I have a symmetric positive definite (SPD) matrix $A\in\mathbb{R}^{n\times n}$ and a full-rank matrix $B\in\mathbb{R}^{m\times n}$.
I know that the pre-conditioned matrix $\begin{bmatrix} A & 0 \\ 0 & BA^{-1}B^T \end{bmatrix}^{-1}\begin{bmatrix} A & B^T \\ B & 0 \end{bmatrix}$ has only three eigenvalues, $1, (1\pm\sqrt{5})/2.$ But how can I find their multiplicity?
I have $\det(M-\lambda I)=\det((\lambda^2-\lambda-1)I)=(\lambda^2-\lambda-1)^{n+m}=0$, but where is the eigenvalue 1 here?
The matrix $$ C:=\begin{bmatrix}A&B^T\\B&0\end{bmatrix} $$ has $n$ positive and $m$ negative eigenvalues. This follows from the congruence relation $$ \begin{bmatrix}A&B^T\\B&0\end{bmatrix} = \begin{bmatrix}I&0\\B^TA^{-1}&I\end{bmatrix} \begin{bmatrix}A&0\\0&S\end{bmatrix} \begin{bmatrix}I&A^{-1}B\\0&I\end{bmatrix} $$ (where $S:=-B^TA^{-1}B$), Sylverster law of inertia, and the fact that $A$ and $S$ are, respectively, positive and negative definite matrices.
Let $$ P:=\begin{bmatrix}A&0\\0&-S\end{bmatrix}. $$ The matrix in question, $P^{-1}C$, is similar to $$ P^{1/2}(P^{-1}C)P^{-1/2}=P^{-1/2}CP^{-1/2}, $$ so $P^{-1}C$ and $C$ are congruent and hence $P^{-1}C$ has the same inertia as $C$.
Assuming that we already know that $\lambda:=1$ and $\lambda_\pm:=(1\pm \sqrt{5})/2$ and in addition that the eigenvalues $\lambda_\pm$ come in pairs, we have that the multiplicity of $\lambda_\pm$ is $m$ and hence the multiplicity of $\lambda$ is $n-m$.