Determinant of $(I+uv^*)$
Let $u,v in R^n$, How can I find $det(I+uv^*)$? This problem was given as preparatory for final exam and I dont know how to approach it. I dont see some nice ways to express this determinant. This is from Numerical Linear algebra course. Could you please give me some hints?
linear-algebra determinant numerical-linear-algebra
asked Nov 21 at 21:33
Studying Optimization
656
add a comment |
Let $u,v in R^n$, How can I find $det(I+uv^*)$? This problem was given as preparatory for final exam and I dont know how to approach it. I dont see some nice ways to express this determinant. This is from Numerical Linear algebra course. Could you please give me some hints?
linear-algebra determinant numerical-linear-algebra
asked Nov 21 at 21:33
Studying Optimization
656
add a comment |
Let $u,v in R^n$, How can I find $det(I+uv^*)$? This problem was given as preparatory for final exam and I dont know how to approach it. I dont see some nice ways to express this determinant. This is from Numerical Linear algebra course. Could you please give me some hints?
linear-algebra determinant numerical-linear-algebra
asked Nov 21 at 21:33
Studying Optimization
656
Let $u,v in R^n$, How can I find $det(I+uv^*)$? This problem was given as preparatory for final exam and I dont know how to approach it. I dont see some nice ways to express this determinant. This is from Numerical Linear algebra course. Could you please give me some hints?
linear-algebra determinant numerical-linear-algebra
linear-algebra determinant numerical-linear-algebra
asked Nov 21 at 21:33
Studying Optimization
656
asked Nov 21 at 21:33
Studying Optimization
656
asked Nov 21 at 21:33
Studying Optimization
656
asked Nov 21 at 21:33
Studying Optimization
656
asked Nov 21 at 21:33
Studying Optimization
656
656
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
Suppose $A$ is an invertible square matrix and $u$, $v$ are column vectors. Then the matrix determinant lemma states that
$$
{displaystyle det left({A} +{uv} ^{textsf {T}}right)
=left(1+{v} ^{textsf {T}}{A} ^{-1}{u} right),det left({A} right),.}
$$
In your case, let $A=I$.
The proof is essentially based on the following observation:
$$
{displaystyle {begin{pmatrix}mathbf {I} &0\mathbf {v} ^{textsf {T}}&1end{pmatrix}}{begin{pmatrix}mathbf {I} +mathbf {uv} ^{textsf {T}}&mathbf {u} \0&1end{pmatrix}}{begin{pmatrix}mathbf {I} &0\-mathbf {v} ^{textsf {T}}&1end{pmatrix}}={begin{pmatrix}mathbf {I} &mathbf {u} \0&1+mathbf {v} ^{textsf {T}}mathbf {u} end{pmatrix}}.}
$$
[Added later:]
Alternatively, in the nontrivial case when both $u$ and $v$ are nonzero vecotrs and $n>1$, note that $B:={uv} ^{textsf {T}}$ is a rank one matrix. Moreover, $lambda = {v} ^{textsf {T}}u$ is an eigenvalue of $B$ since:
$$Bu = (uv^T)u=u(v^Tu)=(v^Tu)u.$$ Since $B$ is of rank one, $0$ must be an eigenvalue as well. It is not difficult to show1 that $0$ has multiplicity $n-1$. So the matrix $B$ is similar to an upper triangle matrix with the diagonal:
$$
(v^Tu,0,cdots,0).
$$
It follows that the matrix $I+B$ is similar to an upper triangle matrix with the diagonal:
$$
(1+v^Tu,1,cdots,1).
$$
But similar matrices has the same determinant. So you have the same result as in the matrix determinant lemma.
1Note: see also answers to this question: Eigenvalues of the rank one matrix $uv^T$
answered Nov 21 at 21:41
user587192
1,486112
add a comment |
Assuming $uv^*ne0$, then $u$ is an eigenvector, as $(uv^*)u=(v^*u)u$. Since the matrix has rank $1$, the characteristic polynomial is $det(uv^*-XI)=(v^*u-X)(0-X)^{n-1}$. Evaluating this at $X=-1$ yields
$$
det(uv^*+I)=v^*u+1
$$
answered Nov 21 at 21:52
egreg
177k1484198
Very clever short proof.
– Jean Marie
Dec 9 at 23:27
1
@JeanMarie And the general formula for $det(A+uv^*)$ follows in the same fashion, because $A^{-1}u$ is an eigenvector of $A^{-1}uv^*$ (which has rank at most $1$) relative to $v^*A^{-1}u$.
– egreg
Dec 9 at 23:47
add a comment |
You can easily convince yourself that one eigenvector is $u$ with eigenvalue $1+v^* u$ and the remaining eigenvalues are 1 (with the eigenspace given by all the vectors $w$ with $v^*w=0$). Thus the determinant is given by $$1+v^*u,.$$
answered Nov 21 at 21:49
Fabian
19.3k3674
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%2f3008399%2fdeterminant-of-iuv%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
Suppose $A$ is an invertible square matrix and $u$, $v$ are column vectors. Then the matrix determinant lemma states that
$$
{displaystyle det left({A} +{uv} ^{textsf {T}}right)
=left(1+{v} ^{textsf {T}}{A} ^{-1}{u} right),det left({A} right),.}
$$
In your case, let $A=I$.
The proof is essentially based on the following observation:
$$
{displaystyle {begin{pmatrix}mathbf {I} &0\mathbf {v} ^{textsf {T}}&1end{pmatrix}}{begin{pmatrix}mathbf {I} +mathbf {uv} ^{textsf {T}}&mathbf {u} \0&1end{pmatrix}}{begin{pmatrix}mathbf {I} &0\-mathbf {v} ^{textsf {T}}&1end{pmatrix}}={begin{pmatrix}mathbf {I} &mathbf {u} \0&1+mathbf {v} ^{textsf {T}}mathbf {u} end{pmatrix}}.}
$$
[Added later:]
Alternatively, in the nontrivial case when both $u$ and $v$ are nonzero vecotrs and $n>1$, note that $B:={uv} ^{textsf {T}}$ is a rank one matrix. Moreover, $lambda = {v} ^{textsf {T}}u$ is an eigenvalue of $B$ since:
$$Bu = (uv^T)u=u(v^Tu)=(v^Tu)u.$$ Since $B$ is of rank one, $0$ must be an eigenvalue as well. It is not difficult to show1 that $0$ has multiplicity $n-1$. So the matrix $B$ is similar to an upper triangle matrix with the diagonal:
$$
(v^Tu,0,cdots,0).
$$
It follows that the matrix $I+B$ is similar to an upper triangle matrix with the diagonal:
$$
(1+v^Tu,1,cdots,1).
$$
But similar matrices has the same determinant. So you have the same result as in the matrix determinant lemma.
1Note: see also answers to this question: Eigenvalues of the rank one matrix $uv^T$
answered Nov 21 at 21:41
user587192
1,486112
add a comment |
Suppose $A$ is an invertible square matrix and $u$, $v$ are column vectors. Then the matrix determinant lemma states that
$$
{displaystyle det left({A} +{uv} ^{textsf {T}}right)
=left(1+{v} ^{textsf {T}}{A} ^{-1}{u} right),det left({A} right),.}
$$
In your case, let $A=I$.
The proof is essentially based on the following observation:
$$
{displaystyle {begin{pmatrix}mathbf {I} &0\mathbf {v} ^{textsf {T}}&1end{pmatrix}}{begin{pmatrix}mathbf {I} +mathbf {uv} ^{textsf {T}}&mathbf {u} \0&1end{pmatrix}}{begin{pmatrix}mathbf {I} &0\-mathbf {v} ^{textsf {T}}&1end{pmatrix}}={begin{pmatrix}mathbf {I} &mathbf {u} \0&1+mathbf {v} ^{textsf {T}}mathbf {u} end{pmatrix}}.}
$$
[Added later:]
Alternatively, in the nontrivial case when both $u$ and $v$ are nonzero vecotrs and $n>1$, note that $B:={uv} ^{textsf {T}}$ is a rank one matrix. Moreover, $lambda = {v} ^{textsf {T}}u$ is an eigenvalue of $B$ since:
$$Bu = (uv^T)u=u(v^Tu)=(v^Tu)u.$$ Since $B$ is of rank one, $0$ must be an eigenvalue as well. It is not difficult to show1 that $0$ has multiplicity $n-1$. So the matrix $B$ is similar to an upper triangle matrix with the diagonal:
$$
(v^Tu,0,cdots,0).
$$
It follows that the matrix $I+B$ is similar to an upper triangle matrix with the diagonal:
$$
(1+v^Tu,1,cdots,1).
$$
But similar matrices has the same determinant. So you have the same result as in the matrix determinant lemma.
1Note: see also answers to this question: Eigenvalues of the rank one matrix $uv^T$
answered Nov 21 at 21:41
user587192
1,486112
add a comment |
Suppose $A$ is an invertible square matrix and $u$, $v$ are column vectors. Then the matrix determinant lemma states that
$$
{displaystyle det left({A} +{uv} ^{textsf {T}}right)
=left(1+{v} ^{textsf {T}}{A} ^{-1}{u} right),det left({A} right),.}
$$
In your case, let $A=I$.
The proof is essentially based on the following observation:
$$
{displaystyle {begin{pmatrix}mathbf {I} &0\mathbf {v} ^{textsf {T}}&1end{pmatrix}}{begin{pmatrix}mathbf {I} +mathbf {uv} ^{textsf {T}}&mathbf {u} \0&1end{pmatrix}}{begin{pmatrix}mathbf {I} &0\-mathbf {v} ^{textsf {T}}&1end{pmatrix}}={begin{pmatrix}mathbf {I} &mathbf {u} \0&1+mathbf {v} ^{textsf {T}}mathbf {u} end{pmatrix}}.}
$$
[Added later:]
Alternatively, in the nontrivial case when both $u$ and $v$ are nonzero vecotrs and $n>1$, note that $B:={uv} ^{textsf {T}}$ is a rank one matrix. Moreover, $lambda = {v} ^{textsf {T}}u$ is an eigenvalue of $B$ since:
$$Bu = (uv^T)u=u(v^Tu)=(v^Tu)u.$$ Since $B$ is of rank one, $0$ must be an eigenvalue as well. It is not difficult to show1 that $0$ has multiplicity $n-1$. So the matrix $B$ is similar to an upper triangle matrix with the diagonal:
$$
(v^Tu,0,cdots,0).
$$
It follows that the matrix $I+B$ is similar to an upper triangle matrix with the diagonal:
$$
(1+v^Tu,1,cdots,1).
$$
But similar matrices has the same determinant. So you have the same result as in the matrix determinant lemma.
1Note: see also answers to this question: Eigenvalues of the rank one matrix $uv^T$
answered Nov 21 at 21:41
user587192
1,486112
Suppose $A$ is an invertible square matrix and $u$, $v$ are column vectors. Then the matrix determinant lemma states that
$$
{displaystyle det left({A} +{uv} ^{textsf {T}}right)
=left(1+{v} ^{textsf {T}}{A} ^{-1}{u} right),det left({A} right),.}
$$
In your case, let $A=I$.
The proof is essentially based on the following observation:
$$
{displaystyle {begin{pmatrix}mathbf {I} &0\mathbf {v} ^{textsf {T}}&1end{pmatrix}}{begin{pmatrix}mathbf {I} +mathbf {uv} ^{textsf {T}}&mathbf {u} \0&1end{pmatrix}}{begin{pmatrix}mathbf {I} &0\-mathbf {v} ^{textsf {T}}&1end{pmatrix}}={begin{pmatrix}mathbf {I} &mathbf {u} \0&1+mathbf {v} ^{textsf {T}}mathbf {u} end{pmatrix}}.}
$$
[Added later:]
Alternatively, in the nontrivial case when both $u$ and $v$ are nonzero vecotrs and $n>1$, note that $B:={uv} ^{textsf {T}}$ is a rank one matrix. Moreover, $lambda = {v} ^{textsf {T}}u$ is an eigenvalue of $B$ since:
$$Bu = (uv^T)u=u(v^Tu)=(v^Tu)u.$$ Since $B$ is of rank one, $0$ must be an eigenvalue as well. It is not difficult to show1 that $0$ has multiplicity $n-1$. So the matrix $B$ is similar to an upper triangle matrix with the diagonal:
$$
(v^Tu,0,cdots,0).
$$
It follows that the matrix $I+B$ is similar to an upper triangle matrix with the diagonal:
$$
(1+v^Tu,1,cdots,1).
$$
But similar matrices has the same determinant. So you have the same result as in the matrix determinant lemma.
1Note: see also answers to this question: Eigenvalues of the rank one matrix $uv^T$
answered Nov 21 at 21:41
user587192
1,486112
edited Nov 21 at 22:06
answered Nov 21 at 21:41
user587192
1,486112
answered Nov 21 at 21:41
user587192
1,486112
answered Nov 21 at 21:41
user587192
1,486112
1,486112
add a comment |
add a comment |
Assuming $uv^*ne0$, then $u$ is an eigenvector, as $(uv^*)u=(v^*u)u$. Since the matrix has rank $1$, the characteristic polynomial is $det(uv^*-XI)=(v^*u-X)(0-X)^{n-1}$. Evaluating this at $X=-1$ yields
$$
det(uv^*+I)=v^*u+1
$$
answered Nov 21 at 21:52
egreg
177k1484198
Very clever short proof.
– Jean Marie
Dec 9 at 23:27
1
@JeanMarie And the general formula for $det(A+uv^*)$ follows in the same fashion, because $A^{-1}u$ is an eigenvector of $A^{-1}uv^*$ (which has rank at most $1$) relative to $v^*A^{-1}u$.
– egreg
Dec 9 at 23:47
add a comment |
Assuming $uv^*ne0$, then $u$ is an eigenvector, as $(uv^*)u=(v^*u)u$. Since the matrix has rank $1$, the characteristic polynomial is $det(uv^*-XI)=(v^*u-X)(0-X)^{n-1}$. Evaluating this at $X=-1$ yields
$$
det(uv^*+I)=v^*u+1
$$
answered Nov 21 at 21:52
egreg
177k1484198
Very clever short proof.
– Jean Marie
Dec 9 at 23:27
1
@JeanMarie And the general formula for $det(A+uv^*)$ follows in the same fashion, because $A^{-1}u$ is an eigenvector of $A^{-1}uv^*$ (which has rank at most $1$) relative to $v^*A^{-1}u$.
– egreg
Dec 9 at 23:47
add a comment |
Assuming $uv^*ne0$, then $u$ is an eigenvector, as $(uv^*)u=(v^*u)u$. Since the matrix has rank $1$, the characteristic polynomial is $det(uv^*-XI)=(v^*u-X)(0-X)^{n-1}$. Evaluating this at $X=-1$ yields
$$
det(uv^*+I)=v^*u+1
$$
answered Nov 21 at 21:52
egreg
177k1484198
Assuming $uv^*ne0$, then $u$ is an eigenvector, as $(uv^*)u=(v^*u)u$. Since the matrix has rank $1$, the characteristic polynomial is $det(uv^*-XI)=(v^*u-X)(0-X)^{n-1}$. Evaluating this at $X=-1$ yields
$$
det(uv^*+I)=v^*u+1
$$
answered Nov 21 at 21:52
egreg
177k1484198
answered Nov 21 at 21:52
egreg
177k1484198
answered Nov 21 at 21:52
egreg
177k1484198
answered Nov 21 at 21:52
egreg
177k1484198
177k1484198
Very clever short proof.
– Jean Marie
Dec 9 at 23:27
1
@JeanMarie And the general formula for $det(A+uv^*)$ follows in the same fashion, because $A^{-1}u$ is an eigenvector of $A^{-1}uv^*$ (which has rank at most $1$) relative to $v^*A^{-1}u$.
– egreg
Dec 9 at 23:47
add a comment |
Very clever short proof.
– Jean Marie
Dec 9 at 23:27
1
@JeanMarie And the general formula for $det(A+uv^*)$ follows in the same fashion, because $A^{-1}u$ is an eigenvector of $A^{-1}uv^*$ (which has rank at most $1$) relative to $v^*A^{-1}u$.
– egreg
Dec 9 at 23:47
Very clever short proof.
– Jean Marie
Dec 9 at 23:27
Very clever short proof.
– Jean Marie
Dec 9 at 23:27
1
1
@JeanMarie And the general formula for $det(A+uv^*)$ follows in the same fashion, because $A^{-1}u$ is an eigenvector of $A^{-1}uv^*$ (which has rank at most $1$) relative to $v^*A^{-1}u$.
– egreg
Dec 9 at 23:47
@JeanMarie And the general formula for $det(A+uv^*)$ follows in the same fashion, because $A^{-1}u$ is an eigenvector of $A^{-1}uv^*$ (which has rank at most $1$) relative to $v^*A^{-1}u$.
– egreg
Dec 9 at 23:47
add a comment |
You can easily convince yourself that one eigenvector is $u$ with eigenvalue $1+v^* u$ and the remaining eigenvalues are 1 (with the eigenspace given by all the vectors $w$ with $v^*w=0$). Thus the determinant is given by $$1+v^*u,.$$
answered Nov 21 at 21:49
Fabian
19.3k3674
add a comment |
You can easily convince yourself that one eigenvector is $u$ with eigenvalue $1+v^* u$ and the remaining eigenvalues are 1 (with the eigenspace given by all the vectors $w$ with $v^*w=0$). Thus the determinant is given by $$1+v^*u,.$$
answered Nov 21 at 21:49
Fabian
19.3k3674
add a comment |
You can easily convince yourself that one eigenvector is $u$ with eigenvalue $1+v^* u$ and the remaining eigenvalues are 1 (with the eigenspace given by all the vectors $w$ with $v^*w=0$). Thus the determinant is given by $$1+v^*u,.$$
answered Nov 21 at 21:49
Fabian
19.3k3674
You can easily convince yourself that one eigenvector is $u$ with eigenvalue $1+v^* u$ and the remaining eigenvalues are 1 (with the eigenspace given by all the vectors $w$ with $v^*w=0$). Thus the determinant is given by $$1+v^*u,.$$
answered Nov 21 at 21:49
Fabian
19.3k3674
answered Nov 21 at 21:49
Fabian
19.3k3674
answered Nov 21 at 21:49
Fabian
19.3k3674
answered Nov 21 at 21:49
Fabian
19.3k3674
19.3k3674
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%2f3008399%2fdeterminant-of-iuv%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
