To find the eigenvalues of a matrix A, we first rearrange the equation Ax = λx by inserting an identity matrix:
Ax = λIx
Then bring both terms to the same side of the relation
Ax – λIx = 0
and factor
(A – λI)x = 0.
This is a square homogeneous linear system with coefficient matrix A – λI. Since the zero vector can't be an eigenvector (by definition) we want to find all λ's for which this equation has non-trivial solutions. The equation has non-trivial solutions whenever its coefficient matrix has determinant 0.
To find eigenvalues: |
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What sort of solutions can you expect? The matrix A – λI looks like the matrix A with λ's subtracted from its diagonal elements. Its determinant will contain sums of powers of λ, so it'll be a polynomial in λ. The highest power of λ will come from the elementary product of the main diagonal terms, so that power will be n.
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| In some texts, the characteristic polynomial is defined to be det(λI – A) and the characteristic equation det(λI – A) = 0. Since the matrix λI – A is the negative of the matrix A – λI, this may change the sign of the characteristic polynomial but doesn't affect the solutions of the characteristic equation or the eigenvectors. |
To find the eigenvalues of a matrix, then, you have first to find its characteristic polynomial and then solve the characteristic equation.
Some information on polynomial equations.
Every polynomial equation of degree n has n solutions, however, some of those solutions may be repeated solutions and some of them may be complex numbers. To see what this means, suppose that
p(t) = t9 + 6t8 + 13t7 + 14t6 + 12t5 + 8t4
This polynomial factors as follows:
p(t) = t4(t + 2)3(t + i)(t – i).
The roots of the polynomial equation p(t) = 0 are the numbers which make one of the factors is 0, i.e. the numbers
0, –2, –i and +i.
Since the factor (t + 2) is repeated 3 times, the root t = –2 is a repeated root – it has algebraic multiplicity 3. Here's a list of all the roots of this equation and their algebraic multiplicities.
| root |
algebraic multiplicity |
| 0 |
4 |
| –2 |
3 |
| –i |
1 |
| +i |
1 |
Notice that all the algebraic multiplicities must add up to the degree of the polynomial, which in this case is 9.
To solve a polynomial equation, then, you must first factor the polynomial down to linear terms – terms of the form "(t – root)". Once you've done so, the "root"s are the solutions of the equation.
When you solve the characteristic equation of a matrix to find its eigenvalues, you can generally expect to find repeated roots and complex roots. How to interpret the complex roots? If A is a matrix with real entries, then for any x in Rn, Ax is also in Rn, i.e. its coordinates are also real. If λ is complex, then λx is an n-tuple of complex numbers, so the equation Ax = λx is impossible. So if we're talking only about real vectors (i.e. vectors in Rn), the complex solutions of the characteristic equation cannot be eigenvalues. However, it is also possible to do most of what you've learned about vectors and matrices when the scalars you use and the components of the vectors are complex numbers – the vectors are then n-tuples of complex numbers. In that case, complex solutions of the characteristic equation do make sense as eigenvalues, since they can correspond to complex eigenvectors.
For now, however, we will only deal with real eigenvalues and real eigenvectors.