Show that A is singular if, only if, $u^t v=-1$
$begingroup$
Let be $u,vinmathbb{R}^n$, the matrix $A = I + u v^t$ with $I$ the identity of order n. Show that A is singular if, only if, $u^t v=-1$.
I found that this demonstration is related to the Sherman-Morrison formula.
Any suggestions on where to read to get solved?
I did not find anything on the internet to help me get an insight.
linear-algebra matrices
asked Dec 2 '18 at 22:59
Juliana de SouzaJuliana de Souza
636
$endgroup$
add a comment |
$begingroup$
Let be $u,vinmathbb{R}^n$, the matrix $A = I + u v^t$ with $I$ the identity of order n. Show that A is singular if, only if, $u^t v=-1$.
I found that this demonstration is related to the Sherman-Morrison formula.
Any suggestions on where to read to get solved?
I did not find anything on the internet to help me get an insight.
linear-algebra matrices
asked Dec 2 '18 at 22:59
Juliana de SouzaJuliana de Souza
636
$endgroup$
add a comment |
$begingroup$
Let be $u,vinmathbb{R}^n$, the matrix $A = I + u v^t$ with $I$ the identity of order n. Show that A is singular if, only if, $u^t v=-1$.
I found that this demonstration is related to the Sherman-Morrison formula.
Any suggestions on where to read to get solved?
I did not find anything on the internet to help me get an insight.
linear-algebra matrices
asked Dec 2 '18 at 22:59
Juliana de SouzaJuliana de Souza
636
$endgroup$
Let be $u,vinmathbb{R}^n$, the matrix $A = I + u v^t$ with $I$ the identity of order n. Show that A is singular if, only if, $u^t v=-1$.
I found that this demonstration is related to the Sherman-Morrison formula.
Any suggestions on where to read to get solved?
I did not find anything on the internet to help me get an insight.
linear-algebra matrices
linear-algebra matrices
asked Dec 2 '18 at 22:59
Juliana de SouzaJuliana de Souza
636
asked Dec 2 '18 at 22:59
Juliana de SouzaJuliana de Souza
636
asked Dec 2 '18 at 22:59
Juliana de SouzaJuliana de Souza
636
asked Dec 2 '18 at 22:59
Juliana de SouzaJuliana de Souza
636
asked Dec 2 '18 at 22:59
Juliana de SouzaJuliana de Souza
636
636
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
This is the standard line of the proof. Assume $u^tv=-1=ucdot v=vcdot u = v^t u$. We can see that
$$
Au = u + u(v^tu) = u-u = 0,
$$ thus $ker (A) neq (0)$ and $A$ is singular.
Assume $u^tv =cneq -1$. Then, formal power series gives us
$$begin{eqnarray}
A^{-1} = (I+uv^t)^{-1} &=& I-uv^t+u(v^tu)v^t -u(v^tuv^tu)v^t+cdots \&=& I-uv^t +cuv^t -c^2uv^t+cdots\
&=&I-frac{uv^t}{1+c}=I-(1+u^tv)^{-1}uv^t.
end{eqnarray}$$
We can actually see that
$$Acdot(I-(1+u^tv)^{-1}uv^t) = I-(1+c)^{-1}uv^t+uv^t-(1+c)^{-1}uv^tuv^t = I,
$$ as we wanted.
Add: I derived an explicit formula for $A^{-1}$, but in fact, for the purpose of showing singularity of $A$, it suffices to prove $ker(A) = (0)$. Note that
$$
A w = w+u(v^tw)=w+(vcdot w)u = 0
$$ implies that $w =alpha u$ for some scalar $alpha$. This means, of course, that $ker(A) leq langle urangle$. If we show that $Auneq 0$, then the claim $ker(A)=(0)$ follows. It is easy to see that if $u^tv neq -1$, then
$$
Au = (1+u^tv)u neq 0.
$$
answered Dec 2 '18 at 23:13
SongSong
10.7k627
$endgroup$
add a comment |
$begingroup$
Adding $I$ to a matrix has the effect of adding $1$ to the eigenvalues. So the eigenvalues of $I+uv^t$ are precisely those of the form $1+lambda$, with $lambda$ an eigenvalue of $uv^t$. As the matrix $uv^t$ is rank-one, zero is always an eigenvalue; its only possible nonzero eigenvalue is $v^tu$ (proof below). Thus the eigenvalues of $I+uv^t$ are $0$ and $1+v^tu$ (if the latter is nonzero). So $I+uv^t$ is invertible precisely when $1+v^tune0$.
Proof that $lambda $ is a nonzero eigenvalue of $uv^t$ if and only if $lambda=v^tu$. Suppose that $uv^tw=lambda w$. Then
$$
lambda w = uv^tw=(v^tw),u.
$$
So $w=alpha u$, where $alpha=(v^tw)/lambda$. Thus (note that $v^twne0$, since $wne0$)
$$
lambdaalpha u=lambda w=uv^tw=uv^t(alpha u)=alpha (v^tu),u.
$$
Thus $v^tu=lambda$.
answered Dec 2 '18 at 23:25
Martin ArgeramiMartin Argerami
126k1182180
$endgroup$
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%2f3023342%2fshow-that-a-is-singular-if-only-if-ut-v-1%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
$begingroup$
This is the standard line of the proof. Assume $u^tv=-1=ucdot v=vcdot u = v^t u$. We can see that
$$
Au = u + u(v^tu) = u-u = 0,
$$ thus $ker (A) neq (0)$ and $A$ is singular.
Assume $u^tv =cneq -1$. Then, formal power series gives us
$$begin{eqnarray}
A^{-1} = (I+uv^t)^{-1} &=& I-uv^t+u(v^tu)v^t -u(v^tuv^tu)v^t+cdots \&=& I-uv^t +cuv^t -c^2uv^t+cdots\
&=&I-frac{uv^t}{1+c}=I-(1+u^tv)^{-1}uv^t.
end{eqnarray}$$
We can actually see that
$$Acdot(I-(1+u^tv)^{-1}uv^t) = I-(1+c)^{-1}uv^t+uv^t-(1+c)^{-1}uv^tuv^t = I,
$$ as we wanted.
Add: I derived an explicit formula for $A^{-1}$, but in fact, for the purpose of showing singularity of $A$, it suffices to prove $ker(A) = (0)$. Note that
$$
A w = w+u(v^tw)=w+(vcdot w)u = 0
$$ implies that $w =alpha u$ for some scalar $alpha$. This means, of course, that $ker(A) leq langle urangle$. If we show that $Auneq 0$, then the claim $ker(A)=(0)$ follows. It is easy to see that if $u^tv neq -1$, then
$$
Au = (1+u^tv)u neq 0.
$$
answered Dec 2 '18 at 23:13
SongSong
10.7k627
$endgroup$
add a comment |
$begingroup$
This is the standard line of the proof. Assume $u^tv=-1=ucdot v=vcdot u = v^t u$. We can see that
$$
Au = u + u(v^tu) = u-u = 0,
$$ thus $ker (A) neq (0)$ and $A$ is singular.
Assume $u^tv =cneq -1$. Then, formal power series gives us
$$begin{eqnarray}
A^{-1} = (I+uv^t)^{-1} &=& I-uv^t+u(v^tu)v^t -u(v^tuv^tu)v^t+cdots \&=& I-uv^t +cuv^t -c^2uv^t+cdots\
&=&I-frac{uv^t}{1+c}=I-(1+u^tv)^{-1}uv^t.
end{eqnarray}$$
We can actually see that
$$Acdot(I-(1+u^tv)^{-1}uv^t) = I-(1+c)^{-1}uv^t+uv^t-(1+c)^{-1}uv^tuv^t = I,
$$ as we wanted.
Add: I derived an explicit formula for $A^{-1}$, but in fact, for the purpose of showing singularity of $A$, it suffices to prove $ker(A) = (0)$. Note that
$$
A w = w+u(v^tw)=w+(vcdot w)u = 0
$$ implies that $w =alpha u$ for some scalar $alpha$. This means, of course, that $ker(A) leq langle urangle$. If we show that $Auneq 0$, then the claim $ker(A)=(0)$ follows. It is easy to see that if $u^tv neq -1$, then
$$
Au = (1+u^tv)u neq 0.
$$
answered Dec 2 '18 at 23:13
SongSong
10.7k627
$endgroup$
add a comment |
$begingroup$
This is the standard line of the proof. Assume $u^tv=-1=ucdot v=vcdot u = v^t u$. We can see that
$$
Au = u + u(v^tu) = u-u = 0,
$$ thus $ker (A) neq (0)$ and $A$ is singular.
Assume $u^tv =cneq -1$. Then, formal power series gives us
$$begin{eqnarray}
A^{-1} = (I+uv^t)^{-1} &=& I-uv^t+u(v^tu)v^t -u(v^tuv^tu)v^t+cdots \&=& I-uv^t +cuv^t -c^2uv^t+cdots\
&=&I-frac{uv^t}{1+c}=I-(1+u^tv)^{-1}uv^t.
end{eqnarray}$$
We can actually see that
$$Acdot(I-(1+u^tv)^{-1}uv^t) = I-(1+c)^{-1}uv^t+uv^t-(1+c)^{-1}uv^tuv^t = I,
$$ as we wanted.
Add: I derived an explicit formula for $A^{-1}$, but in fact, for the purpose of showing singularity of $A$, it suffices to prove $ker(A) = (0)$. Note that
$$
A w = w+u(v^tw)=w+(vcdot w)u = 0
$$ implies that $w =alpha u$ for some scalar $alpha$. This means, of course, that $ker(A) leq langle urangle$. If we show that $Auneq 0$, then the claim $ker(A)=(0)$ follows. It is easy to see that if $u^tv neq -1$, then
$$
Au = (1+u^tv)u neq 0.
$$
answered Dec 2 '18 at 23:13
SongSong
10.7k627
$endgroup$
This is the standard line of the proof. Assume $u^tv=-1=ucdot v=vcdot u = v^t u$. We can see that
$$
Au = u + u(v^tu) = u-u = 0,
$$ thus $ker (A) neq (0)$ and $A$ is singular.
Assume $u^tv =cneq -1$. Then, formal power series gives us
$$begin{eqnarray}
A^{-1} = (I+uv^t)^{-1} &=& I-uv^t+u(v^tu)v^t -u(v^tuv^tu)v^t+cdots \&=& I-uv^t +cuv^t -c^2uv^t+cdots\
&=&I-frac{uv^t}{1+c}=I-(1+u^tv)^{-1}uv^t.
end{eqnarray}$$
We can actually see that
$$Acdot(I-(1+u^tv)^{-1}uv^t) = I-(1+c)^{-1}uv^t+uv^t-(1+c)^{-1}uv^tuv^t = I,
$$ as we wanted.
Add: I derived an explicit formula for $A^{-1}$, but in fact, for the purpose of showing singularity of $A$, it suffices to prove $ker(A) = (0)$. Note that
$$
A w = w+u(v^tw)=w+(vcdot w)u = 0
$$ implies that $w =alpha u$ for some scalar $alpha$. This means, of course, that $ker(A) leq langle urangle$. If we show that $Auneq 0$, then the claim $ker(A)=(0)$ follows. It is easy to see that if $u^tv neq -1$, then
$$
Au = (1+u^tv)u neq 0.
$$
answered Dec 2 '18 at 23:13
SongSong
10.7k627
edited Dec 2 '18 at 23:28
answered Dec 2 '18 at 23:13
SongSong
10.7k627
answered Dec 2 '18 at 23:13
SongSong
10.7k627
answered Dec 2 '18 at 23:13
SongSong
10.7k627
10.7k627
add a comment |
add a comment |
$begingroup$
Adding $I$ to a matrix has the effect of adding $1$ to the eigenvalues. So the eigenvalues of $I+uv^t$ are precisely those of the form $1+lambda$, with $lambda$ an eigenvalue of $uv^t$. As the matrix $uv^t$ is rank-one, zero is always an eigenvalue; its only possible nonzero eigenvalue is $v^tu$ (proof below). Thus the eigenvalues of $I+uv^t$ are $0$ and $1+v^tu$ (if the latter is nonzero). So $I+uv^t$ is invertible precisely when $1+v^tune0$.
Proof that $lambda $ is a nonzero eigenvalue of $uv^t$ if and only if $lambda=v^tu$. Suppose that $uv^tw=lambda w$. Then
$$
lambda w = uv^tw=(v^tw),u.
$$
So $w=alpha u$, where $alpha=(v^tw)/lambda$. Thus (note that $v^twne0$, since $wne0$)
$$
lambdaalpha u=lambda w=uv^tw=uv^t(alpha u)=alpha (v^tu),u.
$$
Thus $v^tu=lambda$.
answered Dec 2 '18 at 23:25
Martin ArgeramiMartin Argerami
126k1182180
$endgroup$
add a comment |
$begingroup$
Adding $I$ to a matrix has the effect of adding $1$ to the eigenvalues. So the eigenvalues of $I+uv^t$ are precisely those of the form $1+lambda$, with $lambda$ an eigenvalue of $uv^t$. As the matrix $uv^t$ is rank-one, zero is always an eigenvalue; its only possible nonzero eigenvalue is $v^tu$ (proof below). Thus the eigenvalues of $I+uv^t$ are $0$ and $1+v^tu$ (if the latter is nonzero). So $I+uv^t$ is invertible precisely when $1+v^tune0$.
Proof that $lambda $ is a nonzero eigenvalue of $uv^t$ if and only if $lambda=v^tu$. Suppose that $uv^tw=lambda w$. Then
$$
lambda w = uv^tw=(v^tw),u.
$$
So $w=alpha u$, where $alpha=(v^tw)/lambda$. Thus (note that $v^twne0$, since $wne0$)
$$
lambdaalpha u=lambda w=uv^tw=uv^t(alpha u)=alpha (v^tu),u.
$$
Thus $v^tu=lambda$.
answered Dec 2 '18 at 23:25
Martin ArgeramiMartin Argerami
126k1182180
$endgroup$
add a comment |
$begingroup$
Adding $I$ to a matrix has the effect of adding $1$ to the eigenvalues. So the eigenvalues of $I+uv^t$ are precisely those of the form $1+lambda$, with $lambda$ an eigenvalue of $uv^t$. As the matrix $uv^t$ is rank-one, zero is always an eigenvalue; its only possible nonzero eigenvalue is $v^tu$ (proof below). Thus the eigenvalues of $I+uv^t$ are $0$ and $1+v^tu$ (if the latter is nonzero). So $I+uv^t$ is invertible precisely when $1+v^tune0$.
Proof that $lambda $ is a nonzero eigenvalue of $uv^t$ if and only if $lambda=v^tu$. Suppose that $uv^tw=lambda w$. Then
$$
lambda w = uv^tw=(v^tw),u.
$$
So $w=alpha u$, where $alpha=(v^tw)/lambda$. Thus (note that $v^twne0$, since $wne0$)
$$
lambdaalpha u=lambda w=uv^tw=uv^t(alpha u)=alpha (v^tu),u.
$$
Thus $v^tu=lambda$.
answered Dec 2 '18 at 23:25
Martin ArgeramiMartin Argerami
126k1182180
$endgroup$
Adding $I$ to a matrix has the effect of adding $1$ to the eigenvalues. So the eigenvalues of $I+uv^t$ are precisely those of the form $1+lambda$, with $lambda$ an eigenvalue of $uv^t$. As the matrix $uv^t$ is rank-one, zero is always an eigenvalue; its only possible nonzero eigenvalue is $v^tu$ (proof below). Thus the eigenvalues of $I+uv^t$ are $0$ and $1+v^tu$ (if the latter is nonzero). So $I+uv^t$ is invertible precisely when $1+v^tune0$.
Proof that $lambda $ is a nonzero eigenvalue of $uv^t$ if and only if $lambda=v^tu$. Suppose that $uv^tw=lambda w$. Then
$$
lambda w = uv^tw=(v^tw),u.
$$
So $w=alpha u$, where $alpha=(v^tw)/lambda$. Thus (note that $v^twne0$, since $wne0$)
$$
lambdaalpha u=lambda w=uv^tw=uv^t(alpha u)=alpha (v^tu),u.
$$
Thus $v^tu=lambda$.
answered Dec 2 '18 at 23:25
Martin ArgeramiMartin Argerami
126k1182180
edited Dec 3 '18 at 15:43
answered Dec 2 '18 at 23:25
Martin ArgeramiMartin Argerami
126k1182180
answered Dec 2 '18 at 23:25
Martin ArgeramiMartin Argerami
126k1182180
answered Dec 2 '18 at 23:25
Martin ArgeramiMartin Argerami
126k1182180
126k1182180
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%2f3023342%2fshow-that-a-is-singular-if-only-if-ut-v-1%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
