Today we finish Appendix C and Section 4.3. Read Section 4.4 for next class. Work through suggested exercises. Exercises and solutions for Appendix C are posted on that page.
Homework 10 is on WeBWorK and is due on Friday.
For Q2 and Q3 on the midterm, the graders have regraded some submissions and increased the grades slightly. Look at Gradescope to see if your total went up.
Math Help Centre: M-F 12:30-5:30 in PAB48/49 and online 6pm-8pm.
My next office hour is today 3:30-4:20 in MC130.
Final exam: Friday, December 13, 9am to noon.
Section 002 (this section) and Section 003: AH 201 (gym/stage).
Section 001 A-SHEI AH 15; SHEN-Z AH 201 (gym/stage).
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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 $i^2 = -1$.
Addition: $(a+bi)+(c+di) = (a+c) + (b+d)i$, like vector addition.
Multiplication: $(a+bi)(c+di) = (ac-bd) + (ad+bc)i$.
The absolute value or modulus of $z = a+bi$ is $$ \kern-4ex |z| = |a+bi| = \sqrt{a^2+b^2}, $$ the distance from the origin. We learned the properties of absolute value. One of them was $|w z| = |w| |z|$.
The conjugate of $z = a+bi$ is $\bar{z} = a-bi$. Reflection in real axis. We learned the properties of conjugation.
Since $z \bar{z} = |z|^2$, we have that $$ \kern-4ex \frac{z \bar{z}}{|z|^2} = 1 \qtext{so} \frac{1}{z} = \frac{\bar{z}}{|z|^2} \qtext{(for $z \neq 0$)} $$ This can be used to divide complex numbers: $$ \kern-4ex \frac{w}{z} = \frac{w}{z} \frac{\bar{z}}{\bar{z}} = \frac{w \bar{z}}{|z|^2}. $$
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 $\Arg (zw) = \Arg((1+i)(-1+i)) = \Arg(-2) = 180^\circ$.
True/false: There is a nonzero complex number $z$ with $z^2 = 0$.
True/false: If $z$ and $w$ are complex numbers in the first quadrant, then so is $zw$.
Concretely, $1 + 2i$ is in the first quadrant, but $(1+2i)(1+2i) = (1-4) + (2+2)i = -3 + 4i$ is in the second quadrant.
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.
If $a$ is a root of a polynomial $f(x)$, then $f(x) = (x-a) g(x)$. 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.
Example: Let $f(x) = x^2 - 2x + 1$. Since $f(1) = 1 - 2 + 1 = 0$, $1$ is a root of $f$. And since $f(x) = (x-1)^2$, $1$ has multiplicity $2$.
Theorem: An $n \times n$ matrix $A$ has at most $n$ distinct eigenvalues. In fact, the sum of the algebraic multiplicities is at most $n$. If we consider complex eigenvalues, the sum of the algebraic multiplicities is exactly $n$.
Example 4.7: Find the eigenvalues of $A = \bmat{rr} 0 & -1 \\ 1 & 0 \emat$ (a) over $\R$ and (b) over $\C$.
Solution: We must solve $$ 0 = \det(A-\lambda I) = \det \bmat{cc} -\lambda & -1 \\ 1 & -\lambda \emat = \lambda^2 + 1 . $$ (a) Over $\R$, there are no solutions, so $A$ has no real eigenvalues. This is why the Theorem above says "at most $n$". (This matrix represents rotation by 90 degrees, and we also saw geometrically that it has no real eigenvectors.)
(b) Over $\C$, the solutions are $\lambda = i$ and $\lambda = -i$. In fact, $\lambda^2 + 1 = (\lambda - i)(\lambda + i)$, so each of these eigenvalues has algebraic multiplicity 1, and the sum of the algebraic multiplicities is $2$, as expected.
Next we find the (complex) eigenvectors in (b). For $\lambda = i$: $$ \kern-8ex \begin{aligned} \bmat{rr|r} -i & -1 & 0 \\ 1 & -i & 0 \emat \lra{\ iR_1 \ } \bmat{rr|r} 1 & -i & 0 \\ 1 & -i & 0 \emat \lra{\ R_2 - R_1 \ } \bmat{rr|r} 1 & -i & 0 \\ 0 & 0 & 0 \emat \end{aligned} $$ So $x_2 = s$ ($s \in \C$!) and $x_1 - i x_2 = 0$, so $x_1 = is$. So $\vx = s \coll i 1$, the complex multiples of $\coll i 1$.
Finding the eigenvectors for $\lambda = -i$ is similar.
Note: Every polynomial with real coefficents factors over the real numbers into a product of polynomials which are either linear or quadratic.
Example: Factor $3x^3 + 5x^2 + 3x + 1$.
Solution: By trial and error we find that $x = -1$ is a root. So we factor: $$3x^3 + 5x^2 + 3x + 1 = (x+1)(3x^2 + 2x + 1).$$ The quadratic has no obvious roots, so we try the quadratic formula: $$ x = {-b \pm \sqrt{b^2-4ac} \over 2a} $$ gives the roots of $a x^2 + b x + c$, and these can be real (if $b^2 - 4ac \geqslant 0$) or complex (if $b^2 - 4ac < 0$). This formula also works if $a$, $b$ and $c$ are complex.
For $3x^2+2x+1$, we get $$ \kern-6ex \begin{aligned} x &= {-2 \pm \sqrt{2^2-4(3)(1)} \over 2(3)} = {-2 \pm \sqrt{-8} \over 6} \\ &= {-2 \pm \sqrt{-1}\sqrt{2}\sqrt{4} \over 6} = {-2 \pm 2\sqrt{2}i \over 6} = -\frac{1}{3} \pm \frac{\sqrt{2}i}{3} \end{aligned} $$ So $3x^3 + 5x^2 + 3x + 1 = (x+1)(x - (-\frac{1}{3} + \frac{\sqrt{2}i}{3}))(x - (-\frac{1}{3} - \frac{\sqrt{2}i}{3}))$.
Note: Suppose $p(x) = a^n x^n + \cdots + a_0$ is a polynomial with real coefficients, and that $z$ is a root. Then the complex conjugate $\bar{z}$ is also a root: $$ \kern-8ex \begin{aligned} p(\bar{z}) &= a_n \bar{z}^n + a_{n-1} \bar{z}^{n-1} + \cdots + a_1 \bar{z} + a_0 \\[5pt] &= \overline{a_n} \overline{z^n} + \overline{a_{n-1}} \overline{z^{n-1}} + \cdots + \overline{a_1} \, \overline{z} + \overline{a_0} \\[5pt] &= \overline{a_n {z}^n + a_{n-1} {z}^{n-1} + \cdots + a_1 {z} + a_0\!\!} = \bar{0} = 0. \end{aligned} $$ (This also follows from the quadratic formula.)
We've shown:
Theorem: The non-real roots of a polynomial with real coefficients occur in complex conjugate pairs.
If the matrix $A$ has only real entries, then the characteristic polynomial has real coefficients. It follows that:
Theorem: The non-real eigenvalues of a real matrix come in conjugate pairs.
Example: If a real matrix has $3 - 5i$ as an eigenvalue, then it must also have $3 + 5i$ as an eigenvalue.
Example: Find the real and complex eigenvalues of $A = \bmat{rrr} 2 & 3 & 0 \\ 1 & 2 & 2 \\ 0 & -2 & 1 \emat$.
Solution: $$ \kern-8ex \begin{aligned} \bdmat{ccc} 2-\lambda & 3 & 0 \\ 1 & 2-\lambda & 2 \\ 0 & -2 & 1-\lambda \edmat \ &= (2 - \lambda) \bdmat{cc} 2 - \lambda & 2 \\ -2 & 1-\lambda \edmat - 3 \bdmat{cc} 1 & 2 \\ 0 & 1-\lambda \edmat \\ &= (2 - \lambda) ( \lambda^2 - 3 \lambda + 6 ) - 3 (1-\lambda) \\ &= - \lambda^3 + 5 \lambda^2 - 9 \lambda + 9 . \end{aligned} $$ By trial and error, $\lambda = 3$ is a root. So we factor: $$ - \lambda^3 + 5 \lambda^2 - 9 \lambda + 9 = (\lambda - 3)(\query{-} \lambda^2 + \query{2} \lambda \toggle{+ \text{?}}{-3}\endtoggle) $$ We don't find any obvious roots for the quadratic factor, so we use the quadratic formula: $$ \kern-6ex \begin{aligned} \lambda\ &= \frac{-2 \pm \sqrt{2^2 - 4(-1)(-3)}}{-2} = \frac{-2 \pm \sqrt{-8}}{-2} \\ &= \frac{-2 \pm 2 \sqrt{2} \, i}{-2} = 1 \pm \sqrt{2} \, i . \end{aligned} $$ So the eigenvalues are $3$, $1 + \sqrt{2} \, i$ and $1 - \sqrt{2} \, i$. The algebraic multiplicities are all $1$, and $1+1+1=3$.
True/false: If $p(x)$ is a polynomial of odd degree with real coefficients, then $p(x)$ has a real root.
Examples: $\bmat{rr} 1 & 2 \\ 0 & i \emat$, $\bmat{rr} 1 & i \\ 0 & 2 \emat$.
We won't usually deal with matrices with non-real entries, but the procedure for handling them is just the same.
True/false: A degree 4 polynomial always has exactly 4 distinct roots over the complex numbers.
Note: Our questions always involve real eigenvalues and real eigenvectors unless we say otherwise. But there will be problems involving complex eigenvalues.