Mixing
The shape of an orthonormal matrix
Let $U$ be an orthonormal matrix. $U$ has $n$ rows and $n$ columns. We know that the first row is given by the following vector
$$bigg ( frac{1}{sqrt{n}}, ldots frac{1}{sqrt{n}} bigg).$$
Let's consider a column vector $X$.
$$X = (mu, ... mu)^{T}.$$
I am to prove that
$$UX = big( sqrt{n} mu, 0, 0, ldots, 0)^{T} tag{1}.$$
Both $U$ and $X$ are such matrices that their dimension fits and they can be multiplied in the way shown above.
How can I prove $(1)$?
linear-algebra
asked Nov 21 '18 at 11:45
Hendrra
1,079516
add a comment |
Let $U$ be an orthonormal matrix. $U$ has $n$ rows and $n$ columns. We know that the first row is given by the following vector
$$bigg ( frac{1}{sqrt{n}}, ldots frac{1}{sqrt{n}} bigg).$$
Let's consider a column vector $X$.
$$X = (mu, ... mu)^{T}.$$
I am to prove that
$$UX = big( sqrt{n} mu, 0, 0, ldots, 0)^{T} tag{1}.$$
Both $U$ and $X$ are such matrices that their dimension fits and they can be multiplied in the way shown above.
How can I prove $(1)$?
linear-algebra
asked Nov 21 '18 at 11:45
Hendrra
1,079516
add a comment |
Let $U$ be an orthonormal matrix. $U$ has $n$ rows and $n$ columns. We know that the first row is given by the following vector
$$bigg ( frac{1}{sqrt{n}}, ldots frac{1}{sqrt{n}} bigg).$$
Let's consider a column vector $X$.
$$X = (mu, ... mu)^{T}.$$
I am to prove that
$$UX = big( sqrt{n} mu, 0, 0, ldots, 0)^{T} tag{1}.$$
Both $U$ and $X$ are such matrices that their dimension fits and they can be multiplied in the way shown above.
How can I prove $(1)$?
linear-algebra
asked Nov 21 '18 at 11:45
Hendrra
1,079516
Let $U$ be an orthonormal matrix. $U$ has $n$ rows and $n$ columns. We know that the first row is given by the following vector
$$bigg ( frac{1}{sqrt{n}}, ldots frac{1}{sqrt{n}} bigg).$$
Let's consider a column vector $X$.
$$X = (mu, ... mu)^{T}.$$
I am to prove that
$$UX = big( sqrt{n} mu, 0, 0, ldots, 0)^{T} tag{1}.$$
Both $U$ and $X$ are such matrices that their dimension fits and they can be multiplied in the way shown above.
How can I prove $(1)$?
linear-algebra
linear-algebra
asked Nov 21 '18 at 11:45
Hendrra
1,079516
asked Nov 21 '18 at 11:45
Hendrra
1,079516
asked Nov 21 '18 at 11:45
Hendrra
1,079516
asked Nov 21 '18 at 11:45
Hendrra
1,079516
asked Nov 21 '18 at 11:45
Hendrra
1,079516
1,079516
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
It should be clear that the first entry in $UX$ is given by
$( frac{1}{sqrt{n}}, ldots frac{1}{sqrt{n}} ) cdot(mu, ... mu)^{T}= sqrt{n} mu$.
Now suppose that the $j -th$ row of $U$ has the form $(a_1,a_2,...,a_n)$. Since $U$ is orthonormal we have
$0=(a_1,a_2,...,a_n) cdot ( frac{1}{sqrt{n}}, ldots frac{1}{sqrt{n}} )^T =frac{1}{sqrt{n}}(a_1+a_2+...+a_n)$, thus $a_1+a_2+...+a_n=0$.
The $j -th$ entry in $UX$ is therefore
$$= mu(a_1+a_2+...+a_n)=0.$$
answered Nov 21 '18 at 12:00
Fred
44.2k1845
add a comment |
Let me give an alternative to @Fred's answer, or at least a more geometric version, based on three observations:
The rows $u_1, u_2, ldots u_n$ of $U$ are orthonormal vectors, i.e., each has length $1$, and they are all perpendicular, i.e., $u_i cdot u_j = 0$ if $i ne j$, and $u_i cdot u_j = 1$ if $i = j$.
The product $UX$ consists of the dot product $u_i cdot X$ of each row with the vector $X$.
The vector $X$ is a scalar multiple $c u_1$ of the first row, with the scalar being $c = sqrt{n} mu$.
So what's the dot product of the first row with $X$? It's just $c u_1 cdot u_1 = c $.
What about the dot product $u_i cdot X$ for any $i ne 1$? Well, because $X$ is parallel to $u_1$, but $u_1$ is perpendicular to all the other $u_i$, we must have $u_i cdot X = 0$.
answered Nov 21 '18 at 12:32
John Hughes
62.4k24090
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%2f3007611%2fthe-shape-of-an-orthonormal-matrix%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
It should be clear that the first entry in $UX$ is given by
$( frac{1}{sqrt{n}}, ldots frac{1}{sqrt{n}} ) cdot(mu, ... mu)^{T}= sqrt{n} mu$.
Now suppose that the $j -th$ row of $U$ has the form $(a_1,a_2,...,a_n)$. Since $U$ is orthonormal we have
$0=(a_1,a_2,...,a_n) cdot ( frac{1}{sqrt{n}}, ldots frac{1}{sqrt{n}} )^T =frac{1}{sqrt{n}}(a_1+a_2+...+a_n)$, thus $a_1+a_2+...+a_n=0$.
The $j -th$ entry in $UX$ is therefore
$$= mu(a_1+a_2+...+a_n)=0.$$
answered Nov 21 '18 at 12:00
Fred
44.2k1845
add a comment |
It should be clear that the first entry in $UX$ is given by
$( frac{1}{sqrt{n}}, ldots frac{1}{sqrt{n}} ) cdot(mu, ... mu)^{T}= sqrt{n} mu$.
Now suppose that the $j -th$ row of $U$ has the form $(a_1,a_2,...,a_n)$. Since $U$ is orthonormal we have
$0=(a_1,a_2,...,a_n) cdot ( frac{1}{sqrt{n}}, ldots frac{1}{sqrt{n}} )^T =frac{1}{sqrt{n}}(a_1+a_2+...+a_n)$, thus $a_1+a_2+...+a_n=0$.
The $j -th$ entry in $UX$ is therefore
$$= mu(a_1+a_2+...+a_n)=0.$$
answered Nov 21 '18 at 12:00
Fred
44.2k1845
add a comment |
It should be clear that the first entry in $UX$ is given by
$( frac{1}{sqrt{n}}, ldots frac{1}{sqrt{n}} ) cdot(mu, ... mu)^{T}= sqrt{n} mu$.
Now suppose that the $j -th$ row of $U$ has the form $(a_1,a_2,...,a_n)$. Since $U$ is orthonormal we have
$0=(a_1,a_2,...,a_n) cdot ( frac{1}{sqrt{n}}, ldots frac{1}{sqrt{n}} )^T =frac{1}{sqrt{n}}(a_1+a_2+...+a_n)$, thus $a_1+a_2+...+a_n=0$.
The $j -th$ entry in $UX$ is therefore
$$= mu(a_1+a_2+...+a_n)=0.$$
answered Nov 21 '18 at 12:00
Fred
44.2k1845
It should be clear that the first entry in $UX$ is given by
$( frac{1}{sqrt{n}}, ldots frac{1}{sqrt{n}} ) cdot(mu, ... mu)^{T}= sqrt{n} mu$.
Now suppose that the $j -th$ row of $U$ has the form $(a_1,a_2,...,a_n)$. Since $U$ is orthonormal we have
$0=(a_1,a_2,...,a_n) cdot ( frac{1}{sqrt{n}}, ldots frac{1}{sqrt{n}} )^T =frac{1}{sqrt{n}}(a_1+a_2+...+a_n)$, thus $a_1+a_2+...+a_n=0$.
The $j -th$ entry in $UX$ is therefore
$$= mu(a_1+a_2+...+a_n)=0.$$
answered Nov 21 '18 at 12:00
Fred
44.2k1845
edited Nov 21 '18 at 12:26
answered Nov 21 '18 at 12:00
Fred
44.2k1845
answered Nov 21 '18 at 12:00
Fred
44.2k1845
answered Nov 21 '18 at 12:00
Fred
44.2k1845
44.2k1845
add a comment |
add a comment |
Let me give an alternative to @Fred's answer, or at least a more geometric version, based on three observations:
The rows $u_1, u_2, ldots u_n$ of $U$ are orthonormal vectors, i.e., each has length $1$, and they are all perpendicular, i.e., $u_i cdot u_j = 0$ if $i ne j$, and $u_i cdot u_j = 1$ if $i = j$.
The product $UX$ consists of the dot product $u_i cdot X$ of each row with the vector $X$.
The vector $X$ is a scalar multiple $c u_1$ of the first row, with the scalar being $c = sqrt{n} mu$.
So what's the dot product of the first row with $X$? It's just $c u_1 cdot u_1 = c $.
What about the dot product $u_i cdot X$ for any $i ne 1$? Well, because $X$ is parallel to $u_1$, but $u_1$ is perpendicular to all the other $u_i$, we must have $u_i cdot X = 0$.
answered Nov 21 '18 at 12:32
John Hughes
62.4k24090
add a comment |
Let me give an alternative to @Fred's answer, or at least a more geometric version, based on three observations:
The rows $u_1, u_2, ldots u_n$ of $U$ are orthonormal vectors, i.e., each has length $1$, and they are all perpendicular, i.e., $u_i cdot u_j = 0$ if $i ne j$, and $u_i cdot u_j = 1$ if $i = j$.
The product $UX$ consists of the dot product $u_i cdot X$ of each row with the vector $X$.
The vector $X$ is a scalar multiple $c u_1$ of the first row, with the scalar being $c = sqrt{n} mu$.
So what's the dot product of the first row with $X$? It's just $c u_1 cdot u_1 = c $.
What about the dot product $u_i cdot X$ for any $i ne 1$? Well, because $X$ is parallel to $u_1$, but $u_1$ is perpendicular to all the other $u_i$, we must have $u_i cdot X = 0$.
answered Nov 21 '18 at 12:32
John Hughes
62.4k24090
add a comment |
Let me give an alternative to @Fred's answer, or at least a more geometric version, based on three observations:
The rows $u_1, u_2, ldots u_n$ of $U$ are orthonormal vectors, i.e., each has length $1$, and they are all perpendicular, i.e., $u_i cdot u_j = 0$ if $i ne j$, and $u_i cdot u_j = 1$ if $i = j$.
The product $UX$ consists of the dot product $u_i cdot X$ of each row with the vector $X$.
The vector $X$ is a scalar multiple $c u_1$ of the first row, with the scalar being $c = sqrt{n} mu$.
So what's the dot product of the first row with $X$? It's just $c u_1 cdot u_1 = c $.
What about the dot product $u_i cdot X$ for any $i ne 1$? Well, because $X$ is parallel to $u_1$, but $u_1$ is perpendicular to all the other $u_i$, we must have $u_i cdot X = 0$.
answered Nov 21 '18 at 12:32
John Hughes
62.4k24090
Let me give an alternative to @Fred's answer, or at least a more geometric version, based on three observations:
The rows $u_1, u_2, ldots u_n$ of $U$ are orthonormal vectors, i.e., each has length $1$, and they are all perpendicular, i.e., $u_i cdot u_j = 0$ if $i ne j$, and $u_i cdot u_j = 1$ if $i = j$.
The product $UX$ consists of the dot product $u_i cdot X$ of each row with the vector $X$.
The vector $X$ is a scalar multiple $c u_1$ of the first row, with the scalar being $c = sqrt{n} mu$.
So what's the dot product of the first row with $X$? It's just $c u_1 cdot u_1 = c $.
What about the dot product $u_i cdot X$ for any $i ne 1$? Well, because $X$ is parallel to $u_1$, but $u_1$ is perpendicular to all the other $u_i$, we must have $u_i cdot X = 0$.
answered Nov 21 '18 at 12:32
John Hughes
62.4k24090
answered Nov 21 '18 at 12:32
John Hughes
62.4k24090
answered Nov 21 '18 at 12:32
John Hughes
62.4k24090
answered Nov 21 '18 at 12:32
John Hughes
62.4k24090
62.4k24090
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%2f3007611%2fthe-shape-of-an-orthonormal-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
