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.)

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 $A$

**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 $\bmat{rr} 0 & 1 \\ 2 & 1 \emat^{10} \coll 5 1$.

**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.

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. $|w+z| \leq |w| + |z|$.

**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:
$$
\kern-8ex
\Arg w = -3\pi/4 \qtext{and} w = \sqrt{2}(\cos (-3\pi/4) + i \sin (-3\pi/4)).
$$

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
$$
\pm \sqrt{r} (\cos (\theta/2) + i (\sin \theta/2))
$$

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

**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
$$
\kern-6ex
2 ( \cos(\pi/3)+i\sin(\pi/3) ) = 2(1/2 + i \sqrt{3}/2) = 1 + \sqrt{3} i
$$
$$
\kern-6ex
2 ( \cos(3\pi/3)+i\sin(3\pi/3) ) = 2(-1 + 0 i ) = -2
$$
$$
\kern-6ex
2 ( \cos(5\pi/3)+i\sin(5\pi/3) ) = 2(1/2 - i \sqrt{3}/2) = 1 - \sqrt{3} i
$$

**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: $$ e^{i \pi} + 1 = 0 $$