Similar to inverse plus rank 1

Question

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?

Answer 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 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.