Generalised eigenvalue problem: form a basis in $\mathbb{R}^N$.

Consider a generalised eigenvalue problem for two real symmetric $N\times N$ matrices $A$ and $B$ $$ (A-\lambda B)\cdot\mathbf{x}_{\lambda}=0\,. $$

Assume that $A$ and $B$ have non-zero determinants. I am interested in the case where neither $A$ or $B$ are positive definite, so that complex eigenvalues $\lambda$ can exist. Consider the case where $n_R$ eigenvalues are real, and $n_C$ are complex, so that $n_R+2 n_C=N$ (the factor of two arises because complex eigenvalues come in pairs). In general. the eigenvectors associated with the $n_C$ complex eigenvalues will be complex. Denote the complex eigenvectors by $\mathbf{y}_{\lambda_C}$ and the eigenvectors associated with the real eigenvalues by $\mathbf{x}_{\lambda_R}$.

Take the set of $N$ vectors $\{\mathbf{x}_{\lambda_R},\mathrm{Re}(\mathbf{y}_{\lambda_C}),\mathrm{Im}(\mathbf{y}_{\lambda_C})\}$. Do these form a basis in $\mathbb{R}^N$? I have tried this with many examples and it appears correct, but so far a general proof seems illusive.

Example:

Consider $$ A=\left( \begin{array}{ccc} 1 & 3 & 4 \\ 3 & 2 & 0 \\ 4 & 0 & 2 \\ \end{array} \right)\,,\quad B = \left( \begin{array}{ccc} 1 & 2 & 0 \\ 2 & -5 & 0 \\ 0 & 0 & 1 \\ \end{array} \right)\,. $$ Both $A$ and $B$ are symmetric real $3\times 3$ matrices and have non-zero determinants. $\lambda$ can be determined via $$ 9 \lambda ^3-33 \lambda ^2-43 \lambda -46=0\,. $$ It is simple to see that two of the roots of the above polynomial are complex. If you form the set of eigevectors as described above, they also form a basis.

Solution 1:

Not an answer, but some observations that are a bit long for a comment.

1. Since $B$ has a nonzero determinant, it is invertible. Thus, $$(A-\lambda B)x = 0 \quad\Leftrightarrow\quad (B^{-1}A-\lambda I)x = 0.$$ So your generalized eigenvalue problem is in fact just an eigenvalue problem.

2. [removed as per the comment below.]

3. The equation $n_R+2n_C=N$ looks quite strange to me. First, although complex eigenvalues do come in pairs, this does not mean you should multiply $n_C$ by 2; it merely means that $n_C$ is even. Second, it looks as though you are treating this equation as a fact. But even in the case $B=I$, and even if we correct to $n_R+n_C=N$, this is a nontrivial assumption! In that case it is precisely the statement that $A$ is diagonalizable.

4. If we do not change the equation, then it is an even more stringent condition. For instance if $B=I$, it implies that precisely $n_C$ of the eigenvalues are repeated. I am curious to see what examples you have in mind.