The characteristic polynomial¶
Definition and Lemma 6.5 (Related exercises: Exercise 6.1, Exercise 6.3, Exercise 6.4, Exercise 6.6, Exercise 6.2)
For \(A \in {\mathrm {Mat}}_{n \times n}\), the function
is a polynomial of degree \(n\). It is called the characteristic polynomial of the matrix \(A\). A real number \(\lambda\) is an eigenvalue of \(A\) if and only if
For example, for \(A = \left ( \begin{array}{cc} 1 & r \\ 0 & 1 \end{array} \right )\), \(\chi(t) = (t-1)^2\).
Example 6.6
We compute the eigenvalues of \(A = \begin{pmatrix} 2 & -1 \\ 4 & -3 \end{pmatrix}\) using the characteristic polynomial:
The equation \(\chi_A(t) = 0\) solves as
i.e., the eigenvalues are \(t_1 = -2\), \(t_2 = 1\).
Example 6.7
We consider the rotation matrix \(A = \left ( \begin{array}{cc} \cos r & -\sin r \\ \sin r & \cos r \end{array} \right )\). We have:
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if \(r = \dots, -4\pi, -2\pi, 0, 2\pi, \dots\) (i.e., a rotation by a multiple of 360\(^\circ\), i.e., no rotation at all), then \(A = {\mathrm {id}}_2\), and the only eigenvalue is \(\lambda = 1\), and any vector \((x,y) \in {\bf R}^2\) is an eigenvector.
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if \(r = \dots, -3\pi, -\pi, \pi, 3\pi, \dots\) (i.e., a rotation by 180\(^\circ\) (plus an irrelevant multiple of 360\(^\circ\))), the only eigenvalue is \(\lambda = -1\), and again any vector in \({\bf R}^2\) is an eigenvector,
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in all other cases, i.e., if the rotation is not by \(0^\circ\) or by \(180^\circ\), there are no eigenvalues (and therefore no eigenvectors).
These statements are clear geometrically: for a rotation other than the special cases, for any vector \(v \in {\bf R}^2\) we have that the rotated vector \(Av\) lies on a different line, so that it cannot be an eigenvector. To confirm this algebraically, we compute its characteristic polynomial:
We solve this for \(t\) using the usual formula:
We have \(0 \le \cos r \le 1\), and \(\cos r = 1\) if and only if \(r\) is a multiple of \(\pi\) (cf. §Chapter B). Thus the term inside the square root is zero in this case, in all other cases it is strictly negative, so that the equation \(\chi(t) = 0\) has no solution.
The non-existence of eigenvalues can be salvaged by working with complex numbers, instead of real numbers.
Theorem 6.8 (Related exercises: Exercise 6.9)
(Fundamental theorem of algebra, cf. also §Section 1.2 for further discussion) For every non-constant polynomial
where the coefficients \(a_0, \dots, a_n\) are complex numbers (for example, they can be real numbers), there exists a complex number \(z \in {{\bf C}}\) such that
This theorem is famous for the number of entirely different proofs. A completely elementary proof fitting on about two pages is given in (Oliveira 2011).
Example 6.9
Consider the rotation matrix
describing a (counter-clockwise) rotation by \(90^\circ\). Its characteristic polynomial
does not have a real zero, i.e., for all real numbers \(r\), \(\chi_A(r) \ne 0\). However, the complex number \(i\), the imaginary unit, which satisfies \(i^2 = -1\) is a complex zero. In addition, \(-i\), which also satisfies \((-i)^2 = -1\) is another complex zero.
It is therefore helpful to consider the concepts of linear algebra not only for real matrices, but for complex matrices. All of the concepts and theorems that we have encountered in this course continue to hold for complex matrices, complex vector spaces etc.
Corollary 6.10
Any complex square matrix \(A \in {\mathrm {Mat}}_{n \times n}\) has at least one (complex) eigenvalue. In particular, any real square matrix has at least one complex eigenvalue (but it may not have a real eigenvalue).