Math 1600 Lecture 30, Section 002, 22 Nov 2024

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Announcements:

Today we review 4.3 and discuss Appendix C. Next class we finish 4.3. Continue reading Section 4.3 for next class. Work through suggested exercises. Exercises and solutions for Appendix C are posted on that page.

Homework 9 is on Gradescope and is due today at 11:55pm.

Homework 10 will be on WeBWorK starting tomorrow.

Midterm grades were low and will be adjusted upwards by five marks.

Math Help Centre: M-F 12:30-5:30 in PAB48/49 and online 6pm-8pm.

My next office hour is today 2:30-3:20 in MC130.

Brief review of Lecture 29:

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. We are only covering the part of Appendix C covered in today's lecture.

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

Examples: $(1+2i) + (3+4i) = 4+6i$ $$ \kern-7ex \begin{aligned} (1+2i)(3+4i)\ &= 1(3+4i)+2i(3+4i) = 3+4i+6i+8i^2 \\ &= (3 - 8) + 10 i = -5+10i \\[5pt] 5(3+4i)\ &= 15+20i\\[5pt] (-1)(c+di)\ &= -c -di \\[5pt] (c + di) + (-c - di) &= 0 \end{aligned} $$ The usual laws hold, e.g. multiplication is commutative and associative, and distributes over addition.

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: $|3| = 3,\, |-3| = 3,\, |\pm i| = 1,\, |3+4i| = \sqrt{3^2+4^2} = 5$.

The conjugate of $z = a+bi$ is $\bar{z} = a-bi$. Reflection in real axis.

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: $$ \kern-8ex \frac{-1+2i}{3+4i} = \frac{-1+2i}{3+4i} \frac{3-4i}{3-4i} = \frac{5+10i}{3^2+4^2} = \frac{5+10i}{25} = \frac{1}{5} + \frac{2}{5}i . $$ Shorter: $$ \kern-8ex \frac{-1+2i}{3+4i} = \frac{(-1+2i)(3-4i)}{|3+4i|^2} = \frac{5+10i}{25} = \frac{1}{5} + \frac{2}{5}i . $$

Theorem 30-1 (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. $|w+z| \leq |w| + |z|$.

Some explanations on board.

Theorem 30-2 (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 $\overline{w/z} = \query{\bar{w} / \bar{z}}$
5. $z$ is real if and only if $\bar{z} = \query{z}$

Some explanations on board.

For a nonzero complex number $z$, the principal argument of $z$ is the angle between the vector $\vec{Oz}$ (from the origin to $z$) and the positive real axis, measured counterclockwise. We write this as $\Arg z$.

Fact: For nonzero $z$ and $w$, $\Arg (zw) = \Arg z + \Arg w$ (up to multiples of $2\pi$). So multiplying complex numbers multiplies the lengths and adds the angles.

Example: For $z = 1 + i$ and $w = -1 + i$, $\Arg z = 45^\circ$ and $\Arg w = 135^\circ$, so $\Arg (zw) = \Arg((1+i)(-1+i)) = \Arg(-2) = 180^\circ$.

Complex roots of polynomials

When finding eigenvalues, we need to find roots of polynomials. But over the real numbers, not every polynomial has a root and so not every polynomial factors as a product of linear factors.

Example: Find the roots of $x^3 - 2x^2 + x - 2$.

Solution: By trial and error we find that $x = 2$ is a root. So we factor: $$x^3 - 2x^2 + x - 2 = (x-2)(x^2 + 1) . $$ However, $x^2 + 1 > 0$ for every real number $x$, so it has no real roots.

But if we work over the complex numbers, then it does have roots. They are $\query{i}$ and $\query{-i}$.

So the original polynomial factors as $$x^3 - 2x^2 + x - 2 = (x-2)(x^2 + 1) = (x-2)\query{(x-i)(x+i).} $$

In fact, every polynomial has a root over the complex numbers, which means that:

Theorem D.4 (The Fundamental Theorem of Algebra, revised): A polynomial of degree $n$ with real or complex coefficients factors into a product of $n$ linear factors over the complex numbers. In particular, the polynomial has exactly $n$ roots over the complex numbers when counted with multiplicity.