, where A is an 
matrix. A solution to this
problem consists of an eigenvalue 
, and an associated eigenvector
. Usually there are n eigenvalue-eigenvector solutions to
.
The problem of finding the eigenvalues and eigenvectors for a square real
matrix A is a very different kind of problem compared to the problems that
have been solved so far in this course, where the primary tool has been
elementary row operations. Finding eigenvalues is equivalent to finding
roots of a special polynomial 
(the characteristic polynomial).
This degree n polynomial can easily be determined for small matrices using
a special determinant, 
. When evaluating the
determinant to find , a determinant expansion formula should be
used. You might be tempted to try to use elementary row
operations, but these can not easily be used because the diagonal elements
in the determinant depend on . Even though the characteristic
polynomial will have real coefficients, the roots (eigenvalues) can in some
cases be complex numbers. There is one type of matrix where eigenvalues can
easily be found; if a matrix is triangular (it has all entries zero below the
main diagonal, or all entries zero above the main diagonal), then the
eigenvalues are the diagonal entries for the matrix. Once an eigvenvalue
has been found, an associated eigenvector can be found by
solving the appropriate homogeneous system, 
. Notice
that if is an eigenvector, then so is 
for any scalar s;
eigenvectors are unique only up to scalar multiples. The eigenspace for
a particular eigenvalue is the same as 
.
This usually has dimension equal to one; but sometimes the dimension can be
larger (for example, if is a repeated eigenvalue).
If the matrix A is transformed, using a nonsingular matrix P, into another
matrix B with 
, then A and B are said to be similar.
Similar matrices have the same eigenvalues. The aim of diagonalization is to
find a matrix P so that B is a diagonal matrix D. In this case, the
eigenvalues of A (and D) are the diagonal entries of D. Diagonalization
is possible if and only if A has n linearly independent eigenvectors,
which can be used for the columns of P, so that AP = PD.