Suppose v is an eigenvector of the matrix M with eigenvalue.
We then have: (M -I) v = M v -v = 0.
This means that v is in the null space of the matrix M -I, its nullity is at least one, and thereforeM -I.
When M is specified this determinental equation can be written out explicitly and it is a polynomial equation in having degree n. Every eigenvalue of M must obey this equation.
Moreover, if a value of obeys the equation M -I= 0, then the matrix M - I with this value is singular and its nullity is at least 1 so that there is a vector v satisfying (M -I) v = 0.
The equation M -I= 0 is called the characteristic equation for the matrix M. We have just seen that any solution to the characteristic equation is an eigenvalue of M corresponding to at least one eigenvector.
The straightforward way to find an eigenvalue is to write down the characteristic equation and find a solution to it.
Once you know the eigenvalue you can use the equation (M -I) v = 0, to determine v. Since (M -I) v is singular, row reduction will lead you to a subspace of solutions of dimension given by the nullity of this matrix.
We now ask, when can we find a basis of eigenvectors?
The characteristic equation is a polynomial equation of degree n.
Such an equation can have up to n real roots, and in general will have exactly n complex roots, some of which can be multiple ones. ((x-1)2 = 0 has two roots of 1.)
If it has n distinct real roots, then we have n real eigenvalues, and can find n eigenvectors to match. Whether or not the roots are distinct, you can always find a basis consisting of eigenvectors if the matrix is symmetric.
A basis is said to be orthonormal, if its elements each have length 1 and they are mutually perpendicular. Only symmetric matrices have real eigenvalues and real orthonormal bases of eigenvectors.
So far we have assumed that all our numbers are real, and we are then unable to find n eigenvalues and eigenvectors if some of the roots of the characteristic equation are not real. With a small change in definitions required to assure that the dot product of a non-zero vector with itself is always real and positive, it is possible to let our numbers be complex. With complex numbers, the class of polynomial equations with n roots includes all of them, and the only problem with having n eigenvalues comes with matrices whose characteristic equation has multiple roots.
There are some matrices just dont have n distinct dimensions of eigenvectors; the simplest example is the two by two matrix with 1 in the 12 place and 0 elsewhere.
The characteristic equation of a matrix C(x) which is A - x I= 0 can be written in terms of its distinct roots,i , as , where mi is the multiplicity of the eigenvaluei . A diagonal matrix obeys its minimal equation which has the same roots but with all multiplicities 1.
One can show that every matrix satisfies its own characteristic equation; a matrix has a basis of eigenvectors over the complex numbers if and only if it obeys its own minimal equation.