# Determining multiplicity of 1 as an eigenvalue for a certain matrix

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: $$$$A= \begin{pmatrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 2 & 1 & 1 \\ 0 & 0 & 1 & 2 & 1 \\ 0 & 0 & 1 & 1 & 2 \\ \end{pmatrix}, B= \begin{pmatrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 2 & 1 \\ 0 & 0 & 0 & 1 & 2 \\ \end{pmatrix}$$$$ Similarly, the eigenvalues of the matrix $$B^{-1}A$$ are 4.56, 0.43, and 1 (multiplicity 4) where $$A$$ and $$B$$ are defined as: $$$$A= \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0\\ 0 & 2 & 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 1 & 1 & 1 \\ 0 & 1 & 1 & 2 & 1 & 1\\ 0 & 1 & 1 & 1 & 2 & 1\\ 0 & 1 & 1 & 1 & 1 & 2\\ \end{pmatrix}, B= \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0 & 2 & 1\\ 0 & 0 & 0 & 0 & 1 & 2\\ \end{pmatrix}$$$$ In general, given $$n$$, the matrices are defined as follows: $$$$A = \begin{pmatrix} I_{n-m_A} & 0 \\ 0 & I_{m_A} + J_{m_A} \\ \end{pmatrix}, B = \begin{pmatrix} I_{n-m_B} & 0 \\ 0 & I_{m_B} + J_{m_B} \\ \end{pmatrix},$$$$ where $$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?

• 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. – BallBoy Aug 5 at 1:00
• 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 $A$? – Theo Bendit Aug 5 at 1:17
• Sorry for the confusion! I just added the general version of the question. – kx526 Aug 5 at 1:37

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 $$n-2$$.

• Beautiful. Thanks a lot for your help! – kx526 Aug 5 at 2:08
• @kenex Thanks for the question! It was a fun problem. – BallBoy Aug 5 at 2:11
• +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 $A - B$. – Theo Bendit Aug 5 at 3:46
• @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. – BallBoy Aug 5 at 15:56