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Characteristic Polynomials

The study of mathematics, like the Nile, begins in minuteness but ends in magnificence. -- Charles Caleb Colton

A characteristic polynomial is a polynomial representing an operator on a vector space. On a complex vector space, this means that it is a polynomial whose roots are equal to the eigenvalues of the operator.

For example, if the operator in question is T on C3, with T(z1, z2, z3) = (0, 2z2, 3z3), its eigenvalues are 0, 2 and 3, since 0(1,0,0) = T(1,0,0) = (0,0,0), 2(0,1,0) = T(0,1,0) = (0,2,0), and 3(0,0,1) = T(0,0,1) = (0,0,3).

Any operator on a vector space has at most n distinct eigenvalues, where n is the dimension of the vector space. Since the vector space in question here is three dimensional and we have three eigenvalues, we don't need to worry about multiplicity: all three eigenvalues have a multiplicity of 1. But we'll get to that in a second.

The characteristic polynomial of an operator is the polynomial (z - λ1)d1(z - λ2)d2...(z - λm)dm for all of the eigenvalues λ1, ..., λm of the operator. Here, d1... dm represent the multiplicities of the eigenvalues.

In our example above, T = (0, 2z2, 3z3), then, the characteristic polynomial is z(z - 2)(z - 3), or z3 - 5z2 + 6z.

Multiplicity corresponds, in a certain sense, to the number of eigenvectors in a particular basis that correspond to a particular eigenvalue. More precisely, when the operator is written as an upper-triangular matrix over some basis, the multiplicity of an eigenvalue corresponds to the number of times that eigenvalue appears on the diagonal of the matrix. Specifically, this refers to the dimension of the subspace of generalized eigenvectors corresponding to the eigenvalue.

So, take the operator S = (0, 2z1, z3). This operator only has two eigenvalues, namely 0 and 1. We know that there must be some higher multiplicities here, since the sum of the multiplicities of the eigenvalues of an operator must total the dimension of the vector space the operator is over. However, when written as a matrix over the basis ((1,0,0), (0,1,0), (0,0,1)), this operator looks like:

[ 0 0 0 ]
[ 2 0 0 ]
[ 0 0 1 ]

which is not upper triangular, however, using the row matrix operations (note that this no longer corresponds to the same basis, but it does correspond to the operator over a basis -- we just don't know what that basis is any more), we can convert it to:
[ 1 0 0 ]
[ 0 0 1 ]
[ 0 0 0 ]

which is upper triangular and still represents the operator over some basis. Thus we have that 0 has multiplicity 2, and 1's is 1, so the characteristic polynomial of S = (0, 2z1, z3) is z2(z - 1), or z3 - z2.

Reference: Linear Algebra Done Right, 2nd edition, by Sheldon Axler, pp 77-78, 171-173.

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