Characteristic polynomial of diagonal matrix with two rank-one updates

Question

Let D be an $n\times n$ diagonal matrix, and u, v, b, k four n-dim vectors. u and k are of the form: $$ (*,*,0,0,0,\cdots)$$ and v and b are of the form: $$ (0,0,*,*,*,\cdots) $$ The diagonal matrix D gets two rank-one updates: $$D + uv^T + bk^T$$ The resulting matrix looks something like$$ \begin{pmatrix} * & 0 & *&* & * & * \\ 0 & * & *&* & * & * \\ * & * & *&0 & 0 & 0 \\ * & * & 0&* & 0 & 0 \\ * & * & 0&0 & * & 0 \\ * & * & 0&0 & 0 & * \end{pmatrix}$$ where the diagonal elements come from D and the first two rows/columns come from the rank-one updates $uv^T$ and $bk^T$. Numerical evaluation of this matrix shows me that the perturbation changes the eigenvalues. However, if I wanted to get an analytical understanding of what is going on, I get into trouble:

Define A := D + u$v^T$. We can find the characteristic polynomial for $A+bk^T$ using Sylvester's determinant theorem \begin{align} \det(\lambda I-A-bk^T)=\det(\lambda I - A)\det(1-k^T(\lambda I-A)^{-1}b)=\det(\lambda I - A) \end{align} Because of their special form, no matter what is in between $k^T$ and b, the respective term will vanish. But what is $\det(\lambda I - A)$? A is just an upper triangular matrix and thus its characteristic polynomial will be the same as for the diagonal matrix D. We find that the characteristic polynomial of this perturbed matrix will give us the exact same eigenvalues as the original diagonal matrix D.

I must have done a mistake somewhere but I don't seem to find it. Can someone maybe tell me what is wrong?

Answer 1

You seem to have assumed that $\det(1-k^T(\lambda I-A)^{-1}b)=1$, i.e. $k^T(\lambda I-A)^{-1}b=0$.

Although $k^Tb=0$, it isn't true that $k^T(\lambda I-A)^{-1}b$ always vanishes. E.g. when $D=0,\,u=k$ and $v=b$, $$ k^T(\lambda I-A)^{-1}b =u^T(\lambda I-uv^T)^{-1}v =u^T\left[\frac{1}{\lambda}\left(I+\frac{uv^T}{\lambda-v^Tu}\right)\right]v =\frac{1}{\lambda}\left(u^Tv+\frac{(u^Tu)(v^Tv)}{\lambda-v^Tu}\right). $$ Since $v^Tu=0$, we obtain $k^T(\lambda I-A)^{-1}b=\frac{(u^Tu)(v^Tv)}{\lambda^2}$, which is nonzero when $u^Tu\ne0$ and $v^Tv\ne0$.