本文为《Linear algebra and its applications》的读书笔记

This section shows that if the characteristic equation of a real matrix A A A has some complex roots, then these roots provide critical information about A A A. The key is to let A A A act on the space C n \mathbb C^n Cn of n n n-tuples of complex numbers.

The matrix eigenvalue–eigenvector theory already developed for R n \mathbb R^n Rn applies equally well to C n \mathbb C^n Cn. So a complex scalar λ \lambda λ satisfies d e t ( A − λ I ) = 0 det(A -\lambda I)= 0 det(A−λI)=0 if and only if there is a nonzero vector x \boldsymbol x x in C n \mathbb C^n Cn such that A x = λ x A\boldsymbol x =\lambda \boldsymbol x Ax=λx. We call λ \lambda λ a (complex) eigenvalue and x \boldsymbol x x a (complex) eigenvector corresponding to λ \lambda λ.

EXAMPLE 2
Let A = [ . 5 − . 6 . 75 1.1 ] A =\begin{bmatrix} .5 &-.6\\.75&1.1\end{bmatrix} A=[.5.75​−.61.1​]. Find the eigenvalues of A A A, and find a basis for each eigenspace.
SOLUTION
The characteristic equation of A A A is

For the eigenvalue λ = . 8 − . 6 i \lambda =.8 - .6i λ=.8−.6i , construct

Row reduction of the usual augmented matrix is quite unpleasant by hand because of the complex arithmetic. However, here is a nice observation that really simplifies matters: Since . 8 − . 6 i .8 - .6i .8−.6i is an eigenvalue, the system

has a nontrivial solution. Therefore, both equations determine the same relationship between x 1 x_1 x1​ and x 2 x_2 x2​, and either equation can be used to express one variable in terms of the other.

The second equation leads to

Choose x 2 = 5 x_2 = 5 x2​=5 to eliminate the decimals, and obtain x 1 = − 2 − 4 i x_1 = -2 - 4i x1​=−2−4i . A basis for the eigenspace corresponding to λ = . 8 − . 6 i \lambda =.8 - .6i λ=.8−.6i is

Analogous calculations for λ = . 8 + . 6 i \lambda=.8 +.6i λ=.8+.6i produce the eigenvector

Surprisingly, the matrix A A A in Example 2 determines a transformation x ↦ A x \boldsymbol x \mapsto A\boldsymbol x x↦Ax that is essentially a rotation.

One way to see how multiplication by the matrix A A A affects points is to plot an arbitrary initial point—say, x 0 = ( 2 , 0 ) \boldsymbol x_0 = (2, 0) x0​=(2,0)—and then to plot successive images of this point under repeated multiplications by A A A.

Of course, Figure 1 does not explain why the rotation occurs. The secret to the rotation is hidden in the real and imaginary parts(实部和虚部) of a complex eigenvector.

Real and Imaginary Parts of Vectors 向量的实部和虚部

The complex conjugate(共轭) of a complex vector x \boldsymbol x x in C n \mathbb C^n Cn is the vector x \boldsymbol x x in C n \mathbb C^n Cn whose entries are the complex conjugates of the entries in x \boldsymbol x x. The real and imaginary parts of a complex vector x \boldsymbol x x are the vectors R e x Re\boldsymbol x Rex and I m x Im\boldsymbol x Imx in R n \mathbb R^n Rn formed from the real and imaginary parts of the entries of x \boldsymbol x x.

EXAMPLE 4
If

, then

If B B B is an m × n m \times n m×n matrix with possibly complex entries, then B ‾ \overline B B denotes the matrix whose entries are the complex conjugates of the entries in B B B. Properties of conjugates for complex numbers carry over to(适用于) complex matrix algebra:

Eigenvalues and Eigenvectors of a Real Matrix That Acts on C n \mathbb C^n Cn

Let A A A be an n × n n \times n n×n matrix whose entries are real. If λ \lambda λ is an eigenvalue of A A A and x \boldsymbol x x is a corresponding eigenvector in C n \mathbb C^n Cn, then

Hence λ ‾ \overline \lambda λ is also an eigenvalue of A A A, with x ‾ \overline \boldsymbol x x a corresponding eigenvector.

This shows that when A A A is real, its complex eigenvalues occur in conjugate pairs(以共轭复数对的形式出现). (Here and elsewhere, we use the term complex eigenvalue to refer to an eigenvalue λ = a + b i \lambda= a + bi λ=a+bi , with b ≠ 0 b \neq 0 b​=0.)

The next example provides the basic “building block” for all real 2 × 2 2 \times 2 2×2 matrices with complex eigenvalues.

EXAMPLE 6
If C = [ a − b b a ] C =\begin{bmatrix} a &-b\\b&a\end{bmatrix} C=[ab​−ba​], where a a a and b b b are real and not both zero, then the eigenvalues of C C C are λ = a ± b i \lambda= a \pm bi λ=a±bi . Also, if r = ∣ λ ∣ = a 2 + b 2 r = |\lambda|=\sqrt{ a^2 + b^2} r=∣λ∣=a2+b2 ​, then

where ϕ \phi ϕ is the angle between the positive x x x-axis and the ray(射线) from ( 0 , 0 ) (0, 0) (0,0) through ( a , b ) (a, b) (a,b). The angle ϕ \phi ϕ is called the a r g u m e n t argument argument (幅角) of λ = a + b i \lambda= a + bi λ=a+bi .

Thus the transformation x ↦ C x \boldsymbol x\mapsto C\boldsymbol x x↦Cx may be viewed as the composition of a rotation through the angle ϕ \phi ϕ and a scaling by ∣ λ ∣ |\lambda| ∣λ∣ (see Figure 3).

Finally, we are ready to uncover the rotation that is hidden within a real matrix having a complex eigenvalue.

EXAMPLE 7
Let A = [ . 5 − . 6 . 75 1.1 ] A =\begin{bmatrix} .5 &-.6\\.75&1.1\end{bmatrix} A=[.5.75​−.61.1​], λ = . 8 − . 6 i \lambda=.8-.6i λ=.8−.6i, and v 1 = [ − 2 − 4 i 5 ] \boldsymbol v_1 =\begin{bmatrix} -2-4i\\5\end{bmatrix} v1​=[−2−4i5​], as in Example 2. Also, let P P P be the 2 × 2 2 \times 2 2×2 real matrix

and let

By Example 6, C C C is a pure rotation because ∣ λ ∣ 2 = ( . 8 ) 2 + ( . 6 ) 2 = 1 |\lambda|^2=(.8)^2+(.6)^2=1 ∣λ∣2=(.8)2+(.6)2=1. From C = P − 1 A P C = P^{-1}AP C=P−1AP, we obtain

Here is the rotation “inside” A A A! The matrix P P P provides a change of variable, say, x = P u \boldsymbol x = P\boldsymbol u x=Pu. The action of A A A amounts to a change of variable from x \boldsymbol x x to u \boldsymbol u u, followed by a rotation, and then a return to the original variable. See Figure 4.

The rotation produces an ellipse, as in Figure 1, instead of a circle, because the coordinate system determined by the columns of P P P is not rectangular and does not have equal unit lengths on the two axes.

PROOF
The proof uses the fact that if the entries in A A A are real, then A ( R e x ) = R e ( A x ) A(Re\boldsymbol x)= Re(A\boldsymbol x) A(Rex)=Re(Ax) and A ( I m x ) = I m ( A x ) A(Im\boldsymbol x)= Im(A\boldsymbol x) A(Imx)=Im(Ax)([Hint]: Write x = R e x + i ( I m x ) \boldsymbol x = Re \boldsymbol x + i(Im \boldsymbol x) x=Rex+i(Imx),), and if x \boldsymbol x x is an eigenvector for a complex eigenvalue, then R e x Re\boldsymbol x Rex and I m x Im\boldsymbol x Imx are linearly independent in R 2 \mathbb R^2 R2.

The phenomenon displayed in Example 7 persists in higher dimensions. For instance, if A A A is a 3 × 3 3 \times 3 3×3 matrix with a complex eigenvalue, then there is a plane in R 3 \mathbb R^3 R3 on which A A A acts as a rotation (possibly combined with scaling). Every vector in that plane is rotated into another point on the same plane. We say that the plane is invariant under A A A.

EXAMPLE 8
The matrix A = [ . 8 − . 6 0 . 6 . 8 0 0 0 1.07 ] A =\begin{bmatrix} .8 &-.6&0\\.6&.8&0\\0&0&1.07\end{bmatrix} A=⎣⎡​.8.60​−.6.80​001.07​⎦⎤​ has eigenvalues . 8 ± . 6 i .8 \pm .6i .8±.6i and 1.07. Any vector w 0 \boldsymbol w_0 w0​ in the x 1 x 2 x_1x_2 x1​x2​-plane (with third coordinate 0) is rotated by A A A into another point in the plane. Any vector x 0 \boldsymbol x_0 x0​ not in the plane has its x 3 x_3 x3​-coordinate multiplied by 1.07.

Supplementary exercises

Chapter 7 will focus on matrices A A A with the property that A T = A A^T = A AT=A. We will show that every eigenvalue of such a matrix is necessarily real.

Let A A A be an n × n n \times n n×n real matrix with the property that A T = A A^T = A AT=A, let x \boldsymbol x x be any vector in C n \mathbb C^n Cn, and let q = x ‾ T A x \boldsymbol q = \overline\boldsymbol x^TA\boldsymbol x q=xTAx. The equalities below show that q \boldsymbol q q is a real number by verifying that q ‾ = q \overline \boldsymbol q =\boldsymbol q q​=q.

Show that if A x = λ x A\boldsymbol x =\lambda \boldsymbol x Ax=λx for some nonzero vector x \boldsymbol x x in C n \mathbb C^n Cn, then, in fact, λ \lambda λ is real and the real part of x \boldsymbol x x is an eigenvector of A A A.
PROOF
x ‾ T A x = λ x ‾ T x \overline\boldsymbol x^TA\boldsymbol x=\lambda \overline\boldsymbol x^T\boldsymbol x xTAx=λxTx

Thus λ x ‾ T x \lambda\overline\boldsymbol x^T\boldsymbol x λxTx is a real number. Since x ‾ T x \overline\boldsymbol x^T\boldsymbol x xTx is clearly a real number, λ \lambda λ is real.

Since the real part of x \boldsymbol x x equals x + x ‾ 2 \frac{\boldsymbol x+\overline\boldsymbol x}{2} 2x+x​ and A x ‾ = λ x ‾ A\overline\boldsymbol x=\lambda\overline\boldsymbol x Ax=λx,
A x + x ‾ 2 = 1 2 ( A x + A x ‾ ) = 1 2 ( λ x + λ x ‾ ) = λ x + x ‾ 2 A\frac{\boldsymbol x+\overline\boldsymbol x}{2}=\frac{1}{2}(A\boldsymbol x+A\overline\boldsymbol x)=\frac{1}{2}(\lambda\boldsymbol x+\lambda\overline\boldsymbol x)=\lambda\frac{\boldsymbol x+\overline\boldsymbol x}{2} A2x+x​=21​(Ax+Ax)=21​(λx+λx)=λ2x+x​