Study Guide: Eigenvalues, Eigenvectors, and
Diagonalization
1. Eigenvalues and Eignevectors
Let A be an n × n matrix, and suppose that
Av = λv
for some nonzero vector v and some scalar λ. In this case, λ is called an eigenvalue for A,
and v is a corresponding eigenvector.
Note that the zero vector 0 is not allowed as an eigenvector.
Typically an n × n matrix has n different eigenvalues. The eigenvectors associated to each
eigenvalue form a subspace, known as the eigenspace of A associated to the eigenvalue. (Note
that you have to include the zero vector in the eigenspace, even though it technically isn’t an
eigenvector.) In most cases, each eigenspace is a line through the origin.
2. Finding Eigenvalues
The quantity det(A − λI), where I is the identity matrix, is called the characteristic polynomial of A. The eigenvalues of A are the roots of the characteristic polynomial.
Sometimes the characteristic polynomial has a pair of complex roots. In this case, these
complex numbers count as eigenvalues.
Sometimes the characteristic polynomial has a repeated root. In this case, the corresponding
eigenspace may have dimension greater than one. In particular:
• If an eigenvalue is a double root, the corresponding eigenspace may be either a line or a
plane.
• If an eigenvalue is a triple root, the corresponding eigenspace may be either a line, a
plane, or a three-dimensional subspace.
3. Finding Eigenvectors
Given an eigenvalue λ, the eigenvectors are just the solutions to the equation
Av = λv.
Think of v as a vector of variables and solve the corresponding linear system.
Equivalently, you can just solve the equation
(A − λI)v = 0
which saves you the step of having to combine like terms.
Sometimes you are asked to find a basis for an eigenspace. In this case, you are simply trying
to find a basis for the null space of the matrix A − λI. In particular, you should use the
following procedure:
1. Reduce A − λI to reduced echelon form.
2. Each column without a pivot corresponds to a free variable.
3. There is one basis vector for each free variable. The value of the free variable for this
basis vector is 1, and the values of the other free variables for this basis vector is 0. Use
the reduced system of equations to compute the remaining entries of this vector.
4. Diagonalization
Almost any matrix A can be written as SDS −1 , where S is an invertible matrix and D is a
diagonal matrix. In particular, the columns of S consist of a basis of eigenvectors for each
eigenspace, and the diagonal entries of D are the corresponding eigenvalues.
We say that A is diagonalizable if there are enough eigenvectors to be the columns of an
n × n matrix S. This always happens if the roots of the characteristic polynomial are distinct,
but if you have a repeated root then one of the eigenvalues may not have enough eigenvectors.
In particular:
• If an eigenvalue is a double root of the characteristic polynomial but the corresponding
eigenspace is only a line, then A is not diagonalizable.
• If an eigenvalue is a triple root of the characteristic polynomial but the corresponding
eigenspace is either a line or a plane, then A is not diagonalizable.