Mixing
matrix algebra and idempotent matrix
$begingroup$
I'm having a little trouble understanding a few derivations in my book for least squares regression.
$textbf{Question 1}$:
If $textbf{M}^0 textbf{i} = [textbf{I} - frac{1}{n}textbf{ii'}]textbf{i} = textbf{i}-frac{1}{n}textbf{i(i'i)}=textbf{0}$
Then how does this imply that $textbf{i'M}^0 = textbf{0'}$? Shouldn't it be $(textbf{M}^0textbf{i})' = textbf{0}'$?
and moreover how do we get that
$$sum_{i=1}^n(x_i -bar x)=textbf{i}'[textbf{M}^0textbf{x}]$$
It's not explicitly stated, but i'm pretty sure $textbf{i}$ is a row vector of 1's.
And also, evidently $textbf{M}^0$ is being defined as $[textbf{I}-frac{1}{n}textbf{i}textbf{i}']$
$textbf{Question 2}$:
Why is it the case that for the expansion of $textbf{e}'_0textbf{e}_0 = (textbf{y-Xb})'(textbf{y - xb})= $
$$textbf{y'y - b'X'y - y'Xb + b'X'Xb} = textbf{y'y - 2y'Xb + b'X'Xb}$$
In particular, how can we group the $textbf{b'X'y}$ and $textbf{y'Xb}$ as the same?
Q1 is from the appendix of William H. Greene - Econometric analysis 7th ed.
linear-algebra matrix-equations least-squares idempotents
asked Jan 21 at 6:52
elbartoelbarto
1,555827
$endgroup$
add a comment |
$begingroup$
I'm having a little trouble understanding a few derivations in my book for least squares regression.
$textbf{Question 1}$:
If $textbf{M}^0 textbf{i} = [textbf{I} - frac{1}{n}textbf{ii'}]textbf{i} = textbf{i}-frac{1}{n}textbf{i(i'i)}=textbf{0}$
Then how does this imply that $textbf{i'M}^0 = textbf{0'}$? Shouldn't it be $(textbf{M}^0textbf{i})' = textbf{0}'$?
and moreover how do we get that
$$sum_{i=1}^n(x_i -bar x)=textbf{i}'[textbf{M}^0textbf{x}]$$
It's not explicitly stated, but i'm pretty sure $textbf{i}$ is a row vector of 1's.
And also, evidently $textbf{M}^0$ is being defined as $[textbf{I}-frac{1}{n}textbf{i}textbf{i}']$
$textbf{Question 2}$:
Why is it the case that for the expansion of $textbf{e}'_0textbf{e}_0 = (textbf{y-Xb})'(textbf{y - xb})= $
$$textbf{y'y - b'X'y - y'Xb + b'X'Xb} = textbf{y'y - 2y'Xb + b'X'Xb}$$
In particular, how can we group the $textbf{b'X'y}$ and $textbf{y'Xb}$ as the same?
Q1 is from the appendix of William H. Greene - Econometric analysis 7th ed.
linear-algebra matrix-equations least-squares idempotents
asked Jan 21 at 6:52
elbartoelbarto
1,555827
$endgroup$
add a comment |
$begingroup$
I'm having a little trouble understanding a few derivations in my book for least squares regression.
$textbf{Question 1}$:
If $textbf{M}^0 textbf{i} = [textbf{I} - frac{1}{n}textbf{ii'}]textbf{i} = textbf{i}-frac{1}{n}textbf{i(i'i)}=textbf{0}$
Then how does this imply that $textbf{i'M}^0 = textbf{0'}$? Shouldn't it be $(textbf{M}^0textbf{i})' = textbf{0}'$?
and moreover how do we get that
$$sum_{i=1}^n(x_i -bar x)=textbf{i}'[textbf{M}^0textbf{x}]$$
It's not explicitly stated, but i'm pretty sure $textbf{i}$ is a row vector of 1's.
And also, evidently $textbf{M}^0$ is being defined as $[textbf{I}-frac{1}{n}textbf{i}textbf{i}']$
$textbf{Question 2}$:
Why is it the case that for the expansion of $textbf{e}'_0textbf{e}_0 = (textbf{y-Xb})'(textbf{y - xb})= $
$$textbf{y'y - b'X'y - y'Xb + b'X'Xb} = textbf{y'y - 2y'Xb + b'X'Xb}$$
In particular, how can we group the $textbf{b'X'y}$ and $textbf{y'Xb}$ as the same?
Q1 is from the appendix of William H. Greene - Econometric analysis 7th ed.
linear-algebra matrix-equations least-squares idempotents
asked Jan 21 at 6:52
elbartoelbarto
1,555827
$endgroup$
I'm having a little trouble understanding a few derivations in my book for least squares regression.
$textbf{Question 1}$:
If $textbf{M}^0 textbf{i} = [textbf{I} - frac{1}{n}textbf{ii'}]textbf{i} = textbf{i}-frac{1}{n}textbf{i(i'i)}=textbf{0}$
Then how does this imply that $textbf{i'M}^0 = textbf{0'}$? Shouldn't it be $(textbf{M}^0textbf{i})' = textbf{0}'$?
and moreover how do we get that
$$sum_{i=1}^n(x_i -bar x)=textbf{i}'[textbf{M}^0textbf{x}]$$
It's not explicitly stated, but i'm pretty sure $textbf{i}$ is a row vector of 1's.
And also, evidently $textbf{M}^0$ is being defined as $[textbf{I}-frac{1}{n}textbf{i}textbf{i}']$
$textbf{Question 2}$:
Why is it the case that for the expansion of $textbf{e}'_0textbf{e}_0 = (textbf{y-Xb})'(textbf{y - xb})= $
$$textbf{y'y - b'X'y - y'Xb + b'X'Xb} = textbf{y'y - 2y'Xb + b'X'Xb}$$
In particular, how can we group the $textbf{b'X'y}$ and $textbf{y'Xb}$ as the same?
Q1 is from the appendix of William H. Greene - Econometric analysis 7th ed.
linear-algebra matrix-equations least-squares idempotents
linear-algebra matrix-equations least-squares idempotents
asked Jan 21 at 6:52
elbartoelbarto
1,555827
asked Jan 21 at 6:52
elbartoelbarto
1,555827
edited Jan 21 at 7:27
elbarto
asked Jan 21 at 6:52
elbartoelbarto
1,555827
asked Jan 21 at 6:52
elbartoelbarto
1,555827
asked Jan 21 at 6:52
elbartoelbarto
1,555827
1,555827
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For Question 1,
Then how does this imply that $textbf{i'M}^0 = textbf{0'}$?
Shouldn't it be $(textbf{M}^0textbf{i})' = textbf{0}'$?
$i'M^0 = (M^0 i)^{'}$, since $M^0$ is symmetric ($(M^0)^T=M^0$).
For Question 2, both $b'X'y$ and $y'Xb$ are scalars and taking the transpose of a scalar doesn't change its value, hence $b'X'y = y'Xb$.
answered Jan 21 at 7:40
pointguard0pointguard0
1,50211021
$endgroup$
$begingroup$
Thank you for that :)
$endgroup$
– elbarto
Jan 21 at 21:42
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%2f3081577%2fmatrix-algebra-and-idempotent-matrix%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
$begingroup$
For Question 1,
Then how does this imply that $textbf{i'M}^0 = textbf{0'}$?
Shouldn't it be $(textbf{M}^0textbf{i})' = textbf{0}'$?
$i'M^0 = (M^0 i)^{'}$, since $M^0$ is symmetric ($(M^0)^T=M^0$).
For Question 2, both $b'X'y$ and $y'Xb$ are scalars and taking the transpose of a scalar doesn't change its value, hence $b'X'y = y'Xb$.
answered Jan 21 at 7:40
pointguard0pointguard0
1,50211021
$endgroup$
$begingroup$
Thank you for that :)
$endgroup$
– elbarto
Jan 21 at 21:42
add a comment |
$begingroup$
For Question 1,
Then how does this imply that $textbf{i'M}^0 = textbf{0'}$?
Shouldn't it be $(textbf{M}^0textbf{i})' = textbf{0}'$?
$i'M^0 = (M^0 i)^{'}$, since $M^0$ is symmetric ($(M^0)^T=M^0$).
For Question 2, both $b'X'y$ and $y'Xb$ are scalars and taking the transpose of a scalar doesn't change its value, hence $b'X'y = y'Xb$.
answered Jan 21 at 7:40
pointguard0pointguard0
1,50211021
$endgroup$
$begingroup$
Thank you for that :)
$endgroup$
– elbarto
Jan 21 at 21:42
add a comment |
$begingroup$
For Question 1,
Then how does this imply that $textbf{i'M}^0 = textbf{0'}$?
Shouldn't it be $(textbf{M}^0textbf{i})' = textbf{0}'$?
$i'M^0 = (M^0 i)^{'}$, since $M^0$ is symmetric ($(M^0)^T=M^0$).
For Question 2, both $b'X'y$ and $y'Xb$ are scalars and taking the transpose of a scalar doesn't change its value, hence $b'X'y = y'Xb$.
answered Jan 21 at 7:40
pointguard0pointguard0
1,50211021
$endgroup$
For Question 1,
Then how does this imply that $textbf{i'M}^0 = textbf{0'}$?
Shouldn't it be $(textbf{M}^0textbf{i})' = textbf{0}'$?
$i'M^0 = (M^0 i)^{'}$, since $M^0$ is symmetric ($(M^0)^T=M^0$).
For Question 2, both $b'X'y$ and $y'Xb$ are scalars and taking the transpose of a scalar doesn't change its value, hence $b'X'y = y'Xb$.
answered Jan 21 at 7:40
pointguard0pointguard0
1,50211021
answered Jan 21 at 7:40
pointguard0pointguard0
1,50211021
answered Jan 21 at 7:40
pointguard0pointguard0
1,50211021
answered Jan 21 at 7:40
pointguard0pointguard0
1,50211021
1,50211021
$begingroup$
Thank you for that :)
$endgroup$
– elbarto
Jan 21 at 21:42
add a comment |
$begingroup$
Thank you for that :)
$endgroup$
– elbarto
Jan 21 at 21:42
$begingroup$
Thank you for that :)
$endgroup$
– elbarto
Jan 21 at 21:42
$begingroup$
Thank you for that :)
$endgroup$
– elbarto
Jan 21 at 21:42
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%2f3081577%2fmatrix-algebra-and-idempotent-matrix%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
