Objectives: Topics
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Complex numbers: addition, multiplication, complex conjugate
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Complex eigenvalues and eigenvectors
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Eigenvalue theorems
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Section 5.5 Complex Eigenvalues
Handout 5.5 Complex Eigenvalues
Objectives: Goals
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Use eigenvalues to identify the rotation and dilation of a linear transform.
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Recognize rotation–dilation matrices.
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Find complex eigenvalues and eigenvectors of a real matrix.
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Apply theorems to characterize matrices with complex eigenvalues.
Example 5.41.
Consider the linear transformation \(T: \mathbb{R}^2 \rightarrow \mathbb{R}^2\) which rotates vectors counterclockwise by 90\(^\circ\text{.}\)
(a)
Find the matrix \(A\) so that \(T(\vec x) = A\vec x\text{.}\)
Solution.
Note that
\begin{equation*}
T\left(\vec e_1\right) = T\left(\begin{bmatrix} 1 \\ 0 \end{bmatrix}\right) = \begin{bmatrix} 0 \\ 1 \end{bmatrix}
\end{equation*}
and
\begin{equation*}
T\left(\vec e_2\right) = T\left(\begin{bmatrix} 0 \\ 1 \end{bmatrix}\right) = \begin{bmatrix} -1 \\ 0 \end{bmatrix}\text{.}
\end{equation*}
Therefore, we have that
\begin{equation*}
A = \begin{bmatrix} T\left(\vec e_1\right) \amp T\left(\vec e_2\right) \end{bmatrix} = \begin{bmatrix} 0 \amp -1 \\ 1 \amp 0 \end{bmatrix}\text{.}
\end{equation*}
(b)
Find the characteristic polynomial of \(A\text{.}\)
Solution.
Note that
\begin{equation*}
\det(A - \lambda I) = \begin{vmatrix} -\lambda \amp -1 \\ 1 \amp -\lambda \end{vmatrix} = \lambda^2 + 1\text{.}
\end{equation*}
(c)
What are the eigenvalues of \(A\text{?}\)
Solution.
Note that the quadratic equation \(\lambda^2 + 1 = 0\) has no real roots since trying to solve the equation yields
\begin{equation*}
\lambda^2 + 1 = 0 \qquad \implies \qquad \lambda^2 = - 1 \qquad \implies \qquad \lambda = \pm\sqrt{-1}
\end{equation*}
and \(\sqrt{-1}\) is not a real number. If we want to "fix" this, we can define \(i = \sqrt{-1}\) and say that \(i\) and \(-i\) are the two Complex roots of the polynomial \(\lambda^2 + 1\text{.}\)
Definition 5.42 Complex Numbers.
The complex numbers are denoted by \(\mathbb{C}\text{,}\) where
\begin{equation*}
\mathbb{C}=\{\,a+bi \mid a,b\in\mathbb{R}\,\}.
\end{equation*}
We can identify \(\mathbb{C}\) with \(\mathbb{R}^2\) via \(a+bi \leftrightarrow (a,b)\text{.}\)
We can add and multiply complex numbers by treating \(i\) as variable and simplifying powers of \(i\) appropriately.
Example 5.43.
Let \(z_1 = 2 - 3i\) and let \(z_2 = -1 + i\text{.}\) Calculate the following:
(a)
Sum of complex numbers: \(z_1 + z_2\)
Solution.
Note that
\begin{equation*}
z_1 + z_2 = (2 - 3i) + (-1 + i) = (2-1) + (-3i + i) = 1 - 2i\text{.}
\end{equation*}
(b)
Product of complex numbers: \(z_1z_2\)
Solution.
Using the distributive property, we have
\begin{equation*}
z_1z_2 = (2 - 3i)(-1+i) = -2 + 2i -3i(-1) -3i(i) = -2 + 5i -3i^2\text{.}
\end{equation*}
Since \(i = \sqrt{-1}\text{,}\) we have that \(i^2 = -1\text{.}\) Therefore,
\begin{equation*}
z_1z_2 = -2 + 5i - 3i^2 = -2 + 5i -3(-1) = 1 + 5i\text{.}
\end{equation*}
Definition 5.44 Complex Conjugate.
The conjugate of \(a+bi\) is \(\overline{a+bi}=a-bi\text{.}\)
Definition 5.45 Absolute Value (Modulus).
The absolute value (modulus) of \(a+bi\) is \(\lvert a+bi\rvert=\sqrt{a^2+b^2}\text{.}\)
Definition 5.46 Polar Form of a Complex Number.
The polar form of a complex number is \(a+bi = r(\cos\phi + i\sin\phi) = re^{i\phi}\text{,}\) where \(r=\lvert a+bi\rvert\) and \(\phi\) is the argument.
Solution.
Note that Euler’s formula was used which says \(e^{ix} = \cos(x) + i\sin(x)\text{.}\) This is a beautiful result that can be proven using the Power Series for the three functions involved.
Example 5.47.
Given \(z=3-4i\text{,}\) compute \(\overline{z}\text{,}\) \(\lvert z\rvert\text{,}\) and the polar form of \(z\text{.}\)
Solution.
Note that the conjuage of \(z\) is
\begin{equation*}
\overline{z} = \overline{3 - 4i} = 3 + 4i\text{.}
\end{equation*}
The modulus of \(z\) is
\begin{equation*}
\|z\| = \sqrt{3^2 + 4^2} = \sqrt{25} = 5\text{.}
\end{equation*}
To find the polar form of \(z\text{,}\) we first find the angle \(\phi\text{.}\) Note that
\begin{equation*}
\phi = \operatorname{arctan}\left(\frac{b}{a}\right) = \operatorname{arctan}\left(\frac{-4}{3}\right)\text{.}
\end{equation*}
Therefore, we have that
\begin{equation*}
z = \|r\|e^{i\phi} = 5e^{i\operatorname{arctan}\left(-\frac{4}{3}\right)}\text{.}
\end{equation*}
Theorem 5.48 Conjugate Properties.
Let \(x,y\in\mathbb{C}\) and \(\vec v\in\mathbb{C}^n\text{.,}\) Then we have the following properties:
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\(\displaystyle \overline{(x+y)}=\overline{x}+\overline{y}\)
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\(\displaystyle \overline{(xy)}=\overline{x} \ \overline{y}\)
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\(\overline{A\vec v}=A\,\overline{\vec v}\) (for a real matrix \(A\))
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\(\displaystyle \operatorname{Im}(x\overline{x})=0\)
Conjugation reflects points across the real axis.
The point \(z=x+iy\) at angle \(\phi\) reflects to \(\overline{z}=x-iy\) at angle \(-\phi\) across the real axis.
Suppose \(z_1\) has angle \(\phi_1\) and \(z_2\) has angle \(\phi_2\text{;}\) that is, let \(z_1 = |z_1|e^{i\phi_1}\) and \(z_2 = |z_2|e^{i\phi_2}\text{.}\) Note that
\begin{equation*}
z_3=z_1z_2=(\lvert z_1\rvert e^{i\phi_1})(\lvert z_2\rvert e^{i\phi_2})
= \lvert z_1\rvert \lvert z_2\rvert e^{i(\phi_1+\phi_2)}.
\end{equation*}
Therefore, the product \(z_1z_2\) has modulus \(|z_1||z_2|\) and has angle \(\phi_1 + \phi_2\text{.}\)
Vectors representing \(z_1\text{,}\) \(z_2\text{,}\) and their product \(z_3=z_1z_2\) with angles adding: \(\phi_3=\phi_1+\phi_2\text{.}\)
Theorem 5.49 Fundamental Theorem of Algebra.
Every polynomial of degree \(n\) has exactly \(n\) complex roots, counting multiplicity.
Theorem 5.50 Conjugate Pairs for Real Coefficients.
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If \(\lambda\in\mathbb{C}\) is a root of a real polynomial \(p(x)\text{,}\) then the conjugate \(\overline{\lambda}\) is also a root of \(p(x)\text{.}\)
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If \(\lambda\) is an eigenvalue of a real matrix \(A\) with eigenvector \(\vec v\text{,}\) then \(\overline{\lambda}\) is an eigenvalue of \(A\) with eigenvector \(\overline{\vec v}\text{.}\)
Example 5.51.
Four eigenvalues of a \(7\times 7\) real matrix \(A\) are \(-2\text{,}\) \(4+i\text{,}\) \(-4-i\text{,}\) and \(i\text{.}\) What are the other eigenvalues?
Solution.
By Theorem 5.49 and Theorem 5.50, we know that \(A\) has 7 eigenvalues, counting multiplicites. We have been told 4 of them. Note that for each complex eigenvalues, it’s complex conjugate must also be an eigenvalue. Therefore, we have that the eigenvalues are \(\lambda_1 = -2\text{,}\) and
\begin{equation*}
\lambda_2 = 4 + i \quad \implies \quad \lambda_3 = \overline{\lambda_2} = 4 - i
\end{equation*}
and
\begin{equation*}
\lambda_4 = -4 - i \quad \implies \quad \lambda_5 = \overline{\lambda_4} = -4 + i
\end{equation*}
and
\begin{equation*}
\lambda_6 = i \quad \implies \quad \lambda_7 = \overline{\lambda_6} = -i
\end{equation*}
are the 7 eigenvalues of \(A\text{.}\)
Example 5.52.
The matrix that rotates vectors by \(\phi=\pi/4\) about the origin and then scales by \(r=\sqrt{2}\) is
\begin{equation*}
A=
\begin{bmatrix}
r \amp 0\\
0 \amp r
\end{bmatrix}
\begin{bmatrix}
\cos\phi \amp -\sin\phi\\
\sin\phi \amp \cos\phi
\end{bmatrix}
=
\begin{bmatrix}
1 \amp -1\\
1 \amp 1
\end{bmatrix}.
\end{equation*}
Find the eigenvalues of \(A\) and one eigenvector for each eigenvalue.
Solution.
Note that
\begin{equation*}
\det(A - \lambda I) = \begin{vmatrix} 1-\lambda \amp -1 \\ 1 \amp 1 - \lambda\end{vmatrix} = (1-\lambda)^2 + 1 = \lambda^2 - 2\lambda + 2\text{.}
\end{equation*}
Using the quadratic formula, we find the eigenvalues of \(A\) are
\begin{equation*}
\lambda = \dfrac{2 \pm \sqrt{4 - 4(1)(2)}}{2} = \dfrac{2 \pm \sqrt{-4}}{2} = \dfrac{2 \pm 2i}{2} = 1 \pm i\text{.}
\end{equation*}
Let \(\lambda_1 = 1+i\) and \(\lambda_2 = 1-i\text{.}\) Note that these are conjuage pairs as suggested by Theorem 5.50.
Note that \(\lambda_1\) can be written in polar form as
\begin{equation*}
\lambda_1 = \|\lambda_1\|e^{i\operatorname{arctan}\left(\frac{1}{1}\right)} = \sqrt{2}e^{i\frac{\pi}{4}}\text{.}
\end{equation*}
We see that \(r = \|\lambda_1\| = \sqrt{2}\) gives the scaling factor of \(A\) and \(\phi = \frac{\pi}{4}\) gives the rotation factor of \(A\text{.}\)
Now, let’s find at least one eigenvector for each eigenvalue. For \(\lambda_1\text{,}\) we have
\begin{equation*}
A - \lambda_1 I = \begin{bmatrix} 1 \amp -1 \\ 1 \amp 1 \end{bmatrix} - \begin{bmatrix} 1 + i \amp 0 \\ 0 \amp 1 + i \end{bmatrix} = \begin{bmatrix} -i \amp -1 \\ 1 \amp -i \end{bmatrix}\text{.}
\end{equation*}
At first glance, \(A - \lambda_1 I\) might not appear to have a free variable. Hopefully it does, or else we made an error somewhere. Note that we can still use row operations on \(A\) even though it contains complex numbers. This, hopefully, will expose the dependency of the columns. Note that
\begin{equation*}
\begin{bmatrix} -i \amp -1 \\ 1 \amp -i \end{bmatrix} \overset{\text{Swap $R_1$ and $R_2$}}{\LARGE \sim} \begin{bmatrix} 1 \amp -i \\ -i \amp -1 \end{bmatrix} \overset{R_2 + iR_1 \rightarrow R_2}{\LARGE \sim} \begin{bmatrix} 1 \amp -i \\ 0 \amp 0 \end{bmatrix}\text{.}
\end{equation*}
Now, we see that
\begin{equation*}
\vec v_1 = \begin{bmatrix} i \\ 1 \end{bmatrix} \in \Nul(A - \lambda_1 I)\text{.}
\end{equation*}
Let’s check that \(A\vec v_1 = \lambda_1 \vec v_1\text{.}\) Note that
\begin{equation*}
A\vec v_1 = \begin{bmatrix} 1 \amp -1 \\ 1 \amp 1 \end{bmatrix} \begin{bmatrix} i \\ 1 \end{bmatrix} = \begin{bmatrix} i - 1 \\ i + 1 \end{bmatrix}
\end{equation*}
and
\begin{equation*}
\lambda_1 \vec v_1 = (1 + i)\begin{bmatrix} i \\ 1 \end{bmatrix} = \begin{bmatrix} i + i^2 \\ 1 + i \end{bmatrix} = \begin{bmatrix} i - 1 \\ i + 1 \end{bmatrix}\text{,}
\end{equation*}
so \(A\vec v_1 = \lambda_1 \vec v_1\text{.}\)
By Theorem 5.50, we have that
\begin{equation*}
\vec v_2 = \overline{\vec v_1} = \begin{bmatrix} \overline{i} \\ \overline{1} \end{bmatrix} = \begin{bmatrix} -i \\ 1 \end{bmatrix}
\end{equation*}
is an eigenvector for \(\lambda_2 = \overline{\lambda_1}\text{.}\)
The previous example is a special case of the real rotation–dilation matrix \(C=\begin{bmatrix} a \amp -b\\ b \amp a \end{bmatrix}\text{.}\)
Example 5.53.
Compute the eigenvalues of
\begin{equation*}
C=\begin{bmatrix} a \amp -b\\ b \amp a \end{bmatrix}
\end{equation*}
and express them in the polar form \(re^{i\phi}\text{.}\)
Solution.
Note that
\begin{equation*}
\det(C - \lambda I) = \begin{vmatrix} a - \lambda \amp -b \\ b \amp a - \lambda \end{vmatrix} = (a - \lambda)^2 + b^2\text{.}
\end{equation*}
Solving for the eigenvalues, we see that
\begin{align*}
(a - \lambda)^2 + b^2 = 0 \amp \implies (a-\lambda)^2 = -b^2 \\
\amp \implies a - \lambda = \pm bi \\
\amp \implies \lambda = a \pm bi\text{.}
\end{align*}
Therefore, \(\lambda_1 = a + bi\) and \(\lambda_2 = a - bi\) are the two eigenvalues of \(C\text{.}\) Writing these in polar form, we have that
\begin{equation*}
\lambda_1 = \sqrt{a^2 + b^2}e^{i\operatorname{arctan}\left(\frac{b}{a}\right)} \quad \text{and} \quad \lambda_2 = \sqrt{a^2 + b^2}e^{i\operatorname{arctan}\left(-\frac{b}{a}\right)}\text{.}
\end{equation*}
Similar to the eigenvalues found in Example 5.52, we see that \(r = \|\lambda_1\| = \sqrt{a^2 + b^2}\) is the scaling factor of \(C\) and \(\phi = \operatorname{arctan}\left(\frac{b}{a}\right)\) is the rotation factor of \(C\text{.}\)
Example 5.54.
Find the complex eigenvalues and an associated complex eigenvector for each eigenvalue of
\begin{equation*}
A=\begin{bmatrix}
1 \amp -2\\
1 \amp 3
\end{bmatrix}.
\end{equation*}
Solution.
Note that
\begin{equation*}
\det(A - \lambda I) = \begin{vmatrix} 1 - \lambda \amp -2 \\ 1 \amp 3-\lambda \end{vmatrix} = (1-\lambda)(3-\lambda) + 2 = \lambda^2 - 4\lambda + 5\text{.}
\end{equation*}
Using the quadratic formula, we see that the eigenvalues of \(A\) are
\begin{equation*}
\lambda = \dfrac{4 \pm \sqrt{16 - 20}}{2} = \dfrac{4 \pm 2i}{2} = 2\pm i\text{.}
\end{equation*}
Let \(\lambda_1 = 2 + i\text{.}\) Then we have that
\begin{equation*}
A - \lambda_1 I = \begin{bmatrix} -1 -i \amp -2 \\ 1 \amp 1-i \end{bmatrix} \sim \begin{bmatrix} 1 \amp 1-i \\ 0 \amp 0 \end{bmatrix}\text{.}
\end{equation*}
It follows that
\begin{equation*}
\vec v_1 = \begin{bmatrix} 1 - i \\ -1 \end{bmatrix}
\end{equation*}
is an eigenvector corresponding to \(\lambda_1\text{.}\) Check this.
Let \(\lambda_2 = 2 - i = \overline{\lambda_1}\text{.}\) By Theorem 5.50, we see that
\begin{equation*}
\vec v_2 = \overline{\vec v_1} = \begin{bmatrix} 1 + i \\ -1 \end{bmatrix}
\end{equation*}
is an eigenvector corresponding to \(\lambda_2\text{.}\)
