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By Matlab, I know that the eigenvalues of the matrix $B^{-1}A$ are 2.457, 0.542, and 1 (multiplicity 3) where $A$ and $B$ are defined as: \begin{equation} A= \begin{pmatrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 2 & 1 & 1 \\ 0 & 0 & 1 & 2 & 1 \\ 0 & 0 & 1 & 1 & 2 \\ \end{pmatrix}, B= \begin{pmatrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 2 & 1 \\ 0 & 0 & 0 & 1 & 2 \\ \end{pmatrix} \end{equation} Similarly, the eigenvalues of the matrix $B^{-1}A$ are 4.56, 0.43, and 1 (multiplicity 4) where $A$ and $B$ are defined as: \begin{equation} A= \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0\\ 0 & 2 & 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 1 & 1 & 1 \\ 0 & 1 & 1 & 2 & 1 & 1\\ 0 & 1 & 1 & 1 & 2 & 1\\ 0 & 1 & 1 & 1 & 1 & 2\\ \end{pmatrix}, B= \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0 & 2 & 1\\ 0 & 0 & 0 & 0 & 1 & 2\\ \end{pmatrix} \end{equation} In general, given $n$, the matrices are defined as follows: \begin{equation} A = \begin{pmatrix} I_{n-m_A} & 0 \\ 0 & I_{m_A} + J_{m_A} \\ \end{pmatrix}, B = \begin{pmatrix} I_{n-m_B} & 0 \\ 0 & I_{m_B} + J_{m_B} \\ \end{pmatrix}, \end{equation} where $m_A \ne m_B$ and they can be $1,...,n-1$ (so it can be that $m_A < m_B$). $J_m$ is a $m \times m$ matrix of ones. Is there any explanation to why the multiplicity of 1 as an eigenvalue is always $n-2$ where $n$ is the dimension of the matrices?

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    $\begingroup$ The general claim that the multiplicity is always $n-2$ makes sense only if the pattern of how to define $A$ and $B$ were generalizable to all $n$, but the generalization is not obvious. $\endgroup$ – BallBoy Aug 5 at 1:00
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    $\begingroup$ Precisely how do $A$ and $B$ vary? They seem to change in dimensions, and they seem to take a some kind of block diagonal form $\begin{pmatrix} I_{n-m} & 0 \\ 0 & I_m + J_m \end{pmatrix}$ where $J_m$ is the matrix containing only $1$s. How do the sizes of these blocks vary with $n$? In $B$, is the lower right block always $2 \times 2$, or can it vary? Does its size depend on the sizes of the blocks in $A$? $\endgroup$ – Theo Bendit Aug 5 at 1:17
  • $\begingroup$ Sorry for the confusion! I just added the general version of the question. $\endgroup$ – kx526 Aug 5 at 1:37
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Here's one.

Since $A$ and $B$ must always be invertible, the following conditions are equivalent:

  • $1$ is an eigenvalue of $B^{-1}A$ for eigenvector $x$
  • $B^{-1}Ax=x$

  • $Ax = Bx$

  • $(A-B)x=0$

  • $x \in \ker(A-B)$

So the multiplicity of the eigenvalue $1$ of $B^{-1}A$ is equal to the dimension of $\ker(A-B)$.

Edit: As pointed out by Theo Bendit, it's worth showing that $B^{-1}A$ must be diagonalizable in order to rule out any issues of a discrepancy between algebraic and geometric multiplicity of the eigenvalue. One way to see that $B^{-1}A$ must be diagonalizable is to note that $A$ and $B$ are clearly symmetric and positive definite and then apply this result.

Let's now assume $m_A>m_B$; if the opposite is true, we can easily switch the roles of $A$ and $B$. If we examine $A-B$, we find that it has $n-m_A$ zero rows, followed by $m_A-m_B$ rows consisting of $n-m_A$ zeros and $m_A$ ones, followed by $m_B$ rows consisting of $n-m_A$ zeros, $m_A-m_B$ ones, and then $m_B$ zeros. Since $m_B$ and $m_A-m_B$ are both nonzero by assumption, $A-B$ has exactly two distinct nonzero rows, so has rank $2$; thus $\ker(A-B)$ has dimension $n-2$.

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    $\begingroup$ Beautiful. Thanks a lot for your help! $\endgroup$ – kx526 Aug 5 at 2:08
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    $\begingroup$ @kenex Thanks for the question! It was a fun problem. $\endgroup$ – BallBoy Aug 5 at 2:11
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    $\begingroup$ +1 Nice approach! I think you should point out that $B^{-1}A$ is Hermitian and hence diagonalisable. That way, you can validly conclude that the multiplicity of $1$ is the dimension of its eigenspace, the kernel of $A - B$. $\endgroup$ – Theo Bendit Aug 5 at 3:46
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    $\begingroup$ @TheoBendit You raise a very good point! And you made me realize that $B^{-1}A$ is actually not in general Hermitian, but still must be diagonalizable. Answer edited. $\endgroup$ – BallBoy Aug 5 at 15:56

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