Gerschgorin's disksGershgorin disk

In the business of working with eigenvalues, it is useful to know how large or how small they can become. Likely as not to be negative, smallness is often a matter of absolute value, a zero eigenvalue implying singularity, while relative smallness of some eigenvalues relative to others could imply near singularity (if they were all tiny, it would likely be that the matrix was small overall). A good, general purpose upper bound for the absolute value arises from taking absolute values of the terms in the equation for a component of an eigenvector; if $M X = \lambda X$,

$$\begin{eqnarray*} \lambda x_i & = & \sum_j{ m_{ij} x_j}, \\ \vert \lambda \ve... ...x_i \vert & \leq & \sum_j{ \vert m_{ij} \vert \vert x_j \vert}. \end{eqnarray*}$$

Not all the components are zero; furthermore there those for which $\vert x_i \vert$ is larger than for any other; let $I$ be the index of one of them. Then

$$\begin{eqnarray*} \vert \lambda \vert & \leq & \sum_j{ \vert m_{Ij} \vert \frac{\vert x_j \vert}{\vert x_I \vert}}. \end{eqnarray*}$$

Every one of these quotients is less than 1, so the inequality can only be enhanced by dropping them and using 1 instead, with the result

$$\begin{eqnarray*} \vert \lambda \vert & \leq & \sum_j{ \vert mIj \vert }. \end{eqnarray*}$$

Quite reasonable; although a matrix is a collection of numbers, products in which they are involved can't get much larger that the factors permit, nor can a sum of such products ever surpass $n$ (the dimension) times the largest summand, a conclusion already evident in this inequality.

Changing the derivation slightly, moving the diagonal term of the sum over to the left hand side of the equation before taking absolute values, tells how far an eigenvalue may deviate from its diagonal element, even when the eigenvalues are complex.

$$\begin{eqnarray*} \vert \lambda - m_{II} \vert & \leq & \sum_{j\neq I}{ \vert m_{Ij} \vert }. \end{eqnarray*}$$

Figure: some Gerschgorin disks

The resulting disks are Gerschgorin disks, so called in honor of the source of their inspiration. Since it is not always obvious from inspection which was the largest component of the eigenvector, all the disks have to be considered, with the assurance that the eigenvalue lies under at least one of them (no reason for them not to intersect, particularly if diagonal elements are equal or close to one another). Continuity arguments (multiply the off-diagonal elements by a factor and vary it from zero to one) require each isolated group of disks to have their quota of eigenvalues.

The derivation works as well for rows as for columns, providing a practical choice of the row sum or the column sum of absolute values, whichever gives the smaller disk. What doesn't matter in the presence of symmetry can still be useful in other contexts.

Finding an eigenvalue at the center of a Gerschgorin disk is possible, but less likely in the presence of additional elenents in the same row or column. To find an eigenvalue on the rim of one of the disks requires a degree of cooperation, since all products of matrix elements by their matching component need a consistent sign to avoid discrepancies between the absolute value and the actual summand. Furthermore all those quotients replaced by ones must actually be ones, requiring the absolute values of all components of the eigenvector to be equal. In turn all row sums, using the sign convention established by the principal row, have to coincide. Under those conditions the eigenvalue lies on the rim, and all disks intersect at that point. The configuration is typical of stochastic matrices.