Eigendecompositions

Define $\Lambda = {\rm diag}(\lambda_1 ,\ldots, \lambda_n )$ and $X = [x_1,\ldots,x_n]$. $X$ is called an eigenvector matrix of $A - \lambda B$. Since the $x_i$ are unit vectors orthogonal with respect to the inner product induced by $B$, we see that $X^*BX=\Lambda_B$, a nonsingular diagonal matrix. The $n$ equalities $A x_i = \lambda_i B x_i$ for $i=1,\ldots,n$ may also be written $AX = BX \Lambda$ or $X^*AX = X^*BX \Lambda = \Lambda_B \Lambda$. Thus $X^*AX \equiv \Lambda_A$ is diagonal too. The factorizations

$$X^* A X = \Lambda_A \quad \mbox{and} \quad X^* B X = \Lambda_B$$

(or $A = X^{-\ast} \Lambda_A X^{-1}$ and $B = X^{-\ast} \Lambda_B X^{-1}$) are called an eigendecomposition of $A - \lambda B$. In other words, $A - \lambda B$ is congruent to the diagonal pencil $\Lambda_A - \lambda \Lambda_B$, with congruence transformation $X$.

If we take a subset of $k$ columns of $X$ (say $\hat{X}= X(:,[2,3,5])$ = columns 2, 3, and 5), then these columns span an eigenspace of $A - \lambda B$. If we take the corresponding submatrix $\hat{\Lambda}_A = {\rm diag}(\hat{\Lambda}_{A,22} , \hat{\Lambda}_{A,33}, \hat{\Lambda}_{A,55} )$ of $\Lambda_A$, and similarly define $\hat{\Lambda}_B$, then we can write the corresponding partial eigendecomposition as $\hat{X}^* A \hat{X}= \hat{\Lambda}_A$ and $\hat{X}^* B \hat{X}= \hat{\Lambda}_B$. If the columns in $\hat{X}$ are replaced by $k$ different vectors spanning the same eigensubspace, then we get a different partial eigendecomposition, where $\hat{\Lambda}_A$ and $\hat{\Lambda}_B$ are replaced by different $k$-by-$k$ matrices ${\check{\Lambda}_A}$ and ${\check{\Lambda}_B}$ such that the eigenvalues of the pencil ${\check{\Lambda}_A} - \lambda {\check{\Lambda}_B}$ are those of $\hat{\Lambda}_A - \lambda \hat{\Lambda}_B$, though the pencil ${\check{\Lambda}_A} - \lambda {\check{\Lambda}_B}$ may not be diagonal.