Eigendecompositions

Define $\Lambda = {\rm diag}(\lambda_1 ,\ldots, \lambda_n )$ and $X = [x_1,\ldots,x_n]$. $X$ is called an eigenvector matrix of $A$. Since the $x_i$ are orthogonal unit vectors, we see that $X^*X=I$; i.e., $X$ is a unitary (orthogonal) matrix. The $n$ equalities $A x_i = \lambda_i x_i$ for $i=1,\ldots,n$ may also be written $AX = X \Lambda$ or $A = X \Lambda X^*$. The factorization

$$A = X \Lambda X^*$$

is called the eigendecomposition of $A$. In other words, $A$ is similar to the diagonal matrix $\Lambda$, with similarity 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 invariant subspace of $A$. If we take the corresponding submatrix $\hat{\Lambda}= {\rm diag}(\lambda_2 , \lambda_3 , \lambda_5 )$ of $\Lambda$, then we can write the corresponding partial eigendecomposition as $A \hat{X}= \hat{X}\hat{\Lambda}$ or $\hat{X}^* A \hat{X}= \hat{\Lambda}$. If the columns in $\hat{X}$ are replaced by $k$ different vectors spanning the same invariant subspace, then we get a different partial eigendecomposition $A \check{X}= \check{X}\check{A}$, where $\check{A}$ is a $k$-by-$k$ matrix whose eigenvalues are those of $\hat{\Lambda}$, though $\check{A}$ may not be diagonal.