Symbolic computation of matrix eigenvalues

An important tool for describing eigenvalues of square matrices is the characteristic polynomial: saying that λ is an eigenvalue of A is equivalent to stating that the system of linear equations (A - λI) v = 0 (where I is the identity matrix) has a non-zero solution v (namely an eigenvector), and so it is equivalent to the determinant det(A - λI) being zero. The function p(λ) = det(A - λI) is a polynomial in λ since determinants are defined as sums of products. This is the characteristic polynomial of A: the eigenvalues of a matrix are the zeros of its characteristic polynomial.

(Sometimes the characteristic polynomial is taken to be det(λI - A) instead, which is the same polynomial if V is even-dimensional but has the opposite sign if V is odd-dimensional. This has the slight advantage that its leading coefficient is always 1 rather than -1.)

It follows that we can compute all the eigenvalues of a matrix A by solving the equation $p_{A}(\lambda )=0$ . If A is an n-by-n matrix, then $p_{A}$ has degree n and A can therefore have at most n eigenvalues. Conversely, the fundamental theorem of algebra says that this equation has exactly n roots (zeroes), counted with multiplicity. All real polynomials of odd degree have a real number as a root, so for odd n, every real matrix has at least one real eigenvalue. In the case of a real matrix, for even and odd n, the non-real eigenvalues come in conjugate pairs.

An example of a matrix with no real eigenvalues is the 90-degree rotation

{\begin{bmatrix}0&1\\-1&0\end{bmatrix}}

whose characteristic polynomial is $x^{2}+1$ and so its eigenvalues are the pair of complex conjugates i, -i.

The Cayley-Hamilton theorem states that every square matrix satisfies its own characteristic polynomial, that is, $p_{A}(A)=0$ .

Eigenvalues of 2×2 matrices

An analytic solution for the eigenvalues of 2×2 matrices can be obtained directly from the quadratic formula: if

A={\begin{bmatrix}a&b\\c&d\end{bmatrix}}

then the characteristic polynomial is

{\rm {det}}{\begin{bmatrix}a-\lambda &b\\c&d-\lambda \end{bmatrix}}=(a-\lambda )(d-\lambda )-bc=\lambda ^{2}-(a+d)\lambda +(ad-bc)

(notice that the coefficients, up to sign, are the trace ${\rm {tr}}(A)=a+d$ and determinant ${\rm {det}}(A)=ad-bc$ ) so the solutions are

\lambda ={\frac {a+d}{2}}\pm {\sqrt {{\frac {(a+d)^{2}}{4}}+bc-ad}}={\frac {a+d}{2}}\pm {\frac {\sqrt {4bc+(a-d)^{2}}}{2}}

A formula for the eigenvalues of a 3x3 or 4x4 matrix could be derived in an analogous way, using the formulae for the roots of a cubic or quartic equation.

Example computation

The computation of eigenvalue/eigenvector can be computed by the following algorithm.

Consider an n-square matrix A

1. Find the roots of the characteristic polynomial of A. These are the eigenvalues.

If n different roots are found, then the matrix can be diagonalized.

2. Find an basis for the kernel of the matrix given by

B-\lambda _{n}I

. For each of the eigenvalues. These are the eigenvectors

The eigenvectors given from different eigenvalues are linear independent.
The eigenvectors given from a root-multiciply are also linear independent.

Let us determine the eigenvalues of the matrix

A={\begin{bmatrix}0&1&-1\\1&1&0\\-1&0&1\end{bmatrix}}

which represents a linear operator R³ → R³.

Identifying eigenvalues

We first compute the characteristic polynomial of A:

p(\lambda )=\det(A-\lambda I)=\det {\begin{bmatrix}-\lambda &1&-1\\1&1-\lambda &0\\-1&0&1-\lambda \end{bmatrix}}=-\lambda ^{3}+2\lambda ^{2}+\lambda -2.

This polynomial factors to $p(\lambda )=-(\lambda -2)(\lambda -1)(\lambda +1)$ . Therefore, the eigenvalues of A are 2, 1 and −1.

Identifying eigenvectors

With the eigenvalues in hand, we can solve sets of simultaneous linear equations to determine the corresponding eigenvectors. For example, one can check that

A{\begin{bmatrix}\;1\\\;1\\-1\end{bmatrix}}={\begin{bmatrix}\;2\\\;2\\-2\end{bmatrix}}=2{\begin{bmatrix}\;1\\\;1\\-1\end{bmatrix}}

which confirms that 2 is an eigenvalue of A and gives us a corresponding eigenvector.

Note that if A is a real matrix, the characteristic polynomial will have real coefficients, but its roots will not necessarily all be real. The complex eigenvalues come in pairs which are conjugates. For a real matrix, the eigenvectors of a non-real eigenvalue z , which are the solutions of $(A-zI)v=0$ , cannot be real.

If v₁, ..., v_m are eigenvectors with different eigenvalues λ₁, ..., λ_m, then the vectors v₁, ..., v_m are necessarily linearly independent.

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 are mutually orthogonal. Symmetric matrices are commonly encountered in engineering.

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

v_{1}={\begin{bmatrix}\;1\\\;1\\-1\end{bmatrix}},\quad v_{2}={\begin{bmatrix}\;0\;\\1\\1\end{bmatrix}},\quad v_{3}={\begin{bmatrix}\;2\\-1\\\;1\end{bmatrix}}.

These three vectors form 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=x_{1}v_{1}+x_{2}v_{2}+x_{3}v_{3}

and then we have

Ax=2x_{1}v_{1}+x_{2}v_{2}-x_{3}v_{3}.

Eigenvalues of 2×2 matrices

Example computation

See also