Complex Eigenvalues

Complex eigenvalues

Brian Krummel April 17, 2020

1 Motivation

So far we have focused on computing real eigenvalues and eigenvectors of square matrices. Consider the following example:

Example 1. Let A be the rotation matrix by 90 degrees clockwise:

0 −1 A = . 1 0

Find the eigenvalues and eigenvectors of A. Answer. Finding eigenvalues of A. Solving the characteristic equation

−λ −1 2 det(A − λI) = = λ + 1 = 0. 1 −λ √ Then λ = ±i, where i = −1. In particular, A has complex eigenvalues.

2 Complex numbers

In order to ﬁnd and understand the eigenvalues and eigenvectors of general square matrices, including the rotation matrix in Example 1, we need to introduce complex numbers and complex vectors. √ Deﬁnition 1. C is the set of all complex numbers z = a + ib, where a, b ∈ R and i = −1. One can add and multiply complex numbers z = a + ib and w = c + id by

z + w = (a + ib) + (c + id) = (a + c) + i (b + d), zw = (a + ib)(c + id) = ac + i2bd + i (ad + bc) = (ac − bd) + i (ad + bc), where when multiplying zw we used i2 = −1. Also, the complex conjugate z of z = a + ib is

z = a − ib.

1 Example 2. If z = 2 + 3i and w = 4 + 5i then

z + w = (2 + 3i) + (4 + 5i) = (2 + 4) + i (3 + 5) = 6 + 8i, zw = (2 + 3i) (4 + 5i) = 2 · 4 − 3 · 5 + i (2 · 5 + 3 · 4) = 12 − 15 + i (10 + 12) = −3 + 22i, z = 2 − 3i.

Complex conjugation has the following important properties. For each pair of complex numbers z, w ∈ C, z + w = z + w, zw = z w. Given a complex number z = a + ib ∈ C, we have that z = a − ib is equal to z if and only if b = 0. That is, z = z if and only if z is a real number. Finally, given a complex number z = a + ib ∈ C z z = |z|2 = a2 + b2. √ We call |z| = a2 + b2 the absolute value or modulus of z. In particular, z z is a non-negative real number.

We can represent a complex number z = a+ib as a point (a, b) in R2. Thus we regard z = a+ib as a point in the complex plane, whose axes are the real and imaginary axes. Alternatively, we can represent z as z = reiθ = r cos(θ) + ir sin(θ) is polar coordinates, where z has length r = |z| ≥ 0 and angle θ ∈ R (typically 0 ≤ θ < 2π).

Imag

z r b θ a Real

We can express a and b in terms of r and θ usingImag the formulas a = r cos(θ), b = r sin(θ). z Moreover, we can express r and θ in terms of a and rb using the formulas √ θ b r = a2 + b2, tan(θ) = . Real a Notice that tan(θ) is deﬁned only up to adding a multiple of π to θ. Hence we need to look at which quadrant of the complex plane z is in to determine the value of θ.

2 √ Example 3. Express the complex number z = −8 + 8 3 i in polar coordinates as z = reiθ. Answer. It is useful to ﬁrst draw a picture with z as a point in the complex plane:

Imag

8 3 π 3 −8 Real

From the picture and the formulas, we see that Imag √ √ √ √ r = a2 + b2 = 82 + 82 ·ω3 =i 8 1 + 3 = 8 4 = 8 · 2 = 16, √ = 8 3 √ tan(θ) = = − 3 ⇒ θ = 2π/3 or 5π/3. −8

Looking at the picture, we seeω that2 z is in 2nd quadrant and thus θ =Real 2π/3. Therefore, z = 16ei2π/3. = − 1 Imag zw1 Geometrically, we can represent complex number addition and multiplication as follows. We add complex vectors z = a + ib and w = c + id asrs ω3 = − i r z z + w = (a + ib) + (c + idϕ) = (a + c) + i (b + d). θ Real Thus we add complex numbers by adding the respective real and imaginary parts. This is just like how we add vectors in R2 by adding their respective components. In fact, C forms a two- dimensional vector space (over the real numbers) with basis {1, i}. Geometrically, we represent complex number addition via the parallelogram rule:

Imag

z z+w

Real w

We multiply complex vectors z = reiθ and w = seiφ as

zw = reiθ · seiφ = rsei (θ+φ).

Thus we multiply complex numbers by multiplying their lengths rs and adding their angles θ + φ.

3 Geometrically, multiplication zw looks like rotating z by φ radians counter-clockwise and scaling the length of z by s:

Imag zw

rs z ϕ r Imagθ Real

a Let k be an integer (or in fact any real number). If weθ multiply z timesReal itself k times to obtain the Imag− k -b k

k-th power z , geometrically this looks like scaling kr times by z to get a length r and rotating k times by z to get an angle kθ. In other words, z

k z k ikθ k ikθ z = r e = rz+ew . Imag Real w z2 z θ θ θ Real z3

√ Example 4. For the complex number z = −8 + 8 3 i, compute z3. Answer. Recall from Example 3 that z = 16ei2π/3. Hence

z3 = 163ei 3·2π/3 = 4096ei 2π.

Notice that ei 2π = 1 since ei 2π has length one and angle 2π. Therefore, z3 = 4096.

Before we move on to complex vectors, let’s discuss the meaning of eiθ, where θ ∈ R. One interpretation is that z = reiθ simply is a shorthand for polar coordinates. Another interpretation is that eiθ is the exponential function of the complex number iθ. We can deﬁne the exponential function ez for a complex number z ∈ C using the Taylor series ∞ X zk ez = . k! k=0 This series converges absolutely for all z ∈ C and thus deﬁnes the exponential function ez. Note that the exponential function has many of the same properties for complex numbers as for real

4 numbers; for instance, ez+w = ez · ew for all z, w ∈ C. Substituting z = iθ into the Taylor series, we get ∞ X (iθ)k eiθ = . k! k=0 Breaking up the sum into two sums where k = 2j is even and k = 2j + 1 is odd, we get

∞ ∞ X (iθ)2j X (iθ)2j+1 eiθ = + . (2j)! (2j + 1)! j=0 j=0

Since i2 = −1, ∞ ∞ X (−1)jθ2j X (−1)jθ)2j+1 eiθ = + i = cos(θ) + i sin(θ), (2j)! (2j + 1)! j=0 j=0 where in the last step we used the Taylor series for cosine and sine. Therefore, eiθ = cos(θ)+i sin(θ) for all θ ∈ R.

3 Complex vectors

Deﬁnition 2. Cn is the set of all n × 1 column vectors whose entries are complex numbers. For instance,  1 + i  Z =  2 − 3i  i is a vector in C3. Each of the entries of Z is a complex number. We can express each complex vector Z ∈ Cn as Z = X + iY for some real vectors X,Y ∈ Rn. We call X the real part of Z and Y the imaginary part of Z. For instance,

 1 + i   1   1   2 − 3i  =  2  + i  −3  i 0 1 with  1   1  real part =  2  imaginary part =  −3  . 0 1 We add and scale complex vectors entry-by-entry like we normally do, except scalars can be complex numbers:

 1 + i   4 + 7i   (1 + i) + (4 + 7i)   5 + 8i   2 − 3i  +  2i  =  (2 − 3i) + 2i  =  2 − i  i 9 i + 9 9 + i  1 + i   (4 + 5i)(1 + i)   −1 + 9i  (4 + 5i)  2 − 3i  =  (4 + 5i)(2 − 3i)  =  23 − 2i  i (4 + 5i) i −5 + 4i

5 Given Z = X + iY ∈ Cn (where X,Y ∈ Rn), the complex conjugate of a Z = X + iY is the vector Z = X − iY in Cn. In particular, the entries of Z are the complex conjugates of the corresponding entries of Z. For instance,  1 + i   1 − i  conjugate of  2 − 3i  is  2 + 3i  . i −i In this course so far we have worked with vector spaces with real number scalars. We could have also developed the theory of vector spaces with complex number scalars. In fact, we could have worked with any scalars in any field. In mathematics, a field F is a set together with addition and multiplication operations that behave just like the real numbers, including subtraction and division; see Wikipedia for the precise definition. One example of a finite field is the integers modulo two Z2 = {0, 1} for which 0+0=0 0+1=1+0=1 1+1=0, 0 · 0 = 0 0 · 1 = 1 · 0 = 0 1 · 1 = 1.

Notice in particular that 1+1 = 0. Also note that we could have developed the theory of matrices and linear systems in the complex numbers, or any other ﬁeld.

Deﬁnition 3. Let A be an n × n matrix with real or complex entries. We say that λ ∈ C is an eigenvalue of A if there is a non-zero vector X ∈ Cn such that AX = λX.

We call any such non-zero vector X an eigenvector. The set of all solutions X ∈ Cn to (A−λI) X = 0 is called the eigenspace corresponding to the eigenvalue λ.

Notice that λ ∈ C is an eigenvalue and X ∈ Cn is an eigenvector if and only if (A − λI) X = 0. In particular, A − λI is a singular n × n matrix with complex entries. Hence we can ﬁnd complex eigenvalues by solving the characteristic equation

det(A − λI) = 0 like we have been doing. Moreover, when ﬁnding complex eigenvectors, we want to ﬁnd a basis for each eigenspace by solving the homogeneous linear system

(A − λI) X = 0 like we have been doing. We note that: if A is an n × n matrix with real entires and if λ ∈ C is a complex eigenvalue and X ∈ Cn is a complex eigenvector such that AX = λX, then the complex conjugates λ is also a complex eigenvalue and X is also a complex eigenvector such that AX = AX = λX = λ X.

6 Example 1 continued. Let A be the rotation matrix by 90 degrees clockwise 0 −1 A = 1 0 √ and recall that λ = ±i, where i = −1. Find the complex eigenvectors of A. Finding eigenvectors corresponding to λ = i. To find the complex eigenvectors of A, we have to row reduce −i −1 A − i I = . 1 −i which has complex number entries, making life complicated. What we can do is swap rows so that the real number 1 is in the first pivot position. Since A − i I is a 2 × 2 non-zero singular matrix, it must have rank one and thus its second row must be a complex constant multiple of the first row. In this case we add R2 + i R1 7→ R2: −i −1 R1 ↔ R2 1 −i R2+i R1 7→ R2 1 −i A − i I = −−−−−→ −−−−−−−−→ . 1 −i −i −1 0 0

Thus x2 is a free variable and x1 is a basic variable with x1 = i x2. By setting x2 = 1 we obtain the eigenvector corresponding to λ = i i . 1 We could do this a bit differently: −i −1 R2+i R1 7→ R2 −i −1 −R1 7→ R1 i 1 A − i I = −−−−−−−−→ −−−−−−→ 1 −i 0 0 0 0 so that x2 = −ix1. Setting x1 = 1 we obtain the eigenvector corresponding to λ = i 1 . −i Why are we getting two eigenvectors? Recall that finding eigenvectors means finding a basis for the eigenspace. In this case, the eigenspace is 1-dimensional, and thus eigenvectors are defined up to scaling by a complex number scalar. In this case 1 i = −i −i 1 so we really obtained the same eigenvector up to scaling. When finding complex eigenvectors of 2 × 2 matrices, typically people find one of two possible complex eigenvectors up to scaling by a real number. These are not wrong answers, simply the complex eigenvalues are complex number multiples of one another. Finding eigenvectors corresponding to λ = −i. The eigenvector corresponding to λ = −i is just the complex conjugate of the eigenvectors corresponding to i, −i . 1

7 Example 5. Find the eigenvalues and eigenvectors of 2 5 A = . −2 4 Answer. Finding eigenvalues of A. Solving the characteristic equation

2 − λ 5 2 2 det(A − λI) = = (2 − λ)(4 − λ) + 10 = λ − 6λ + 18 = (λ − 3) + 9 = 0. −2 4 − λ Thus λ = 3 ± 3i. Finding eigenvector of A corresponding to λ = 3−3i. We again want to apply Gaussian elimination to −1 + 3i 5 A − (3 − 3i) I = . −2 1 + 3i Since λ = 3 − 3i is a complex eigenvalue, we again know that A − (3 − 3i) I is a rank one 2 × 2 matrix and thus its second row is a multiple of its first row. Of course, it does not seem obvious that the rows are multiples of one another. Let’s multiply the first row by the conjugate of the (1, 1)-entry −1 + 3i, i.e. (−1 − 3i) R1 7→ R1: −1 + 3i 5 (−1−3i) R1 7→ R1 10 −5(1 + 3i) A − (3 − 3i) I = −−−−−−−−−→ −2 1 + 3i −2 1 + 3i Now it is clear that R1 = −5 R2. Verifying this is not absolutely necessary; if we found the complex eigenvalue λ correctly then A − λI is always a rank one matrix and in particular its first row is always a multiple of its second row. However, verifying that the rows are multiples of one another is a good sanity check to make sure we got the right eigenvalue. Since the rows of the matrices are multiples of one another, we can subtract some multiple of the first row from the second row to get −1 + 3i 5 2 −1 − 3i A − (3 − 3i) I = −→ . −2 1 + 3i 0 0

(1+3i) Thus x2 is a free variable and x1 is a basic variable such that x1 = 2 x2. Setting x2 = 2, we get the eigenvector of A corresponding to λ = 3 − 3i 1 + 3i . 2 Like before, we could have done this a bit diﬀerently as −1 + 3i 5 (1−3i) R2 7→ R2 −1 + 3i 5 A − (3 − 3i) I = −−−−−−−−−→ −2 1 + 3i −2(1 − 3i) 10 R2−2 R1 7→ R2 −1 + 3i 5 −−−−−−−−→ . 0 0

−1+3i Thus x2 = 5 x1. Setting x1 = 5, we obtain the eigenvector of A corresponding to λ = 3 − 3i 5 . −1 + 3i

8 Both eigenvectors might look like very diﬀerent answers, but as we mentioned before these eigenvectors are complex scalar multiples of one another:

5 1 − 3i 1 + 3i = . 1 − 3i 2 2

Finding eigenvector of A corresponding to λ = 3 + 3i. Taking the complex conjugate of the eigenvector corresponding to 3 − 3i: 5 . 1 + 3i Notice that this gives us a way to ﬁnd complex eigenvalues and eigenvectors of an n×n matrix A with real entries. However, we would like to understand how the matrix A is behaving in terms of real numbers and real vectors. Let A be a 2 × 2 matrix with real number entries. Suppose that A has a complex eigenvalue λ = a − ib, where a, b ∈ R and b 6= 0, and corresponding eigenvector Z = X + iY , where X,Y ∈ Rn. The reason for looking at the conjugate a − ib will come clear in a moment. Since Z is an eigenvector corresponding to the eigenvalue λ,

AZ = λZ.

Since AZ = AX + iAY, λZ = (a − ib)(X + iY ) = (aX + bY ) + i (−bX + aY ), Hence AZ = λZ means that

AX = aX + bY AY = −bX + aY.

In other words, after converting to {X,Y }-coordinates, A transforms into the matrix

a −b M = . b a

Equivalently, a −b A = PMP −1 = P P −1 where P = XY b a To understand the geometric interpretation of M, let λ = re−iθ = r cos(θ) − i r sin(θ). Then

a −b r cos(θ) −r sin(θ) cos(θ) − sin(θ) M = = = r b a r sin(θ) r cos(θ) sin(θ) cos(θ)

9 In other words, M consists of scaling by r and rotating θ radians counter-clockwise:

x2 MX

θ X x θ X 1

In terms of the basis {X,Y }, multiplication by A consists of scaling by r and rotating θ radians (in {X,Y }-coordinates) rotating from the real part X to the imaginary part Y : x2 x 2 AX AX X X

Y x1 Y x1

AY AY

Example 5 continued. Given 2 5 A = , −2 4 write A = PMP −1. Answer. Recall that we found the eigenvalues and eigenvectors

1 + 3i 1 3 λ = a − bi = 3 − 3i X + iY = = + i . 2 2 0

Thus A = PMP −1 where a −b 3 −3 1 3 M = = P = XY = . (1) b a 3 3 2 0

10 x2 MX

θ X

√ √ Since λ = 3 − 3i = 3 2e−iπ/4, multiplication by A is given by scaling by 3 2 and rotating π/4 radians in {X,Y }-coordinates:

AY AX Y X x1

Example 6. Find A14 where 2 5 A = . −2 4 −1 18 18 −1 Answer. Recall√ that we expressed A as A = PMP √in (??). Hence A = PM P . Since λ = 3 − 3i = 3 2e−iπ/4, M consists of scaling by 3 2 and rotating by π/4 radians counter- clockwise: 3 −3 √ cos(π/4) − sin(π/4) M = = 3 2 . 3 3 sin(π/4) cos(π/4) √ Hence M 14 consists of scaling by 3 218 = 187 and rotating by 14 · π/4 = 7π/2 = 3π + π/2 radians counter-clockwise: cos(π/2) − sin(π/2) 0 −1 M 14 = 187 = 187 . sin(π/2) cos(π/2) 1 0 Therefore, 1 3 0 −1 1 3 −1 A = PMP −1 = · 187 2 0 1 0 2 0 1 3 0 −1 1 0 3 = · 187 · 2 0 1 0 6 2 −1 187 1 3 −2 1 = 6 2 0 0 3 187 −2 10 187 −1 5 = = . 6 −4 2 3 −2 1 Note that if A is an n × n matrix with a mix of real and complex eigenvalues, we can represent A as A = PMP −1 in a manner similar to diagonal matrices and 2 × 2 matrices with complex eigenvalues. Example 7. Find the eigenvalues and eigenvectors of  1 0 −2  A =  1 3 1  . 2 0 1

11 Answer. Finding eigenvalues of A. Solving the characteristic equation

1 − λ 0 −2 1 − λ −2 det(A − λI) = 1 3 − λ 1 = (3 − λ) 2 1 − λ 2 0 1 − λ = (3 − λ) ((1 − λ)2 + 4) = (3 − λ) ((λ − 1)2 + 4) = 0. Then λ = 3, 1 + 2i, 1 − 2i. Finding eigenvector of A corresponding to λ = 3. We have  −2 0 −2   1 0 1   1 0 0  A − 3 I =  1 0 1  −→  1 0 −1  −→  0 0 1  2 0 −2 0 0 0 0 0 0

Thus x2 is a free variable and x1, x3 are basic variables such that x1 = x3 = 0. An eigenvector of A corresponding to λ = 3 is  0   1  0 Finding eigenvector of A corresponding to λ = 1 − 2i. We have  2i 0 −2   i 0 −2  (1/2) R1 7→ R1 A − (1 − 2i) I =  1 2 + 2i 1  −−−−−−−−→  1 2 + 2i 1  . 2 0 2i (1/2) R3 7→ R3 1 0 i Multiplying the ﬁrst row by the conjugate of the ﬁrst entry i:  i 0 −2   1 0 i  −i R1 7→ R1 A − (1 − 2i) I −→  1 2 + 2i 1  −−−−−−→  1 2 + 2i 1  1 0 i 1 0 i  1 0 i  R2−R1 7→ R2 −−−−−−−→  0 2 + 2i 1 − i  . R3−R1 7→ R3 0 0 0 Multiplying the second row by 1 − i to get (2 + 2i)(1 − i) = 2(1 + i)(1 − i) = 4 and noting that (1 − i)2 = −2i:  1 0 i   1 0 i  (1−i) R1 7→ R1 A − (1 − 2i) I −→  0 2 + 2i 1 − i  −−−−−−−−→  1 4 −2i  0 0 0 1 0 i  1 0 i  (1/2) R2 7→ R2 −−−−−−−−→  0 2 −i  0 0 0

Thus x3 is a free variable and x1, x2 are basic variables such that x1 = −i x3 and x2 = (i/2) x3. Setting x3 = 2 we obtain the eigenvectors of A corresponding to λ = 1 − 2i  −2i   −2   0   i  =  1  + i  0  2 0 2

12 Change of basis. Let’s write A = PMP −1. For the real eigenvalue λ = 3, we simply have the diagonal entry 3 in M and corresponding eigenvector in P :

 0 ∗ ∗   3 0 0   0 ∗ ∗ −1 A =  1 ∗ ∗   0 ∗ ∗   1 ∗ ∗  . 0 ∗ ∗ 0 ∗ ∗ 0 ∗ ∗

For the complex eigenvalue λ = −2 + i, one has a 2 × 2 block in M as above and corresponding real and imaginary parts of the complex eigenvector in P :

 ∗ −2 0   ∗ 0 0   ∗ −2 0 −1 A =  ∗ 1 0   0 1 −2   ∗ 1 0  . ∗ 0 2 0 2 1 ∗ 0 2

Putting these two things together gives us:

 0 −2 0   3 0 0   0 −2 0 −1 A =  1 1 0   0 1 −2   1 1 0  . 0 0 2 0 2 1 0 0 2