Insightful proof for $UV^T = sum_{i=1}^n u_iv_i^T $
$begingroup$
Let $U = [u_1, u_2,dots ,u_n] in mathbb{R}^{mtimes n}$ with $u_i in mathbb{R}^m$ and $V = [v_1, v_2,dots ,v_n] in mathbb{R}^{ptimes n}$ with $v_i in mathbb{R}^p$. Then
$$UV^T = sum_{i=1}^n u_iv_i^T in mathbb{R}^{mtimes n} $$
I am trying to get more intuitive understanding of matrix multiplications (say thinking in terms of linear mapping) and for the above I know how to prove it using 'tedious' algebra but I am looking for more insightful proof that will illuminate the essence of the above identity.
linear-algebra matrices linear-transformations
asked Dec 13 '18 at 4:38
hi15hi15
1456
$endgroup$
add a comment |
$begingroup$
Let $U = [u_1, u_2,dots ,u_n] in mathbb{R}^{mtimes n}$ with $u_i in mathbb{R}^m$ and $V = [v_1, v_2,dots ,v_n] in mathbb{R}^{ptimes n}$ with $v_i in mathbb{R}^p$. Then
$$UV^T = sum_{i=1}^n u_iv_i^T in mathbb{R}^{mtimes n} $$
I am trying to get more intuitive understanding of matrix multiplications (say thinking in terms of linear mapping) and for the above I know how to prove it using 'tedious' algebra but I am looking for more insightful proof that will illuminate the essence of the above identity.
linear-algebra matrices linear-transformations
asked Dec 13 '18 at 4:38
hi15hi15
1456
$endgroup$
$begingroup$
the matrix is in $mathbb R^{mtimes p}$
$endgroup$
– daw
Dec 13 '18 at 7:31
add a comment |
$begingroup$
Let $U = [u_1, u_2,dots ,u_n] in mathbb{R}^{mtimes n}$ with $u_i in mathbb{R}^m$ and $V = [v_1, v_2,dots ,v_n] in mathbb{R}^{ptimes n}$ with $v_i in mathbb{R}^p$. Then
$$UV^T = sum_{i=1}^n u_iv_i^T in mathbb{R}^{mtimes n} $$
I am trying to get more intuitive understanding of matrix multiplications (say thinking in terms of linear mapping) and for the above I know how to prove it using 'tedious' algebra but I am looking for more insightful proof that will illuminate the essence of the above identity.
linear-algebra matrices linear-transformations
asked Dec 13 '18 at 4:38
hi15hi15
1456
$endgroup$
Let $U = [u_1, u_2,dots ,u_n] in mathbb{R}^{mtimes n}$ with $u_i in mathbb{R}^m$ and $V = [v_1, v_2,dots ,v_n] in mathbb{R}^{ptimes n}$ with $v_i in mathbb{R}^p$. Then
$$UV^T = sum_{i=1}^n u_iv_i^T in mathbb{R}^{mtimes n} $$
I am trying to get more intuitive understanding of matrix multiplications (say thinking in terms of linear mapping) and for the above I know how to prove it using 'tedious' algebra but I am looking for more insightful proof that will illuminate the essence of the above identity.
linear-algebra matrices linear-transformations
linear-algebra matrices linear-transformations
asked Dec 13 '18 at 4:38
hi15hi15
1456
asked Dec 13 '18 at 4:38
hi15hi15
1456
asked Dec 13 '18 at 4:38
hi15hi15
1456
asked Dec 13 '18 at 4:38
hi15hi15
1456
asked Dec 13 '18 at 4:38
hi15hi15
1456
1456
$begingroup$
the matrix is in $mathbb R^{mtimes p}$
$endgroup$
– daw
Dec 13 '18 at 7:31
add a comment |
$begingroup$
the matrix is in $mathbb R^{mtimes p}$
$endgroup$
– daw
Dec 13 '18 at 7:31
$begingroup$
the matrix is in $mathbb R^{mtimes p}$
$endgroup$
– daw
Dec 13 '18 at 7:31
$begingroup$
the matrix is in $mathbb R^{mtimes p}$
$endgroup$
– daw
Dec 13 '18 at 7:31
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Think of it like this. You are constructing a matrix $Ainmathbb{R}^{mtimes n}$ via the product $A = UV^T$. $A$ is a linear map from $mathbb{R}^n$ to $mathbb{R}^m$, but by using this product, you are sort of taking a detour through $mathbb{R}^p$. The result of multiplying $UV^T$ times a vector $w$ will be a linear combination of the $p$ columns of $U$ (so the range of $A$ has dimension at most $p$) and the mapping from $w$ to which linear combination we take is controlled by $V^T$.
answered Dec 13 '18 at 4:53
whpowell96whpowell96
58115
$endgroup$
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%2f3037605%2finsightful-proof-for-uvt-sum-i-1n-u-iv-it%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
$begingroup$
Think of it like this. You are constructing a matrix $Ainmathbb{R}^{mtimes n}$ via the product $A = UV^T$. $A$ is a linear map from $mathbb{R}^n$ to $mathbb{R}^m$, but by using this product, you are sort of taking a detour through $mathbb{R}^p$. The result of multiplying $UV^T$ times a vector $w$ will be a linear combination of the $p$ columns of $U$ (so the range of $A$ has dimension at most $p$) and the mapping from $w$ to which linear combination we take is controlled by $V^T$.
answered Dec 13 '18 at 4:53
whpowell96whpowell96
58115
$endgroup$
add a comment |
$begingroup$
Think of it like this. You are constructing a matrix $Ainmathbb{R}^{mtimes n}$ via the product $A = UV^T$. $A$ is a linear map from $mathbb{R}^n$ to $mathbb{R}^m$, but by using this product, you are sort of taking a detour through $mathbb{R}^p$. The result of multiplying $UV^T$ times a vector $w$ will be a linear combination of the $p$ columns of $U$ (so the range of $A$ has dimension at most $p$) and the mapping from $w$ to which linear combination we take is controlled by $V^T$.
answered Dec 13 '18 at 4:53
whpowell96whpowell96
58115
$endgroup$
add a comment |
$begingroup$
Think of it like this. You are constructing a matrix $Ainmathbb{R}^{mtimes n}$ via the product $A = UV^T$. $A$ is a linear map from $mathbb{R}^n$ to $mathbb{R}^m$, but by using this product, you are sort of taking a detour through $mathbb{R}^p$. The result of multiplying $UV^T$ times a vector $w$ will be a linear combination of the $p$ columns of $U$ (so the range of $A$ has dimension at most $p$) and the mapping from $w$ to which linear combination we take is controlled by $V^T$.
answered Dec 13 '18 at 4:53
whpowell96whpowell96
58115
$endgroup$
Think of it like this. You are constructing a matrix $Ainmathbb{R}^{mtimes n}$ via the product $A = UV^T$. $A$ is a linear map from $mathbb{R}^n$ to $mathbb{R}^m$, but by using this product, you are sort of taking a detour through $mathbb{R}^p$. The result of multiplying $UV^T$ times a vector $w$ will be a linear combination of the $p$ columns of $U$ (so the range of $A$ has dimension at most $p$) and the mapping from $w$ to which linear combination we take is controlled by $V^T$.
answered Dec 13 '18 at 4:53
whpowell96whpowell96
58115
answered Dec 13 '18 at 4:53
whpowell96whpowell96
58115
answered Dec 13 '18 at 4:53
whpowell96whpowell96
58115
answered Dec 13 '18 at 4:53
whpowell96whpowell96
58115
58115
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%2f3037605%2finsightful-proof-for-uvt-sum-i-1n-u-iv-it%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
$begingroup$
the matrix is in $mathbb R^{mtimes p}$
$endgroup$
– daw
Dec 13 '18 at 7:31