Given a real, invertible matrix $A$. For which vectors $b$ and $c$ is
$$
A^{-1} + bc^T
$$
similar to $A$? And is the rank-1 matrix $bc^T$ unique?



Similar to inverse plus rank 1
Let's assume for simplicity that $A$ has distinct eigenvalues, so similarity to $A$
just means having the same characteristic polynomial.


The [Matrix determinant lemma](https://en.wikipedia.org/wiki/Matrix_determinant_lemma)
says that the characteristic polynomial
$$ \eqalign{\det(A^{-1} + b c^T - \lambda I) &= (1 + c^T (A^{-1}-\lambda I)^{-1} b) \det(A^{-1}-\lambda I)\cr
&= (1 + c^T A (I - \lambda A)^{-1} b) \det(A)^{-1} \det(I - \lambda A)}$$
which we want to be the same as the characteristic polynomial $\det(A - \lambda I)$ of $A$.
Thus we want $$
1 + c^T A (I - \lambda A)^{-1} b = \frac{\det(A) \det(A - \lambda I)}{\det(I-\lambda A)}$$
Writing $b$ as a linear combination $\sum\_j b\_j v\_j$ of the eigenvectors of $A$
and $c$ as a linear combination $\sum\_j c\_j u\_j$ of the left eigenvectors of $A$
(where $Av\_j = \lambda\_j v\_j$, $u\_j^T A = \lambda\_j u\_j^T$, and $u\_i^T v\_j = \delta\_{ij}$), the left side is
$$ 1 + \sum\_j c\_j b\_j \frac{\lambda\_j}{1 - \lambda \lambda\_j}$$
and the equation will be true as long as $-c\_j b\_j $ is the residue of the 
right side at $\lambda = \lambda\_j^{-1}$. 


We will not have uniqueness because only the products $c\_j b\_j$ matter.


