Math 1600B Lecture 28, Section 2, 17 Mar 2014

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Announcements:

Today we finish 4.3 and discuss Appendix C. Read Section 4.4 for next class. Work through recommended homework questions.

Tutorials: Quiz 8 covers 4.2, 4.3, and the parts of Appendix D that we covered in class.

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

Office hour: Wednesday, 10:30-11:15, MC103B. (Today's hour is cancelled.)

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 $A$

New Material: 4.3: Eigenvalues of powers and inverses

Suppose $\vx$ is an eigenvector of $A$ with eigenvalue $\lambda$. What can we say about $A^2$ or $A^3$? If $A$ is invertible, how about the eigenvalues/vectors of $A^{-1}$? On board.

We've shown:

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

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

Solution: By finding the eigenspaces of the matrix, we can show that $$ \kern-6ex \bmat{rr} 0 & 1 \\ 2 & 1 \emat \coll 1 {-1} = - \coll 1 {-1} \qtext{and} \bmat{rr} 0 & 1 \\ 2 & 1 \emat \coll 1 2 = 2 \coll 1 2 $$ Write $A = \bmat{rr} 0 & 1 \\ 2 & 1 \emat$, $\vx = \coll 5 1$, $\vv_1 = \coll 1 {-1}$ and $\vv_2 = \coll 1 2$. Since $\vx = 3 \vv_1 + 2 \vv_2$ we have $$ \begin{aligned} A^{10} \vx &= A^{10} (3 \vv_1 + 2 \vv_2) = 3 A^{10} \vv_1 + 2 A^{10} \vv_2 \\ &= 3 (-1)^{10} \vv_1 + 2(2^{10}) \vv_2 = \coll {3+2^{11}}{-3+2^{12}} \end{aligned} $$ Much faster than repeated matrix multiplication, especially if $10$ is replaced with $100$.

This raises an interesting question. In the example, the eigenvectors were a basis for $\R^2$, so we could use this method to compute $A^k \vx$ for any $\vx$. However, last class we saw a $3 \times 3$ matrix with two one-dimensional eigenspaces, so the eigenvectors didn't span $\R^3$. We will study this further in Section 4.4, but right now we can answer a related question about linear independence.

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.

Proof in case $m = 2$: If $\vv_1$ and $\vv_2$ are linearly dependent, then $\vv_1 = c \vv_2$ for some $c$. Therefore $$ A \vv_1 = A \, c \vv_2 = c A \vv_2 $$ so $$ \lambda_1 \vv_1 = c \lambda_2 \vv_2 = \lambda_2 \vv_1 $$ Since $\vv_1 \neq \vec 0$, this forces $\lambda_1 = \lambda_2$, a contradiction.$\quad\Box$

The general case is very similar; see text.

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 $$ \kern-4ex |z| = |a+bi| = \sqrt{a^2+b^2}, \qtext{the distance from the origin.} $$ 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 $$ \kern-9ex \begin{aligned} z_1 z_2 &= r_1 r_2 (\cos \theta_1 + i \sin\theta_1) (\cos \theta_2 + i \sin\theta_2) \\ &= r_1 r_2 [(\cos \theta_1 \cos\theta_2 - \sin\theta_1\sin\theta_2) + i(\sin\theta_1 \cos\theta_2 + \cos\theta_1\sin\theta_2)] \\ &= r_1 r_2 [\cos(\theta_1 + \theta_2) +i \sin(\theta_1+\theta_2)] \end{aligned} $$ 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: $$ e^{i \pi} + 1 = 0 $$