Mixing
what is Matrix of a linear transformation?
$begingroup$
I am a student and I'm studying linear algebra. in the Sheldon Axler book in the part "The Spectral Theorem" and in this video he mentions the operator $T$ as
$$T=begin{pmatrix}2&-3\3&2end{pmatrix}$$
then after finding it's eigenvectors $2 + 3i$ and $2 - 3i$ (in the video I've linked), he says: "with respect to this basis, the matrix of $T$ is the diagonal matrix":
$$
begin{pmatrix}
2 + 3i&0 \
0&2 - 3i
end{pmatrix}
$$
I am confused. $T$ already mentions as $T=begin{pmatrix}2&-3\3&2end{pmatrix}$ so the matrix of $T$ must be $begin{pmatrix}2&-3\3&2end{pmatrix}$ so I think that I don't know the meaning of the matrix of $T$.
could you help me figure it out?
linear-algebra matrices operator-theory spectral-theory
asked Jan 5 at 22:31
Peyman mohseni kiasariPeyman mohseni kiasari
1089
$endgroup$
add a comment |
$begingroup$
I am a student and I'm studying linear algebra. in the Sheldon Axler book in the part "The Spectral Theorem" and in this video he mentions the operator $T$ as
$$T=begin{pmatrix}2&-3\3&2end{pmatrix}$$
then after finding it's eigenvectors $2 + 3i$ and $2 - 3i$ (in the video I've linked), he says: "with respect to this basis, the matrix of $T$ is the diagonal matrix":
$$
begin{pmatrix}
2 + 3i&0 \
0&2 - 3i
end{pmatrix}
$$
I am confused. $T$ already mentions as $T=begin{pmatrix}2&-3\3&2end{pmatrix}$ so the matrix of $T$ must be $begin{pmatrix}2&-3\3&2end{pmatrix}$ so I think that I don't know the meaning of the matrix of $T$.
could you help me figure it out?
linear-algebra matrices operator-theory spectral-theory
asked Jan 5 at 22:31
Peyman mohseni kiasariPeyman mohseni kiasari
1089
$endgroup$
add a comment |
$begingroup$
I am a student and I'm studying linear algebra. in the Sheldon Axler book in the part "The Spectral Theorem" and in this video he mentions the operator $T$ as
$$T=begin{pmatrix}2&-3\3&2end{pmatrix}$$
then after finding it's eigenvectors $2 + 3i$ and $2 - 3i$ (in the video I've linked), he says: "with respect to this basis, the matrix of $T$ is the diagonal matrix":
$$
begin{pmatrix}
2 + 3i&0 \
0&2 - 3i
end{pmatrix}
$$
I am confused. $T$ already mentions as $T=begin{pmatrix}2&-3\3&2end{pmatrix}$ so the matrix of $T$ must be $begin{pmatrix}2&-3\3&2end{pmatrix}$ so I think that I don't know the meaning of the matrix of $T$.
could you help me figure it out?
linear-algebra matrices operator-theory spectral-theory
asked Jan 5 at 22:31
Peyman mohseni kiasariPeyman mohseni kiasari
1089
$endgroup$
I am a student and I'm studying linear algebra. in the Sheldon Axler book in the part "The Spectral Theorem" and in this video he mentions the operator $T$ as
$$T=begin{pmatrix}2&-3\3&2end{pmatrix}$$
then after finding it's eigenvectors $2 + 3i$ and $2 - 3i$ (in the video I've linked), he says: "with respect to this basis, the matrix of $T$ is the diagonal matrix":
$$
begin{pmatrix}
2 + 3i&0 \
0&2 - 3i
end{pmatrix}
$$
I am confused. $T$ already mentions as $T=begin{pmatrix}2&-3\3&2end{pmatrix}$ so the matrix of $T$ must be $begin{pmatrix}2&-3\3&2end{pmatrix}$ so I think that I don't know the meaning of the matrix of $T$.
could you help me figure it out?
linear-algebra matrices operator-theory spectral-theory
linear-algebra matrices operator-theory spectral-theory
asked Jan 5 at 22:31
Peyman mohseni kiasariPeyman mohseni kiasari
1089
asked Jan 5 at 22:31
Peyman mohseni kiasariPeyman mohseni kiasari
1089
edited Jan 5 at 22:52
Peyman mohseni kiasari
asked Jan 5 at 22:31
Peyman mohseni kiasariPeyman mohseni kiasari
1089
asked Jan 5 at 22:31
Peyman mohseni kiasariPeyman mohseni kiasari
1089
asked Jan 5 at 22:31
Peyman mohseni kiasariPeyman mohseni kiasari
1089
1089
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Matrices only define linear transformations relative to some basis. They don't describe a linear transformation on their own. Thus implicit in $$T=begin{pmatrix} 2&-3 \ 3 & 2 end{pmatrix}$$
is the statement that the vector space has some basis already and that this is the matrix for $T$ with respect to that basis. E.g., perhaps it's $k^2$, so it has the standard basis vectors $begin{pmatrix} 1 \ 0end{pmatrix}$ and $begin{pmatrix} 0 \ 1end{pmatrix}$ already, and $T$ has that matrix with respect to that basis.
If $V$ is a vector space, then a linear transformation $T:Vto V$ is not a matrix, but rather a function with nice properties that respect the vector space structure. We can then describe it using bases and a matrix, but that's only a description, and the description depends on the basis used to compute the matrix.
As for how to compute the matrix of a linear transformation with respect to some basis, look here for some random notes I found online, or in any decent linear algebra textbook, like presumably in Axler's book somewhere (It appears to be section 3C). For an example of how to do this in a particular case, you can look at this question.
answered Jan 5 at 22:42
jgonjgon
13.6k22041
$endgroup$
$begingroup$
great answer, thanks.
$endgroup$
– Peyman mohseni kiasari
Jan 5 at 22:46
$begingroup$
@Peyman Glad it was helpful, also the matrix for a linear transformation appears to be defined in section 3C of Axler's book based on a table of contents I found online.
$endgroup$
– jgon
Jan 5 at 22:48
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%2f3063266%2fwhat-is-matrix-of-a-linear-transformation%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
$begingroup$
Matrices only define linear transformations relative to some basis. They don't describe a linear transformation on their own. Thus implicit in $$T=begin{pmatrix} 2&-3 \ 3 & 2 end{pmatrix}$$
is the statement that the vector space has some basis already and that this is the matrix for $T$ with respect to that basis. E.g., perhaps it's $k^2$, so it has the standard basis vectors $begin{pmatrix} 1 \ 0end{pmatrix}$ and $begin{pmatrix} 0 \ 1end{pmatrix}$ already, and $T$ has that matrix with respect to that basis.
If $V$ is a vector space, then a linear transformation $T:Vto V$ is not a matrix, but rather a function with nice properties that respect the vector space structure. We can then describe it using bases and a matrix, but that's only a description, and the description depends on the basis used to compute the matrix.
As for how to compute the matrix of a linear transformation with respect to some basis, look here for some random notes I found online, or in any decent linear algebra textbook, like presumably in Axler's book somewhere (It appears to be section 3C). For an example of how to do this in a particular case, you can look at this question.
answered Jan 5 at 22:42
jgonjgon
13.6k22041
$endgroup$
$begingroup$
great answer, thanks.
$endgroup$
– Peyman mohseni kiasari
Jan 5 at 22:46
$begingroup$
@Peyman Glad it was helpful, also the matrix for a linear transformation appears to be defined in section 3C of Axler's book based on a table of contents I found online.
$endgroup$
– jgon
Jan 5 at 22:48
add a comment |
$begingroup$
Matrices only define linear transformations relative to some basis. They don't describe a linear transformation on their own. Thus implicit in $$T=begin{pmatrix} 2&-3 \ 3 & 2 end{pmatrix}$$
is the statement that the vector space has some basis already and that this is the matrix for $T$ with respect to that basis. E.g., perhaps it's $k^2$, so it has the standard basis vectors $begin{pmatrix} 1 \ 0end{pmatrix}$ and $begin{pmatrix} 0 \ 1end{pmatrix}$ already, and $T$ has that matrix with respect to that basis.
If $V$ is a vector space, then a linear transformation $T:Vto V$ is not a matrix, but rather a function with nice properties that respect the vector space structure. We can then describe it using bases and a matrix, but that's only a description, and the description depends on the basis used to compute the matrix.
As for how to compute the matrix of a linear transformation with respect to some basis, look here for some random notes I found online, or in any decent linear algebra textbook, like presumably in Axler's book somewhere (It appears to be section 3C). For an example of how to do this in a particular case, you can look at this question.
answered Jan 5 at 22:42
jgonjgon
13.6k22041
$endgroup$
$begingroup$
great answer, thanks.
$endgroup$
– Peyman mohseni kiasari
Jan 5 at 22:46
$begingroup$
@Peyman Glad it was helpful, also the matrix for a linear transformation appears to be defined in section 3C of Axler's book based on a table of contents I found online.
$endgroup$
– jgon
Jan 5 at 22:48
add a comment |
$begingroup$
Matrices only define linear transformations relative to some basis. They don't describe a linear transformation on their own. Thus implicit in $$T=begin{pmatrix} 2&-3 \ 3 & 2 end{pmatrix}$$
is the statement that the vector space has some basis already and that this is the matrix for $T$ with respect to that basis. E.g., perhaps it's $k^2$, so it has the standard basis vectors $begin{pmatrix} 1 \ 0end{pmatrix}$ and $begin{pmatrix} 0 \ 1end{pmatrix}$ already, and $T$ has that matrix with respect to that basis.
If $V$ is a vector space, then a linear transformation $T:Vto V$ is not a matrix, but rather a function with nice properties that respect the vector space structure. We can then describe it using bases and a matrix, but that's only a description, and the description depends on the basis used to compute the matrix.
As for how to compute the matrix of a linear transformation with respect to some basis, look here for some random notes I found online, or in any decent linear algebra textbook, like presumably in Axler's book somewhere (It appears to be section 3C). For an example of how to do this in a particular case, you can look at this question.
answered Jan 5 at 22:42
jgonjgon
13.6k22041
$endgroup$
Matrices only define linear transformations relative to some basis. They don't describe a linear transformation on their own. Thus implicit in $$T=begin{pmatrix} 2&-3 \ 3 & 2 end{pmatrix}$$
is the statement that the vector space has some basis already and that this is the matrix for $T$ with respect to that basis. E.g., perhaps it's $k^2$, so it has the standard basis vectors $begin{pmatrix} 1 \ 0end{pmatrix}$ and $begin{pmatrix} 0 \ 1end{pmatrix}$ already, and $T$ has that matrix with respect to that basis.
If $V$ is a vector space, then a linear transformation $T:Vto V$ is not a matrix, but rather a function with nice properties that respect the vector space structure. We can then describe it using bases and a matrix, but that's only a description, and the description depends on the basis used to compute the matrix.
As for how to compute the matrix of a linear transformation with respect to some basis, look here for some random notes I found online, or in any decent linear algebra textbook, like presumably in Axler's book somewhere (It appears to be section 3C). For an example of how to do this in a particular case, you can look at this question.
answered Jan 5 at 22:42
jgonjgon
13.6k22041
edited Jan 5 at 22:48
answered Jan 5 at 22:42
jgonjgon
13.6k22041
answered Jan 5 at 22:42
jgonjgon
13.6k22041
answered Jan 5 at 22:42
jgonjgon
13.6k22041
13.6k22041
$begingroup$
great answer, thanks.
$endgroup$
– Peyman mohseni kiasari
Jan 5 at 22:46
$begingroup$
@Peyman Glad it was helpful, also the matrix for a linear transformation appears to be defined in section 3C of Axler's book based on a table of contents I found online.
$endgroup$
– jgon
Jan 5 at 22:48
add a comment |
$begingroup$
great answer, thanks.
$endgroup$
– Peyman mohseni kiasari
Jan 5 at 22:46
$begingroup$
@Peyman Glad it was helpful, also the matrix for a linear transformation appears to be defined in section 3C of Axler's book based on a table of contents I found online.
$endgroup$
– jgon
Jan 5 at 22:48
$begingroup$
great answer, thanks.
$endgroup$
– Peyman mohseni kiasari
Jan 5 at 22:46
$begingroup$
great answer, thanks.
$endgroup$
– Peyman mohseni kiasari
Jan 5 at 22:46
$begingroup$
@Peyman Glad it was helpful, also the matrix for a linear transformation appears to be defined in section 3C of Axler's book based on a table of contents I found online.
$endgroup$
– jgon
Jan 5 at 22:48
$begingroup$
@Peyman Glad it was helpful, also the matrix for a linear transformation appears to be defined in section 3C of Axler's book based on a table of contents I found online.
$endgroup$
– jgon
Jan 5 at 22:48
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%2f3063266%2fwhat-is-matrix-of-a-linear-transformation%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
