Symbolic computation of matrix eigenvalues

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In mathematics, and in particular in linear algebra, 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 $p A (λ) = 0$ . 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, $p A (A) = 0$ .

1 Eigenvalues of 2×2 matrices
2 Example computation
- 2.1 Identifying eigenvalues
- 2.2 Identifying eigenvectors
3 See also

[edit] 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 $tr(A) = a + d$ and determinant $det(A) = a d - b c$ 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 matrix or 4x4 matrix could be derived in an analogous way, using the formulae for the roots of a cubic or quartic equation.

[edit] 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 a basis for the kernel of the matrix given by

B - λ n I

. For each of the eigenvalues. These are the eigenvectors

The eigenvectors given from different eigenvalues are linearly independent.
The eigenvectors given from a root-multiplicity are also linearly 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.

[edit] 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 $p (λ) = - (λ - 2)(λ - 1)(λ + 1)$ . Therefore, the eigenvalues of A are 2, 1 and −1.

[edit] Identifying eigenvectors

With the eigenvalues in hand, we can solve sets of simultaneous linear equations to determine the corresponding eigenvectors. Since we are solving for the system $(A - \lambda I)v = 0\,$ , if $λ = 2$ then,

$\begin{bmatrix} -2 & 1 & -1 \\ 1 & -1 & 0 \\ -1 & 0 & -1 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix} = 0.$

Now, reducing $(A - 2I)\,$ to row echelon form:

$\begin{bmatrix} -2 & 1 & -1 \\ 1 & -1 & 0 \\ -1 & 0 & -1 \end{bmatrix} \to \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{bmatrix}$

allows us to solve easily for the eigenspace $E 2$ :

$\begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix} = 0 \to \begin{cases} v_1 + v_3 = 0 \\ v_2 + v_3 = 0 \end{cases}$

$\to E_2 = {\rm span}\begin{bmatrix}1 \\ 1 \\ -1\end{bmatrix}$ .

We can confirm that a simple example vector chosen from eigenspace $E 2$ is a valid eigenvector with eigenvalue $λ = 2$ :

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

Note that we can determine the degrees of freedom of the solution by the number of pivots.

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 $(A - z I) v = 0$ , 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

A x = 2 x 1 v 1 + x 2 v 2 - x 3 v 3 .

[edit] See also

For a more general point of view, see Eigenvalue, eigenvector, and eigenspace
For numerical methods for computing eigenvalues of matrices, see eigenvalue algorithm

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Categories: Articles to be merged since December 2006 | Linear algebra