Show that A is singular if, only if, $u^t v=-1$ Let be $u,v\in\mathbb{R}^n$, 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.
 A: This is the standard line of the proof. Assume $u^tv=-1=u\cdot v=v\cdot u = v^t u$. We can see that
$$
Au = u + u(v^tu) = u-u = 0,
$$ thus $\ker (A) \neq (0)$ and $A$ is singular.
Assume $u^tv =c\neq -1$. 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
$$A\cdot(I-(1+u^tv)^{-1}uv^t) = I-(1+c)^{-1}uv^t+uv^t-(1+c)^{-1}uv^tuv^t = I,
$$ 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 
$$
A w = w+u(v^tw)=w+(v\cdot w)u = 0
$$ implies that $w =\alpha u$ for some scalar $\alpha$. This means, of course, that $\ker(A) \leq \langle u\rangle$. If we show that $Au\neq 0$, then the claim $\ker(A)=(0)$ follows. It is easy to see that if $u^tv \neq -1$, then
$$
Au = (1+u^tv)u \neq 0.
$$
A: 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 $1+v^tu\ne0$. 
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 $v^tw\ne0$, since $w\ne0$)
$$
\lambda\alpha u=\lambda w=uv^tw=uv^t(\alpha u)=\alpha (v^tu)\,u.
$$
Thus $v^tu=\lambda$. 
