Mixing
Is there any connection between QR and SVD of a matrix?
$begingroup$
Is it possible to draw any parallels between the SVD and QR decomposition of a matrix?
Moreover, for a given matrix $mathbf{A}inmathbb{R}^{ntimes m}$, under what conditions, the $mathbf{U}$ matrix coming from singular value decomposition of $mathbf{A}$ is equal to $mathbf{Q}$ matrix obtained via QR-decomposition of $mathbf{A}$
matrix-decomposition
asked Jul 6 '17 at 19:27
NAASINAASI
494414
$endgroup$
add a comment |
$begingroup$
Is it possible to draw any parallels between the SVD and QR decomposition of a matrix?
Moreover, for a given matrix $mathbf{A}inmathbb{R}^{ntimes m}$, under what conditions, the $mathbf{U}$ matrix coming from singular value decomposition of $mathbf{A}$ is equal to $mathbf{Q}$ matrix obtained via QR-decomposition of $mathbf{A}$
matrix-decomposition
asked Jul 6 '17 at 19:27
NAASINAASI
494414
$endgroup$
$begingroup$
Why does your title differ from the question ?
$endgroup$
– Yves Daoust
Jul 6 '17 at 19:29
$begingroup$
I realized that difference after posting. Anyways let me come up with a better title
$endgroup$
– NAASI
Jul 6 '17 at 19:39
add a comment |
$begingroup$
Is it possible to draw any parallels between the SVD and QR decomposition of a matrix?
Moreover, for a given matrix $mathbf{A}inmathbb{R}^{ntimes m}$, under what conditions, the $mathbf{U}$ matrix coming from singular value decomposition of $mathbf{A}$ is equal to $mathbf{Q}$ matrix obtained via QR-decomposition of $mathbf{A}$
matrix-decomposition
asked Jul 6 '17 at 19:27
NAASINAASI
494414
$endgroup$
Is it possible to draw any parallels between the SVD and QR decomposition of a matrix?
Moreover, for a given matrix $mathbf{A}inmathbb{R}^{ntimes m}$, under what conditions, the $mathbf{U}$ matrix coming from singular value decomposition of $mathbf{A}$ is equal to $mathbf{Q}$ matrix obtained via QR-decomposition of $mathbf{A}$
matrix-decomposition
matrix-decomposition
asked Jul 6 '17 at 19:27
NAASINAASI
494414
asked Jul 6 '17 at 19:27
NAASINAASI
494414
edited Jul 6 '17 at 19:41
NAASI
asked Jul 6 '17 at 19:27
NAASINAASI
494414
asked Jul 6 '17 at 19:27
NAASINAASI
494414
asked Jul 6 '17 at 19:27
NAASINAASI
494414
494414
$begingroup$
Why does your title differ from the question ?
$endgroup$
– Yves Daoust
Jul 6 '17 at 19:29
$begingroup$
I realized that difference after posting. Anyways let me come up with a better title
$endgroup$
– NAASI
Jul 6 '17 at 19:39
add a comment |
$begingroup$
Why does your title differ from the question ?
$endgroup$
– Yves Daoust
Jul 6 '17 at 19:29
$begingroup$
I realized that difference after posting. Anyways let me come up with a better title
$endgroup$
– NAASI
Jul 6 '17 at 19:39
$begingroup$
Why does your title differ from the question ?
$endgroup$
– Yves Daoust
Jul 6 '17 at 19:29
$begingroup$
Why does your title differ from the question ?
$endgroup$
– Yves Daoust
Jul 6 '17 at 19:29
$begingroup$
I realized that difference after posting. Anyways let me come up with a better title
$endgroup$
– NAASI
Jul 6 '17 at 19:39
$begingroup$
I realized that difference after posting. Anyways let me come up with a better title
$endgroup$
– NAASI
Jul 6 '17 at 19:39
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I don't know if there really is a link between them... Notice that those decompositions are not unique, so the question is a bit unclear. But here is some thoughts on your second question. Maybe it can help you ...
Consider $A in Bbb C^{m times n}$, $m geq n$. (In the other case, just take the transpose...)
We first fix some notations :
1) A $QR$ decomposition of $A$ is any decomposition $A = QR = Q_1R_1$, where $Q = begin{pmatrix} Q_1 & Q_2 end{pmatrix} in Bbb C^{m times m}$ is a unitary matrix, $Q_1 in Bbb C^{m times n}$, and $R = begin{pmatrix} R_1 \ 0end{pmatrix} in Bbb C^{m times n}$, $R_1 in Bbb C^{n times n}$ is upper triangular (we say such a $R$ is upper triangular).
2) A $SVD$ of $A$ is any decomposition $A = U Sigma V^*$, where $Sigma = begin{pmatrix} Sigma_1 \ 0end{pmatrix} in Bbb C^{m times n}$, $Sigma_1 in Bbb C^{n times n}$ is diagonal with non negative entries (we say such a $Sigma$ is diagonal), and $U in Bbb C^{m times m}$, $V in Bbb C^{n times n}$ are unitary.
Suppose that $A$ has full column rank and let $A = QR$ be a $QR$ decomposition for $A$. There exists a $SVD$ of $A$ such that $Q = U$ if and only if $A^*A$ is diagonal.
Suppose $A^*A$ is diagonal. Then, $A^*A = R^*Q^*QR = R^*R = R_1^*R_1$ so $R_1$ must be diagonal (using the fact that $A$ has full column rank). Then $A = QR = QRI$ and we can multiply some columns of $R$ and the corresponding rows of $I$ by $-1$ so that we get $A = QR'J = U Sigma V$, where $R'$ is diagonal with non negative entries, $Q$ and $J$ are unitary. This is a $SVD$ decomposition for $A$ with $Q = U$.
Now, suppose $Q = U$ for some $SVD$ of $A$. Then we have $Sigma_1$ is diagonal with strictly positive entries (since $A$ has full column rank) and $Sigma V^* = R Rightarrow Sigma_1 V^* = R_1$ so $V^*$ is upper triangular. Since $V^*$ is unitary and upper triangular, it must be diagonal. But the columns of $V$ form a basis of eigenvectors of $A^*A$. This means $e_i in Bbb R^n$ is an eigenvector of $A^*A$ for all $i = 1, ..., n$. Therefore, $A^*A$ is diagonal.
answered Jul 8 '17 at 18:46
DesuraDesura
901514
$endgroup$
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%2f2348807%2fis-there-any-connection-between-qr-and-svd-of-a-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
$begingroup$
I don't know if there really is a link between them... Notice that those decompositions are not unique, so the question is a bit unclear. But here is some thoughts on your second question. Maybe it can help you ...
Consider $A in Bbb C^{m times n}$, $m geq n$. (In the other case, just take the transpose...)
We first fix some notations :
1) A $QR$ decomposition of $A$ is any decomposition $A = QR = Q_1R_1$, where $Q = begin{pmatrix} Q_1 & Q_2 end{pmatrix} in Bbb C^{m times m}$ is a unitary matrix, $Q_1 in Bbb C^{m times n}$, and $R = begin{pmatrix} R_1 \ 0end{pmatrix} in Bbb C^{m times n}$, $R_1 in Bbb C^{n times n}$ is upper triangular (we say such a $R$ is upper triangular).
2) A $SVD$ of $A$ is any decomposition $A = U Sigma V^*$, where $Sigma = begin{pmatrix} Sigma_1 \ 0end{pmatrix} in Bbb C^{m times n}$, $Sigma_1 in Bbb C^{n times n}$ is diagonal with non negative entries (we say such a $Sigma$ is diagonal), and $U in Bbb C^{m times m}$, $V in Bbb C^{n times n}$ are unitary.
Suppose that $A$ has full column rank and let $A = QR$ be a $QR$ decomposition for $A$. There exists a $SVD$ of $A$ such that $Q = U$ if and only if $A^*A$ is diagonal.
Suppose $A^*A$ is diagonal. Then, $A^*A = R^*Q^*QR = R^*R = R_1^*R_1$ so $R_1$ must be diagonal (using the fact that $A$ has full column rank). Then $A = QR = QRI$ and we can multiply some columns of $R$ and the corresponding rows of $I$ by $-1$ so that we get $A = QR'J = U Sigma V$, where $R'$ is diagonal with non negative entries, $Q$ and $J$ are unitary. This is a $SVD$ decomposition for $A$ with $Q = U$.
Now, suppose $Q = U$ for some $SVD$ of $A$. Then we have $Sigma_1$ is diagonal with strictly positive entries (since $A$ has full column rank) and $Sigma V^* = R Rightarrow Sigma_1 V^* = R_1$ so $V^*$ is upper triangular. Since $V^*$ is unitary and upper triangular, it must be diagonal. But the columns of $V$ form a basis of eigenvectors of $A^*A$. This means $e_i in Bbb R^n$ is an eigenvector of $A^*A$ for all $i = 1, ..., n$. Therefore, $A^*A$ is diagonal.
answered Jul 8 '17 at 18:46
DesuraDesura
901514
$endgroup$
add a comment |
$begingroup$
I don't know if there really is a link between them... Notice that those decompositions are not unique, so the question is a bit unclear. But here is some thoughts on your second question. Maybe it can help you ...
Consider $A in Bbb C^{m times n}$, $m geq n$. (In the other case, just take the transpose...)
We first fix some notations :
1) A $QR$ decomposition of $A$ is any decomposition $A = QR = Q_1R_1$, where $Q = begin{pmatrix} Q_1 & Q_2 end{pmatrix} in Bbb C^{m times m}$ is a unitary matrix, $Q_1 in Bbb C^{m times n}$, and $R = begin{pmatrix} R_1 \ 0end{pmatrix} in Bbb C^{m times n}$, $R_1 in Bbb C^{n times n}$ is upper triangular (we say such a $R$ is upper triangular).
2) A $SVD$ of $A$ is any decomposition $A = U Sigma V^*$, where $Sigma = begin{pmatrix} Sigma_1 \ 0end{pmatrix} in Bbb C^{m times n}$, $Sigma_1 in Bbb C^{n times n}$ is diagonal with non negative entries (we say such a $Sigma$ is diagonal), and $U in Bbb C^{m times m}$, $V in Bbb C^{n times n}$ are unitary.
Suppose that $A$ has full column rank and let $A = QR$ be a $QR$ decomposition for $A$. There exists a $SVD$ of $A$ such that $Q = U$ if and only if $A^*A$ is diagonal.
Suppose $A^*A$ is diagonal. Then, $A^*A = R^*Q^*QR = R^*R = R_1^*R_1$ so $R_1$ must be diagonal (using the fact that $A$ has full column rank). Then $A = QR = QRI$ and we can multiply some columns of $R$ and the corresponding rows of $I$ by $-1$ so that we get $A = QR'J = U Sigma V$, where $R'$ is diagonal with non negative entries, $Q$ and $J$ are unitary. This is a $SVD$ decomposition for $A$ with $Q = U$.
Now, suppose $Q = U$ for some $SVD$ of $A$. Then we have $Sigma_1$ is diagonal with strictly positive entries (since $A$ has full column rank) and $Sigma V^* = R Rightarrow Sigma_1 V^* = R_1$ so $V^*$ is upper triangular. Since $V^*$ is unitary and upper triangular, it must be diagonal. But the columns of $V$ form a basis of eigenvectors of $A^*A$. This means $e_i in Bbb R^n$ is an eigenvector of $A^*A$ for all $i = 1, ..., n$. Therefore, $A^*A$ is diagonal.
answered Jul 8 '17 at 18:46
DesuraDesura
901514
$endgroup$
add a comment |
$begingroup$
I don't know if there really is a link between them... Notice that those decompositions are not unique, so the question is a bit unclear. But here is some thoughts on your second question. Maybe it can help you ...
Consider $A in Bbb C^{m times n}$, $m geq n$. (In the other case, just take the transpose...)
We first fix some notations :
1) A $QR$ decomposition of $A$ is any decomposition $A = QR = Q_1R_1$, where $Q = begin{pmatrix} Q_1 & Q_2 end{pmatrix} in Bbb C^{m times m}$ is a unitary matrix, $Q_1 in Bbb C^{m times n}$, and $R = begin{pmatrix} R_1 \ 0end{pmatrix} in Bbb C^{m times n}$, $R_1 in Bbb C^{n times n}$ is upper triangular (we say such a $R$ is upper triangular).
2) A $SVD$ of $A$ is any decomposition $A = U Sigma V^*$, where $Sigma = begin{pmatrix} Sigma_1 \ 0end{pmatrix} in Bbb C^{m times n}$, $Sigma_1 in Bbb C^{n times n}$ is diagonal with non negative entries (we say such a $Sigma$ is diagonal), and $U in Bbb C^{m times m}$, $V in Bbb C^{n times n}$ are unitary.
Suppose that $A$ has full column rank and let $A = QR$ be a $QR$ decomposition for $A$. There exists a $SVD$ of $A$ such that $Q = U$ if and only if $A^*A$ is diagonal.
Suppose $A^*A$ is diagonal. Then, $A^*A = R^*Q^*QR = R^*R = R_1^*R_1$ so $R_1$ must be diagonal (using the fact that $A$ has full column rank). Then $A = QR = QRI$ and we can multiply some columns of $R$ and the corresponding rows of $I$ by $-1$ so that we get $A = QR'J = U Sigma V$, where $R'$ is diagonal with non negative entries, $Q$ and $J$ are unitary. This is a $SVD$ decomposition for $A$ with $Q = U$.
Now, suppose $Q = U$ for some $SVD$ of $A$. Then we have $Sigma_1$ is diagonal with strictly positive entries (since $A$ has full column rank) and $Sigma V^* = R Rightarrow Sigma_1 V^* = R_1$ so $V^*$ is upper triangular. Since $V^*$ is unitary and upper triangular, it must be diagonal. But the columns of $V$ form a basis of eigenvectors of $A^*A$. This means $e_i in Bbb R^n$ is an eigenvector of $A^*A$ for all $i = 1, ..., n$. Therefore, $A^*A$ is diagonal.
answered Jul 8 '17 at 18:46
DesuraDesura
901514
$endgroup$
I don't know if there really is a link between them... Notice that those decompositions are not unique, so the question is a bit unclear. But here is some thoughts on your second question. Maybe it can help you ...
Consider $A in Bbb C^{m times n}$, $m geq n$. (In the other case, just take the transpose...)
We first fix some notations :
1) A $QR$ decomposition of $A$ is any decomposition $A = QR = Q_1R_1$, where $Q = begin{pmatrix} Q_1 & Q_2 end{pmatrix} in Bbb C^{m times m}$ is a unitary matrix, $Q_1 in Bbb C^{m times n}$, and $R = begin{pmatrix} R_1 \ 0end{pmatrix} in Bbb C^{m times n}$, $R_1 in Bbb C^{n times n}$ is upper triangular (we say such a $R$ is upper triangular).
2) A $SVD$ of $A$ is any decomposition $A = U Sigma V^*$, where $Sigma = begin{pmatrix} Sigma_1 \ 0end{pmatrix} in Bbb C^{m times n}$, $Sigma_1 in Bbb C^{n times n}$ is diagonal with non negative entries (we say such a $Sigma$ is diagonal), and $U in Bbb C^{m times m}$, $V in Bbb C^{n times n}$ are unitary.
Suppose that $A$ has full column rank and let $A = QR$ be a $QR$ decomposition for $A$. There exists a $SVD$ of $A$ such that $Q = U$ if and only if $A^*A$ is diagonal.
Suppose $A^*A$ is diagonal. Then, $A^*A = R^*Q^*QR = R^*R = R_1^*R_1$ so $R_1$ must be diagonal (using the fact that $A$ has full column rank). Then $A = QR = QRI$ and we can multiply some columns of $R$ and the corresponding rows of $I$ by $-1$ so that we get $A = QR'J = U Sigma V$, where $R'$ is diagonal with non negative entries, $Q$ and $J$ are unitary. This is a $SVD$ decomposition for $A$ with $Q = U$.
Now, suppose $Q = U$ for some $SVD$ of $A$. Then we have $Sigma_1$ is diagonal with strictly positive entries (since $A$ has full column rank) and $Sigma V^* = R Rightarrow Sigma_1 V^* = R_1$ so $V^*$ is upper triangular. Since $V^*$ is unitary and upper triangular, it must be diagonal. But the columns of $V$ form a basis of eigenvectors of $A^*A$. This means $e_i in Bbb R^n$ is an eigenvector of $A^*A$ for all $i = 1, ..., n$. Therefore, $A^*A$ is diagonal.
answered Jul 8 '17 at 18:46
DesuraDesura
901514
edited Jul 8 '17 at 19:02
answered Jul 8 '17 at 18:46
DesuraDesura
901514
answered Jul 8 '17 at 18:46
DesuraDesura
901514
answered Jul 8 '17 at 18:46
DesuraDesura
901514
901514
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%2f2348807%2fis-there-any-connection-between-qr-and-svd-of-a-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
$begingroup$
Why does your title differ from the question ?
$endgroup$
– Yves Daoust
Jul 6 '17 at 19:29
$begingroup$
I realized that difference after posting. Anyways let me come up with a better title
$endgroup$
– NAASI
Jul 6 '17 at 19:39