Math 1600 Lecture 28, Section 2, 12 Nov 2014

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Announcements:

Today we review 4.3 and discuss Appendix C. Next class we finish 4.3 and start 4.4. Read Section 4.4 for next class. Work through recommended homework questions. Exercises and solutions for Appendix C are posted on that page.

Next class: course evaluations at start.

Final exam: Monday, December 8, 9am to noon. See the course home page for final exam conflict policy. You should immediately notify the registrar or your Dean's office (and your instructor) of any conflicts! (Deadline Nov 21.)

Tutorials: Quiz 7 covers 4.2, the parts of Appendix D that we covered, and the part of 4.3 we finished Monday. No complex eigenvalues/roots.

Help Centers: Monday-Friday 2:30-6:30 in MC 106.

Office hour: Wednesday, 11:30-noon, MC103B.

Brief review of last lecture:

The characteristic polynomial of a square matrix $A$ is $\det(A - \lambda I)$, which is a polynomial in $\lambda$. The roots/zeros of this polynomial are the eigenvalues of $A$.

A root $a$ of a polynomial $f$ implies that $f(x) = (x-a) g(x)$. Sometimes, $a$ is also a root of $g(x)$. Then $f(x) = (x-a)^2 h(x)$. The largest $k$ such that $(x-a)^k$ is a factor of $f$ is called the multiplicity of the root $a$ in $f$.

In the case of an eigenvalue, we call its multiplicity in the characteristic polynomial the algebraic multiplicity of this eigenvalue.

For example, if $\det(A - \lambda I) = -(\lambda - 1)^2(\lambda -2)$, then $\lambda = 1$ is an eigenvalue with algebraic multiplicity 2, and $\lambda = 2$ is an eigenvalue with algebraic multiplicity 1.

We also define the geometric multiplicity of an eigenvalue $\lambda$ to be the dimension of the corresponding eigenspace.

Theorem 4.15: The eigenvalues of a triangular matrix are the entries on its main diagonal (repeated according to their algebraic multiplicity).

Theorem 4.17: Let $A$ be an $n \times n$ matrix. The following are equivalent:
a. $A$ is invertible.
b. $A \vx = \vb$ has a unique solution for every $\vb \in \R^n$.
c. $A \vx = \vec 0$ has only the trivial (zero) solution.
d. The reduced row echelon form of $A$ is $I_n$.
f. $\rank(A) = n$
g. $\nullity(A) = 0$
h. The columns of $A$ are linearly independent.
i. The columns of $A$ span $\R^n$.
j. The columns of $A$ are a basis for $\R^n$.
k. The rows of $A$ are linearly independent.
l. The rows of $A$ span $\R^n$.
m. The rows of $A$ are a basis for $\R^n$.
n. $\det A \neq 0$
o. $0$ is not an eigenvalue of $A$

Eigenvalues of powers and inverses

Theorem 4.18: If $\vx$ is an eigenvector of $A$ with eigenvalue $\lambda$, then $\vx$ is an eigenvector of $A^k$ with eigenvalue $\lambda^k$. This holds for each integer $k \geq 0$, and also for $k < 0$ if $A$ is invertible.

In contrast to some other recent results, this one is very useful computationally:

Example 4.21: Compute $\bmat{rr} 0 & 1 \\ 2 & 1 \emat^{10} \coll 5 1$.

See last lecture for the method used, which is much faster than repeated matrix multiplication.

Theorem: If $\vv_1, \vv_2, \ldots, \vv_m$ are eigenvectors of $A$ corresponding to distinct eigenvalues $\lambda_1, \lambda_2, \ldots, \lambda_m$, then $\vv_1, \vv_2, \ldots, \vv_m$ are linearly independent.

New material: Appendix C: Complex numbers

Sometimes a polynomial has complex numbers as its roots, so we need to learn a bit about them.

A complex number is a number of the form $a + bi$, where $a$ and $b$ are real numbers and $i$ is a symbol such that $i^2 = -1$.

If $z = a + bi$, we call $a$ the real part of $z$, written $\Re z$, and $b$ the imaginary part of $z$, written $\Im z$.

Complex numbers $a+bi$ and $c+di$ are equal if $a=c$ and $b=d$.

On board: sketch complex plane and various points.

Addition: $(a+bi)+(c+di) = (a+c) + (b+d)i$, like vector addition.

Multiplication: $(a+bi)(c+di) = (ac-bd) + (ad+bc)i$. (Explain.)

Examples: $(1+2i) + (3+4i) = 4+6i$ $$ \kern-7ex \begin{aligned} (1+2i)(3+4i)\ &= 1(3+4i)+2i(3+4i) = 3+4i+6i+8i^2 \\ &= (3 - 8) + 10 i = -5+10i \\[5pt] 5(3+4i)\ &= 15+20i\\[5pt] (-1)(c+di)\ &= -c -di \end{aligned} $$ The conjugate of $z = a+bi$ is $\bar{z} = a-bi$. Reflection in real axis. We'll use this for division of complex numbers in a moment.

Theorem (Properties of conjugates): Let $w$ and $z$ be complex numbers. Then:
1. $\bar{\bar{z}} = \query{z}$
2. $\overline{w+z} = \query{\bar{w} + \bar{z}}$
3. $\overline{w z} = \query{\bar{w} \bar{z}}$ (typo in text) (good exercise)
4. If $z \neq 0$, then $\overline{w/z} = \query{\bar{w} / \bar{z}}$ (see below for division)
5. $z$ is real if and only if $\bar{z} = \query{z}$

The absolute value or modulus $|z|$ of $z = a+bi$ is $$ \kern-4ex |z| = |a+bi| = \sqrt{a^2+b^2}, \qtext{the distance from the origin.} $$ Examples: $|3| = 3,\, |-3| = 3,\, |\pm i| = 1,\, |3+4i| = \sqrt{3^2+4^2} = 5$.

Note that $$ \kern-7ex z \bar{z} = (a+bi)(a-bi) = a^2 -abi+abi-b^2 i^2 = a^2 + b^2 = |z|^2 $$ This means that for $z \neq 0$ $$ \kern-4ex \frac{z \bar{z}}{|z|^2} = 1 \qtext{so} z^{-1} = \frac{\bar{z}}{|z|^2} $$ This can be used to compute quotients of complex numbers: $$ \kern-4ex \frac{w}{z} = \frac{w}{z} \frac{\bar{z}}{\bar{z}} = \frac{w \bar{z}}{|z|^2}. $$ Example: $$ \kern-8ex \frac{-1+2i}{3+4i} = \frac{-1+2i}{3+4i} \frac{3-4i}{3-4i} = \frac{5+10i}{3^2+4^2} = \frac{5+10i}{25} = \frac{1}{5} + \frac{2}{5}i $$

Theorem (Properties of absolute value): Let $w$ and $z$ be complex numbers. Then:
1. $|z| = 0$ if and only if $z = 0$.
2. $|\bar{z}| = |z|$
3. $|w z| = |w| |z|$  (good exercise!)
4. If $z \neq 0$, then $|w/z| = |w|/|z|$. In particular, $|1/z| = 1/|z|$.
5. $|w+z| \leq |w| + |z|$.

Polar Form

A complex number $z = a + bi$ can also be expressed in polar coordinates $(r, \theta)$, where $r = |z| \geq 0$ and $\theta$ is such that $$ \kern-6ex a = r \cos \theta \qqtext{and} b = r \sin \theta \qqtext{(sketch)} $$ Then $$ \kern-6ex z = r \cos \theta + (r \sin \theta) i = r(\cos \theta + i \sin \theta) $$ To compute $\theta$, note that $$ \kern-6ex \tan \theta = \sin\theta/\cos\theta = b/a . $$ But this doesn't pin down $\theta$, since $\tan(\theta+\pi) = \tan\theta$. You must choose $\theta$ based on what quadrant $z$ is in. There is a unique correct $\theta$ with $-\pi < \theta \leq \pi$, and this is called the principal argument of $z$ and is written $\Arg z$ (or $\arg z$).

Examples: If $z = 1 + i$, then $r = |z| = \sqrt{1^2+1^2} = \sqrt{2}$. By inspection, $\theta = \pi/4 = 45^\circ$. We also know that $\tan \theta = 1/1 = 1$, which gives $\theta = \pi/4 + k \pi$, and $k = 0$ gives the right quadrant.

We write $\Arg z = \pi/4$ and $z = \sqrt{2}(\cos \pi/4 + i \sin \pi/4)$.

If $w = -1 - i$, then $r = \sqrt{2}$ and by inspection $\theta = -3\pi/4 = -135^\circ$. We still have $\tan \theta = -1/-1 = 1$, which gives $\theta = \pi/4 + k \pi$, but now we must take $k$ odd to land in the right quadrant. Taking $k=-1$ gives the principal argument: $$ \kern-8ex \Arg w = -3\pi/4 \qtext{and} w = \sqrt{2}(\cos (-3\pi/4) + i \sin (-3\pi/4)). $$

Multiplication and division in polar form

Let $$ \kern-7ex z_1 = r_1(\cos \theta_1 + i \sin\theta_1) \qtext{and} z_2 = r_2(\cos \theta_2 + i \sin\theta_2) . $$ Then $$ \kern-8ex \begin{aligned} z_1 z_2\ &= r_1 r_2 (\cos \theta_1 + i \sin\theta_1) (\cos \theta_2 + i \sin\theta_2) \\ &= r_1 r_2 [(\cos \theta_1 \cos\theta_2 - \sin\theta_1\sin\theta_2) + i(\sin\theta_1 \cos\theta_2 + \cos\theta_1\sin\theta_2)] \\ &= r_1 r_2 [\cos(\theta_1 + \theta_2) +i \sin(\theta_1+\theta_2)] \end{aligned} $$ So $$ \kern-8ex |z_1 z_2| = |z_1| |z_2| \qtext{and} \Arg(z_1 z_2) = \Arg z_1 + \Arg z_2 $$ (up to multiples of $2\pi$). Sketch on board. See also Example C.4.

In particular, if $z = r (\cos \theta + i \sin \theta)$, then $z^2 = r^2 (\cos (2 \theta) + i \sin (2 \theta))$. It follows that the two square roots of $z$ are $$ \pm \sqrt{r} (\cos (\theta/2) + i (\sin \theta/2)) $$

The remaining material is for your interest only:

Repeating this argument gives:

Theorem (De Moivre's Theorem): If $z = r(\cos\theta+i\sin\theta)$ and $n$ is a positive integer, then $$ z^n = r^n (\cos(n\theta) + i \sin(n\theta)) $$ When $r \neq 0$, this also holds for $n$ negative. In particular, $$ \frac{1}{z} = \frac{1}{r} (\cos\theta - i\sin\theta). $$

Example C.5: Find $(1+i)^6$.

Solution: We saw that $1+i = \sqrt{2}(\cos(\pi/4)+i\sin(\pi/4))$. So $$ \kern-4ex \begin{aligned} (1+i)^6\ &= (\sqrt{2})^6 (\cos(6\pi/4)+i\sin(6\pi/4)) \\ &= 8 (\cos(3\pi/2) + i\sin(3\pi/2)) \\ &= 8 (0 + i(-1)) = -8i \end{aligned} $$

$n$th roots

De Moivre's Theorem also lets us compute $n$th roots:

Theorem: Let $z = r(\cos\theta + i\sin\theta)$ and let $n$ be a positive integer. Then $z$ has exactly $n$ distinct $n$th roots, given by $$ r^{1/n} \left[ \cos\left(\frac{\theta+2k\pi}{n}\right) + i\sin\left(\frac{\theta+2k\pi}{n}\right) \right] $$ for $k = 0, 1, \ldots, n-1$.

These are equally spaced points on the circle of radius $r^{1/n}$.

Example: The cube roots of $-8$: Since $-8 = 8(\cos(\pi)+i\sin(\pi))$, we have $$ \kern-6ex (-8)^{1/3} = 8^{1/3} \left[ \cos\left(\frac{\pi+2k\pi}{3}\right) + i\sin\left(\frac{\pi+2k\pi}{3}\right) \right] $$ for $k = 0, 1, 2$. We get $$ \kern-6ex 2 ( \cos(\pi/3)+i\sin(\pi/3) ) = 2(1/2 + i \sqrt{3}/2) = 1 + \sqrt{3} i $$ $$ \kern-6ex 2 ( \cos(3\pi/3)+i\sin(3\pi/3) ) = 2(-1 + 0 i ) = -2 $$ $$ \kern-6ex 2 ( \cos(5\pi/3)+i\sin(5\pi/3) ) = 2(1/2 - i \sqrt{3}/2) = 1 - \sqrt{3} i $$

Euler's formula

Using some Calculus, one can prove:

Theorem (Euler's formula): For any real number $x$, $$ e^{ix} = \cos x + i \sin x $$

Thus $e^{ix}$ is a complex number on the unit circle. This is most often used as a shorthand: $$ z = r (\cos\theta + i\sin\theta) = re^{i\theta} $$ It also leads to one of the most remarkable formulas in mathematics, which combines 5 of the most important numbers: $$ e^{i \pi} + 1 = 0 $$