Canonical forms

Whether there is a sufficiency or insufficiency of eigenvectors eventually depends on cases, but some progress can be made by rewriting the defining equation as

$$\begin{eqnarray*} (M - \lambda I) X & = & 0. \end{eqnarray*}$$

Were $(M - \lambda I)$ invertible, $X$ would have to be zero, which is not a desirable conclusion. For finite vector spaces, at least, there is a very simple numerical criterion for noninvertibility, namely

$$\begin{eqnarray*} \vert M - \lambda I\vert & = & 0. \end{eqnarray*}$$

This determinant works out to be a polynomial of degree equal to the dimension of $M$; usually called the characteristic polynomialcharacteristic polynomial of M.In practice, however, it is convenient to work with monicmonic polynomial polynomials, which is the term applied to polynomials whose leading coefficient is $+1$. Consequently it would be preferable to use the definition

$$\begin{eqnarray*} \chi(\lambda) & = & \vert\lambda I - M\vert. \end{eqnarray*}$$

Vanishing of the characteristic determinant is a requirement for the existence of an eigenvector. Conversely, if the determinant vanishes, the columns of $M$ are dependent, whereupon the coefficients of the linear combination responsible generate a vector satisfying the eigenvalue equation. Therefore there must be at least one eigenvector for every number which is an eigenvalue; sometimes there are more if the equation has multiple roots.

Additionally, whatever the multiplicity of the roots of its characteristic equation, there always has to be at least one root, and so at least one eigenvector for whatever matrix. Such a definitive statement supposes that the coefficients of the matrix belong to an algebraically closed field; otherwise there may be no roots at all. Rotations in the real plane move all points save the origin; but their characteristic equation has only complex roots (unless the angle of rotation is zero and hence done by the identity matrix).

If all the roots are distinct, there have to be enough eigenvectors for a basis and a dual basis as well. Questions of insufficiency of eigenvectors hinge on the existence of multiple roots for the polynomial Since slight alterations in the matrix elements could split the roots, it may be suspected that some sort of limiting process would account for matrices which lack their full complement of eigenvectors. Due to the wide variety of possible limits, however, some other approach to a general theory is preferable.