Eigenvectors

HomePage | Recent changes | View source | Discuss this page | Page history | Log in |

Printable version | Disclaimers | Privacy policy

In Linear Algebra, the eigenvectors (from the German eigen meaning own) of a linear operator are non-zero vectors which, when operated on by the operator, result in a scalar multiple of themselves. The scalar is then called the eigenvalue associated with the eigenvector.

In applied mathematics and physics the eigenvectors of a matrix or a differential operator often have important physical significance. In classical mechanics the eigenvectors of the governing equations typically correspond to natural modes of vibration in a body, and the eigenvalues to their frequencies. In quantum mechanics, operators correspond to observable variables, eigenvectors are also called eigenstates, and the eigenvalues of an operator represent those values of the corresponding variable that have non-zero probability of occuring.

Formally, we define eigenvectors and eigenvalues as follows: If A : V -> V is a linear operator on some vector space V, v is a non-zero vector in V and c is a scalar (possibly zero) such that

A v = c v,

then we say that v is an eigenvector of the operator A, and its associated eigenvalue is c. Note that if v is an eigenvector with eigenvalue c, then any non-zero multiple of v is also an eigenvector with eigenvalue c. In fact, all the eigenvectors with associated eigenvalue c, together with 0, form a subspace of V, the eigenspace for the eigenvalue c.

For example, consider the matrix

       / 0  1 -1 \ 
 A  =  | 1  1  0 |
       \-1  0  1 /

which represents a linear operator R3 -> R3. One can check that

   / 1 \       / 1 \ 
 A | 1 |  =  2 | 1 |
   \-1 /       \-1 /

and therefore 2 is an eigenvalue of A and we have found a corresponding eigenvector.

An important tool for computing eigenvalues of square matrices is the characteristic polynomial: saying that c is an eigenvalue of A is equivalent to stating that the system of linear equations (A - cI) x = 0 (where I is the identity matrix) has a non-zero solution x (namely an eigenvector), and so it is equivalent to the determinant det(A - cI) being zero. The function p(c) = det(A - cI) is a polynomial in c since determinants are defined as sums of products. This is the characteristic polynomial of A; its zeros are precisely the eigenvalues of A. If A is an n-by-n matrix, then its characteristic polynomial has degree n and A can therefore have at most n eigenvalues.

Returning to the example above, if we wanted to compute all of A's eigenvalues, we could determine the characteristic polynomial first:

                          /-x  1  -1 \ 
 p(x) = det(A - xI) = det | 1 1-x  0 | = -x3 + 2x2 + x - 2
                          \-1  0  1-x/

and because of p(x) = -(x - 2) (x - 1) (x + 1) we see that the eigenvalues of A are 2, 1 and -1.

Cayley's theorem states that if p(x) is the characteristic polynomial of A, then replacing x by A in the expressionfor p yields the zero matrix: p(A) = 0.

For our matrix A above, we therefore have -A3 + A2 + A - 2I = O.

Note that if A is a real matrix, the characteristic polynomial will have real coefficients, but not all its roots will necessarily be real. The complex eigenvalues will all be associated to complex eigenvectors.

The spectral theorem for symmetric matrices states that, if A is a real symmetric n-by-n matrix, then all its eigenvalues are real, and there exist n linearly independent eigenvectors for A which all have length 1 and are mutually orthogonal.

Our example matrix from above is symmetric, and three mutually orthogonal eigenvectors of A are

       / 1 \        / 0 \         / 2 \ 
  v1 = | 1  | , v2 = | 1  | , v3 = | -1 |
       \-1 /        \ 1 /         \ 1 /

These three vectors from a basis of R3. With respect to this basis, the linear map represented by A takes a particularly simple form: every vector x in R3 can be written uniquely as x = x1 v1 + x2 v2 + x3 v3 and then we have

Ax = 2x1 v1 + x2 v2 - x3 v3.

/Talk