Rank-one perturbation proof

Question

I wrote a proof for a problem in my textbook. Can someone please verify it or offer suggestions for improvement?

$\textbf{Problem:} $If $u$ and $v$ are $m$-vectors, the matrix $A = I+uv^*$ is known as the rank-one perturbation of the identity. Show that if $A$ is nonsingular, then its inverse has the form $A^{-1}=I+\alpha u v^*$ for some scalar $\alpha$, and give an expression for $\alpha$. For which $u$ and $v$ is $A$ singular? If it is singular, what is null($A$)?

Suppose $A$ is singular. Then, there exists a nonzero $x \in \mathbb{C}^m$ such that $Ax=0$. That is, $(I+uv^*)x=0$. This implies that $u(v^*x) = -x$. For this to be true, $x$ has to be a multiple of $u$. So, let $x = c_1u$ for some nonzero scalar $c_1$. Then, $Ax = c_1u(v^*u) = -c_1u$. Note that $u = 0$ if $v^*u = 0$. Therefore, $A$ is singular if and only if $v^*u \neq 0$, and null($A$) $=$ span($u$) if it is singular.

Now, set $\alpha = -1$. By the reasoning in the above paragraph, $v^*u = 0$ for $A$ to be invertible. Now, $(I+uv^*)(I-uv^*) = I - uv^*uv^* = I - u(v^*u)v^* = I$. Therefore, $A^{-1} = I - uv^*$.

Answer 1

There is an error in your argument. $Ax$ is equal to $c_1u+c_1u(v^*u)$, not $c_1u(v^*u)$.

You can actually simplify your argument by considering $A-I$ instead of $A$. That is, $A$ is singular if and only if $-1$ is an eigenvalue of $A-I=uv^\ast$. However, as $uv^\ast$ has at most rank 1, its only possibly nonzero eigenvalue is $v^\ast u$.

Answer 2

These are very nice considerations, but the easiest way to solve the question is to take the hint and first compute the product and equate to the identity $$ (I+αuv^∗)(I+uv^∗)=I $$ Then you will find that there is also a solution if $u$ and $v$ are not orthogonal.