix $mathcal{M}(T)$ of $T$ (when $T$ maps between Euclidean spaces), the range is the columnspace of $mathcal{M}(T)$, meaning the span of the columns of $mathcal{M}(T)$. This cannot be deduced purely from the reduced row echelon form of the matrix, as elementary row operations do not preserve the columnspace.
However, if you take note of which columns in the RREF contain pivots (leading $1$s), the corresponding columns of the standard matrix will form a basis for the columnspace of $mathcal{M}(T)$, and hence the image of $T$. Alternatively, row reducing $mathcal{M}(T)^top$ will also yield a basis for the image.
answered Dec 16 '18 at 0:33
Theo BenditTheo Bendit
18.4k12152
$endgroup$
add a comment |
$begingroup$
The image of a linear transformation $T : V to W$ (or indeed any function) is a synonym for the range of the transformation. It's the set
$$operatorname{Im} T = operatorname{Range} T = { Tv : v in V }.$$
Given the standard matrix $mathcal{M}(T)$ of $T$ (when $T$ maps between Euclidean spaces), the range is the columnspace of $mathcal{M}(T)$, meaning the span of the columns of $mathcal{M}(T)$. This cannot be deduced purely from the reduced row echelon form of the matrix, as elementary row operations do not preserve the columnspace.
However, if you take note of which columns in the RREF contain pivots (leading $1$s), the corresponding columns of the standard matrix will form a basis for the columnspace of $mathcal{M}(T)$, and hence the image of $T$. Alternatively, row reducing $mathcal{M}(T)^top$ will also yield a basis for the image.
answered Dec 16 '18 at 0:33
Theo BenditTheo Bendit
18.4k12152
$endgroup$
The image of a linear transformation $T : V to W$ (or indeed any function) is a synonym for the range of the transformation. It's the set
$$operatorname{Im} T = operatorname{Range} T = { Tv : v in V }.$$
Given the standard matrix $mathcal{M}(T)$ of $T$ (when $T$ maps between Euclidean spaces), the range is the columnspace of $mathcal{M}(T)$, meaning the span of the columns of $mathcal{M}(T)$. This cannot be deduced purely from the reduced row echelon form of the matrix, as elementary row operations do not preserve the columnspace.
However, if you take note of which columns in the RREF contain pivots (leading $1$s), the corresponding columns of the standard matrix will form a basis for the columnspace of $mathcal{M}(T)$, and hence the image of $T$. Alternatively, row reducing $mathcal{M}(T)^top$ will also yield a basis for the image.
answered Dec 16 '18 at 0:33
Theo BenditTheo Bendit
18.4k12152
answered Dec 16 '18 at 0:33
Theo BenditTheo Bendit
18.4k12152
answered Dec 16 '18 at 0:33
Theo BenditTheo Bendit
18.4k12152
answered Dec 16 '18 at 0:33
Theo BenditTheo Bendit
18.4k12152
18.4k12152
add a comment |
add a comment |
$begingroup$
The image of a linear transformation is actually the span of the columns of the matrix rel a basis (see this answer of mine).
Thus, you just need a basis for the column space. To do that, you can read off which columns form a basis for the column space of the RREF. The corresponding columns in the original matrix are a basis (for the original column space, and hence the image).
answered Dec 16 '18 at 0:28
Chris CusterChris Custer
13.2k3827
$endgroup$
add a comment |
$begingroup$
The image of a linear transformation is actually the span of the columns of the matrix rel a basis (see this answer of mine).
Thus, you just need a basis for the column space. To do that, you can read off which columns form a basis for the column space of the RREF. The corresponding columns in the original matrix are a basis (for the original column space, and hence the image).
answered Dec 16 '18 at 0:28
Chris CusterChris Custer
13.2k3827
$endgroup$
add a comment |
$begingroup$
The image of a linear transformation is actually the span of the columns of the matrix rel a basis (see this answer of mine).
Thus, you just need a basis for the column space. To do that, you can read off which columns form a basis for the column space of the RREF. The corresponding columns in the original matrix are a basis (for the original column space, and hence the image).
answered Dec 16 '18 at 0:28
Chris CusterChris Custer
13.2k3827
$endgroup$
The image of a linear transformation is actually the span of the columns of the matrix rel a basis (see this answer of mine).
Thus, you just need a basis for the column space. To do that, you can read off which columns form a basis for the column space of the RREF. The corresponding columns in the original matrix are a basis (for the original column space, and hence the image).
answered Dec 16 '18 at 0:28
Chris CusterChris Custer
13.2k3827
edited Dec 16 '18 at 1:00
answered Dec 16 '18 at 0:28
Chris CusterChris Custer
13.2k3827
answered Dec 16 '18 at 0:28
Chris CusterChris Custer
13.2k3827
answered Dec 16 '18 at 0:28
Chris CusterChris Custer
13.2k3827
13.2k3827
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%2f3042091%2ffind-a-basis-for-the-image-of-t-where-t-is-a-linear-transformation-how-do-you%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
$begingroup$
$V$ is tacitly assumed to be the entire space on which $T$ is defined.
$endgroup$
– amd
Dec 16 '18 at 0:11