Perturbation

Continuing to speculate on the eigenvectors and eigenvalues of a sum of two matrices, consider the case where the second is small relative to the first, perhaps on account of multiplying it by a small parameter. Maybe a small change of coordinates, depending on the same parameter, could account for the changed matrix; suppose then that

$$\begin{eqnarray*} (I - \varepsilon P) A (I + \varepsilon P) & = & A + \varepsilon B, \end{eqnarray*}$$

with the intention of disregarding anything multipying $\varepsilon^2$. First,

$$\begin{eqnarray*} A + \varepsilon (A P - P A) + \ldots & = & A + \varepsilon B, \end{eqnarray*}$$

so the task becomes solving for the commutator $A P - P A = B$, which is a special case of a more general first order (not linear) equation involving an unknown $P$ and given matrices $A$ and $B$, or even $A$, $B$, and $C$ (instead of the second $A$).

As usual, the first step is to diagonalize $A$, but it is reasonable to suppose that that has already been done, since it is only a question of the coordinate system. Once that is done, and the equation reduced to components, we find

$$\begin{eqnarray*} \lambda_i p_{ij} - p_{ij} \lambda_j & = & b_{ij}, \end{eqnarray*}$$

or for reference,

$$\begin{eqnarray*} p_{ij} & = & \frac{b_{ij}}{(\lambda_i - \lambda_j)}. \end{eqnarray*}$$

As always, a problem arises when $\lambda_i = \lambda_j$, which is traditionally resolved by making still further preparations, namely choosing coordinates for which $b_{ij}$ vanishes, avoiding the necessity of division by zero - just leave that part of the equation alone. The new eigenvectors can now be read off from the columns of $I + \varepsilon P$ and the rows of $I - \varepsilon P$.

Not only is degeneracy an obstacle to this derivation, there is the impicit assumption that $B$ has no diagonal elements, avoiding $(\lambda_i - \lambda_i)$ as a divisor. Consequently this procedure cannot change the eigenvalues of $A$, just its eigenvectors. If it is necessary to change the eigenvalues of $A$ as well, that has to be done independently of applying the operator $P$. Why is such a subterfuge necessary? Because

$$\begin{eqnarray*} (I - \varepsilon P) (I + \varepsilon P) & = & I - \varepsilon^2 P, \end{eqnarray*}$$

making the transformation orthogonal to first order. That is a rotation, which will not change the lengths of semiaxes of an ellipsiod, which are eigenvalues.

This whole scheme is cometimes called Primas' method.