Change of Coordinate Matrix Collary Proof
Let $Ain M_{n times n}(Bbb F)$, and let $gamma$ be the basis for $Bbb F^n$.
Show that if $[L_A]_γ=Q^{-1}AQ$, then the j-th column of Q is the j-th vector in $γ$.
I start by applying the theorem:
$[L_A]_γ = Q^{-1}[L_A]_{α} Q$
And let $[L_A]_α=A$, $α$ is another unknown basis of $Bbb F^n$.
Then we have $Q=[I_n]_{γ→α}$, $Q$ is an $n times n$ matrix.
But I still can't figure out why the $j$-th column of $Q$ is the same as the $j$-th vector in $γ$.
Hope someone can give me some hints.
Thanks a lot.
linear-algebra matrices linear-transformations
edited Nov 26 at 19:12
Omnomnomnom
126k788176
asked Nov 26 at 15:31
Ann P.
133
add a comment |
Let $Ain M_{n times n}(Bbb F)$, and let $gamma$ be the basis for $Bbb F^n$.
Show that if $[L_A]_γ=Q^{-1}AQ$, then the j-th column of Q is the j-th vector in $γ$.
I start by applying the theorem:
$[L_A]_γ = Q^{-1}[L_A]_{α} Q$
And let $[L_A]_α=A$, $α$ is another unknown basis of $Bbb F^n$.
Then we have $Q=[I_n]_{γ→α}$, $Q$ is an $n times n$ matrix.
But I still can't figure out why the $j$-th column of $Q$ is the same as the $j$-th vector in $γ$.
Hope someone can give me some hints.
Thanks a lot.
linear-algebra matrices linear-transformations
edited Nov 26 at 19:12
Omnomnomnom
126k788176
asked Nov 26 at 15:31
Ann P.
133
The statement is not correct as stated: if $[L_A]_{gamma} = Q^{-1}AQ$, then there are many matrices that $Q$ could be. For instance: for any such matrix $Q$ and non-zero $k in Bbb F$, $kQ$ will satisfy $(kQ)^{-1}A(kQ) = [L_A]_{gamma}$. I suspect that you are meant to prove the converse, that is: if $Q$ is the matrix whose columns are the elements of $gamma$, then $[L_A]_gamma = Q^{-1}AQ$.
– Omnomnomnom
Nov 26 at 19:17
@Omnomnomnom Thanks, I've revisited the original question and found that you're right. English is not my first language, so sometimes it's hard for me to conveyed the question properly. I'll try considering $alpha$ as standard basis!! Thanks again.
– Ann P.
Nov 27 at 0:00
add a comment |
Let $Ain M_{n times n}(Bbb F)$, and let $gamma$ be the basis for $Bbb F^n$.
Show that if $[L_A]_γ=Q^{-1}AQ$, then the j-th column of Q is the j-th vector in $γ$.
I start by applying the theorem:
$[L_A]_γ = Q^{-1}[L_A]_{α} Q$
And let $[L_A]_α=A$, $α$ is another unknown basis of $Bbb F^n$.
Then we have $Q=[I_n]_{γ→α}$, $Q$ is an $n times n$ matrix.
But I still can't figure out why the $j$-th column of $Q$ is the same as the $j$-th vector in $γ$.
Hope someone can give me some hints.
Thanks a lot.
linear-algebra matrices linear-transformations
edited Nov 26 at 19:12
Omnomnomnom
126k788176
asked Nov 26 at 15:31
Ann P.
133
Let $Ain M_{n times n}(Bbb F)$, and let $gamma$ be the basis for $Bbb F^n$.
Show that if $[L_A]_γ=Q^{-1}AQ$, then the j-th column of Q is the j-th vector in $γ$.
I start by applying the theorem:
$[L_A]_γ = Q^{-1}[L_A]_{α} Q$
And let $[L_A]_α=A$, $α$ is another unknown basis of $Bbb F^n$.
Then we have $Q=[I_n]_{γ→α}$, $Q$ is an $n times n$ matrix.
But I still can't figure out why the $j$-th column of $Q$ is the same as the $j$-th vector in $γ$.
Hope someone can give me some hints.
Thanks a lot.
linear-algebra matrices linear-transformations
linear-algebra matrices linear-transformations
edited Nov 26 at 19:12
Omnomnomnom
126k788176
asked Nov 26 at 15:31
Ann P.
133
edited Nov 26 at 19:12
Omnomnomnom
126k788176
asked Nov 26 at 15:31
Ann P.
133
edited Nov 26 at 19:12
Omnomnomnom
126k788176
edited Nov 26 at 19:12
Omnomnomnom
126k788176
edited Nov 26 at 19:12
Omnomnomnom
126k788176
126k788176
asked Nov 26 at 15:31
Ann P.
133
asked Nov 26 at 15:31
Ann P.
133
asked Nov 26 at 15:31
Ann P.
133
133
The statement is not correct as stated: if $[L_A]_{gamma} = Q^{-1}AQ$, then there are many matrices that $Q$ could be. For instance: for any such matrix $Q$ and non-zero $k in Bbb F$, $kQ$ will satisfy $(kQ)^{-1}A(kQ) = [L_A]_{gamma}$. I suspect that you are meant to prove the converse, that is: if $Q$ is the matrix whose columns are the elements of $gamma$, then $[L_A]_gamma = Q^{-1}AQ$.
– Omnomnomnom
Nov 26 at 19:17
@Omnomnomnom Thanks, I've revisited the original question and found that you're right. English is not my first language, so sometimes it's hard for me to conveyed the question properly. I'll try considering $alpha$ as standard basis!! Thanks again.
– Ann P.
Nov 27 at 0:00
add a comment |
The statement is not correct as stated: if $[L_A]_{gamma} = Q^{-1}AQ$, then there are many matrices that $Q$ could be. For instance: for any such matrix $Q$ and non-zero $k in Bbb F$, $kQ$ will satisfy $(kQ)^{-1}A(kQ) = [L_A]_{gamma}$. I suspect that you are meant to prove the converse, that is: if $Q$ is the matrix whose columns are the elements of $gamma$, then $[L_A]_gamma = Q^{-1}AQ$.
– Omnomnomnom
Nov 26 at 19:17
@Omnomnomnom Thanks, I've revisited the original question and found that you're right. English is not my first language, so sometimes it's hard for me to conveyed the question properly. I'll try considering $alpha$ as standard basis!! Thanks again.
– Ann P.
Nov 27 at 0:00
The statement is not correct as stated: if $[L_A]_{gamma} = Q^{-1}AQ$, then there are many matrices that $Q$ could be. For instance: for any such matrix $Q$ and non-zero $k in Bbb F$, $kQ$ will satisfy $(kQ)^{-1}A(kQ) = [L_A]_{gamma}$. I suspect that you are meant to prove the converse, that is: if $Q$ is the matrix whose columns are the elements of $gamma$, then $[L_A]_gamma = Q^{-1}AQ$.
– Omnomnomnom
Nov 26 at 19:17
The statement is not correct as stated: if $[L_A]_{gamma} = Q^{-1}AQ$, then there are many matrices that $Q$ could be. For instance: for any such matrix $Q$ and non-zero $k in Bbb F$, $kQ$ will satisfy $(kQ)^{-1}A(kQ) = [L_A]_{gamma}$. I suspect that you are meant to prove the converse, that is: if $Q$ is the matrix whose columns are the elements of $gamma$, then $[L_A]_gamma = Q^{-1}AQ$.
– Omnomnomnom
Nov 26 at 19:17
@Omnomnomnom Thanks, I've revisited the original question and found that you're right. English is not my first language, so sometimes it's hard for me to conveyed the question properly. I'll try considering $alpha$ as standard basis!! Thanks again.
– Ann P.
Nov 27 at 0:00
@Omnomnomnom Thanks, I've revisited the original question and found that you're right. English is not my first language, so sometimes it's hard for me to conveyed the question properly. I'll try considering $alpha$ as standard basis!! Thanks again.
– Ann P.
Nov 27 at 0:00
add a comment |
1 Answer
1
active
oldest
votes
Hint: Let $alpha$ denote the standard basis (that is, $e_1 = (1,0,dots,0)$ is the first vector, $e_2 = (0,1,0,dots,0)$ is the second, and so on). Then if we take $Q$ to be the matrix whose $i$th column is the $i$th vector in $gamma$, we have
$$
Q = [I_n]_{gamma to alpha}
$$
We can that this is the case as follows: if $v_1,dots,v_n$ are the elements of $gamma$, then we note that the $i$th column of $[I_n]_{gamma to alpha}$ is given by
$$
[I_n]_{gamma to alpha} e_i =
[I_n]_{gamma to alpha} [v_i]_{gamma} = [I_n v_i]_{alpha} = [v_i]_{alpha} = v_i
$$
answered Nov 26 at 19:22
Omnomnomnom
126k788176
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%2f3014470%2fchange-of-coordinate-matrix-collary-proof%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
Hint: Let $alpha$ denote the standard basis (that is, $e_1 = (1,0,dots,0)$ is the first vector, $e_2 = (0,1,0,dots,0)$ is the second, and so on). Then if we take $Q$ to be the matrix whose $i$th column is the $i$th vector in $gamma$, we have
$$
Q = [I_n]_{gamma to alpha}
$$
We can that this is the case as follows: if $v_1,dots,v_n$ are the elements of $gamma$, then we note that the $i$th column of $[I_n]_{gamma to alpha}$ is given by
$$
[I_n]_{gamma to alpha} e_i =
[I_n]_{gamma to alpha} [v_i]_{gamma} = [I_n v_i]_{alpha} = [v_i]_{alpha} = v_i
$$
answered Nov 26 at 19:22
Omnomnomnom
126k788176
add a comment |
Hint: Let $alpha$ denote the standard basis (that is, $e_1 = (1,0,dots,0)$ is the first vector, $e_2 = (0,1,0,dots,0)$ is the second, and so on). Then if we take $Q$ to be the matrix whose $i$th column is the $i$th vector in $gamma$, we have
$$
Q = [I_n]_{gamma to alpha}
$$
We can that this is the case as follows: if $v_1,dots,v_n$ are the elements of $gamma$, then we note that the $i$th column of $[I_n]_{gamma to alpha}$ is given by
$$
[I_n]_{gamma to alpha} e_i =
[I_n]_{gamma to alpha} [v_i]_{gamma} = [I_n v_i]_{alpha} = [v_i]_{alpha} = v_i
$$
answered Nov 26 at 19:22
Omnomnomnom
126k788176
add a comment |
Hint: Let $alpha$ denote the standard basis (that is, $e_1 = (1,0,dots,0)$ is the first vector, $e_2 = (0,1,0,dots,0)$ is the second, and so on). Then if we take $Q$ to be the matrix whose $i$th column is the $i$th vector in $gamma$, we have
$$
Q = [I_n]_{gamma to alpha}
$$
We can that this is the case as follows: if $v_1,dots,v_n$ are the elements of $gamma$, then we note that the $i$th column of $[I_n]_{gamma to alpha}$ is given by
$$
[I_n]_{gamma to alpha} e_i =
[I_n]_{gamma to alpha} [v_i]_{gamma} = [I_n v_i]_{alpha} = [v_i]_{alpha} = v_i
$$
answered Nov 26 at 19:22
Omnomnomnom
126k788176
Hint: Let $alpha$ denote the standard basis (that is, $e_1 = (1,0,dots,0)$ is the first vector, $e_2 = (0,1,0,dots,0)$ is the second, and so on). Then if we take $Q$ to be the matrix whose $i$th column is the $i$th vector in $gamma$, we have
$$
Q = [I_n]_{gamma to alpha}
$$
We can that this is the case as follows: if $v_1,dots,v_n$ are the elements of $gamma$, then we note that the $i$th column of $[I_n]_{gamma to alpha}$ is given by
$$
[I_n]_{gamma to alpha} e_i =
[I_n]_{gamma to alpha} [v_i]_{gamma} = [I_n v_i]_{alpha} = [v_i]_{alpha} = v_i
$$
answered Nov 26 at 19:22
Omnomnomnom
126k788176
answered Nov 26 at 19:22
Omnomnomnom
126k788176
answered Nov 26 at 19:22
Omnomnomnom
126k788176
answered Nov 26 at 19:22
Omnomnomnom
126k788176
126k788176
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3014470%2fchange-of-coordinate-matrix-collary-proof%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
The statement is not correct as stated: if $[L_A]_{gamma} = Q^{-1}AQ$, then there are many matrices that $Q$ could be. For instance: for any such matrix $Q$ and non-zero $k in Bbb F$, $kQ$ will satisfy $(kQ)^{-1}A(kQ) = [L_A]_{gamma}$. I suspect that you are meant to prove the converse, that is: if $Q$ is the matrix whose columns are the elements of $gamma$, then $[L_A]_gamma = Q^{-1}AQ$.
– Omnomnomnom
Nov 26 at 19:17
@Omnomnomnom Thanks, I've revisited the original question and found that you're right. English is not my first language, so sometimes it's hard for me to conveyed the question properly. I'll try considering $alpha$ as standard basis!! Thanks again.
– Ann P.
Nov 27 at 0:00