Show that A is singular if, only if, $u^t v=-1$
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
add a comment |
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
add a comment |
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
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 Souza
656
656
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2 Answers
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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
$$
Au = (1+u^tv)u neq 0.
$$
add a comment |
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 $v^tu=lambda$.
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
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active
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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
$$
Au = (1+u^tv)u neq 0.
$$
add a comment |
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
$$
Au = (1+u^tv)u neq 0.
$$
add a comment |
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
$$
Au = (1+u^tv)u neq 0.
$$
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
$$
Au = (1+u^tv)u neq 0.
$$
edited Dec 2 '18 at 23:28
answered Dec 2 '18 at 23:13
Song
4,630317
4,630317
add a comment |
add a comment |
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 $v^tu=lambda$.
add a comment |
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 $v^tu=lambda$.
add a comment |
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 $v^tu=lambda$.
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 $v^tu=lambda$.
edited Dec 3 '18 at 15:43
answered Dec 2 '18 at 23:25
Martin Argerami
124k1176174
124k1176174
add a comment |
add a comment |
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