Today we finish 4.2 and start 4.3. Continue **reading** Section 4.3 for next class
and also read **Appendices C and D** on complex numbers and polynomials.
Work through recommended homework questions.

**Tutorials:** Quiz 7 covers 3.7 (just Markov chains), 4.1,
and 4.2 up to and including Example 4.13.

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

**Midterm average:** 53/70 = 76%

**True/false:** $\det(A B) = (\det A)(\det B)$.

**True/false:** $\det(A+B) = \det A + \det B$.

**Question:** $\det (3 I_2) = \query{3^2 \det I_2 = 3^2 = 9}$

**Question:** $\bdmat{rrr} 0 & 0 & a \\ 0 & b & c \\ d & e & f \edmat
= \query{-\bdmat{rrr} d & e & f \\ 0 & b & c \\ 0 & 0 & a \edmat = -abd \qtext{(not triangular!)}}$

**Notation:** If $A$ is an $n \times n$ matrix and $\vb \in \R^n$, we write
$A_i(\vb)$ for the matrix obtained from $A$ by replacing the $i$th column with
the vector $\vb$:
$$
A_i(\vb) = [ \va_1 \cdots \va_{i-1} \, \vb \,\, \va_{i+1} \cdots \va_n \,]
$$

**Theorem 4.11:** Let $A$ be an invertible $n \times n$
matrix and let $\vb$ be in $\R^n$. Then the unique solution $\vx$ of the
system $A \vx = \vb$ has components
$$
x_i = \frac{\det(A_i(\vb))}{\det A},\quad \text{for } i = 1, \ldots, n
$$

**Theorem 4.12:** If $A$ is an invertible matrix, then
$$
A^{-1} = \frac{1}{\det A} \adj A
$$

**Example:** If $A = \bmat{rr} a & b \\ c & d \emat$, then
the cofactors are
$$
\begin{aligned}
C_{11} &= + \det [d] = +d & C_{12} &= - \det [c] = -c \\
C_{21} &= - \det [b] = -b & C_{22} &= + \det [a] = +a \\
\end{aligned}
$$
so the adjoint matrix is
$$
\adj A = \bmat{rr} d & -b \\ -c & a \emat
$$
and
$$
A^{-1} = \frac{1}{\det A} \adj A = \frac{1}{\det A} \bmat{rr} d & -b \\ -c & a \emat
$$
as we saw before.

See Example 4.17 in the text for a $3 \times 3$ example. This is not generally a good computational approach. It's importance is theoretical.

A **polynomial** is a function $p$ of a single variable $x$ that
can be written in the form
$$
p(x) = a_0 + a_1 x + a_2 x^2 + \cdots + a_n x^n
$$
where the **coefficients** $a_i$ are constants. The highest power
of $x$ appearing with a non-zero coefficient is called
the **degree** of $p$.

**Examples:** $2 - 0.5 x + \sqrt{2} x^3$,
$\ \ln \left(\Large\frac{e^{5x^3}}{e^{3x}}\right) = \query{\ln(e^{5x^3-3x}) = 5x^3-3x}$

**Non-examples:** $\sqrt{x}$, $\ 1/x$, $\ \cos(x)$, $\ \ln(x)$.

(The text gives more examples, non-examples and explanations.)

**Addition** of polynomials is easy:
$$
(1 + 2x - 4x^3) + (3 - 3x^2 + 6x^3) = 4 + 2x -3x^2+2x^3
$$

To **multiply** polynomials, you use the distributive law and collect terms:
$$
\kern-4ex
\begin{aligned}
(x + 3)(1 + 2x + 4x^2) &= x ( 1 + 2x + 4x^2) + 3 ( 1 + 2x + 4x^2) \\
&= x + 2x^2 + 4x^3 + 3 + 6x + 12x^2 \\
&= 3 + 7x + 14x^2 + 4x^3
\end{aligned}
$$
Note that $\deg(f(x)g(x)) = \deg(f(x)) + \deg(g(x))$.

If $f$ and $g$ are polynomials, sometimes you can find a polynomial $q$
such that $f(x) = g(x) q(x)$, and sometimes you can't. If you can, then we say
that $g$ is a **factor** of $f$.

**Example:** Is $(x-2)$ a factor of $x^2 - x - 2$?

**Solution**: If it is, then the quotient has degree 1.
So suppose $x^2 - x - 2 = (x-2)(ax+b)$. Then $ax^2 = x^2$, so $a=1$.
And $-x = -2ax+bx = -2x+bx$, so $b=1$. Check the constant term: $-2 = -2b$.
It works, so $x^2 - x - 2 = (x-2)(x+1)$, and the answer is "yes".

**Example:** Is $(x-2)$ a factor of $x^2 + x - 2$?

**Solution**: If it is, then the quotient has degree 1.
So suppose $x^2 + x - 2 = (x-2)(ax+b)$. Then $ax^2 = x^2$, so $a=1$.
And $x = -2ax+bx = -2x+bx$, so $b=3$. Check the constant term: $-2 = -2b$.
Nope, so the answer is "no".

The above ad hoc method works for a degree 1 polynomial. For higher degrees, one can use long division (see Example D.4). But the degree 1 case will be most important to us, and is made even simpler by the following result:

**Theorem D.2 (The Factor Theorem):** Let $f$ be a polynomial and
let $a$ be a constant.
Then $f(a) = 0$ if and only if $x - a$ is a factor of $f(x)$.

When $f(a) = 0$, we say that $a$ is a **zero** of $f$ or a **root** of $f$.

It is clear that if $f(x) = (x - a) q(x)$, then $f(a) = 0$. The book explains the other direction.

Once you find a zero, you can use the ad hoc method shown above to find the other factor $q$. We'll see more examples soon.

Our interest will be in finding all zeros of a polynomial $f$ of degree $n$. By the above, if you find a zero $a$, then $f(x) = (x-a) q(x)$, where $q$ has degree $n-1$. If there is another root $b$ of $f$, it must be a root of $q$, and so $q$ will factor as $q(x) = (x-b) r(x)$, where $r$ has degree $n-2$. Since the degrees are going down by one, there can be at most $n$ distinct roots in total:

**Theorem D.4 (The Fundamental Theorem of Algebra):**
A polynomial of degree $n$ has at most $n$ distinct roots.

**Definition:** Let $A$ be an $n \times n$ matrix.
A scalar $\lambda$ (lambda) is called an **eigenvalue** of $A$ if
there is a nonzero vector $\vx$ such that $A \vx = \lambda \vx$.
Such a vector $\vx$ is called an **eigenvector** of $A$ corresponding to $\lambda$.

The eigenvectors for a given eigenvalue $\lambda$ are
the **nonzero** solutions to $(A - \lambda I) \vx = \vec 0$.

**Definition:** The collection of **all** solutions to
$(A - \lambda I) \vx = \vec 0$ is a subspace called the
**eigenspace** of $\lambda$ and is denoted $E_\lambda$. In other words,
$$ E_\lambda = \null(A - \lambda I) . $$
It consists of the eigenvectors plus the zero vector.

By the fundamental theorem of invertible matrices, $A - \lambda I$ has a nontrivial null space if and only if it is not invertible, and we now know that this is the case if and only if $\det (A - \lambda I) = 0$.

The expression $\det (A - \lambda I)$ is always a polynomial in $\lambda$. For example, when $A = \bmat{rr} a & b \\ c & d \emat$, $$ \kern-8ex \begin{aligned} \det(A- \lambda I) &= \bdmat{cc} a-\lambda & b \\ c & d-\lambda \edmat = (a - \lambda)(d-\lambda) - bc \\ &= \lambda^2 - (a+d)\lambda + (ad - bc) \end{aligned} $$ If $A$ is $3 \times 3$, then $\det(A - \lambda I)$ is equal to $$ \kern-8ex (a_{11} - \lambda) \bdmat{cc} a_{22} - \lambda\! & a_{23} \\ a_{32} & \!a_{33} - \lambda \edmat - a_{12} \bdmat{cc} a_{21} & a_{23} \\ a_{31} & \!a_{33} - \lambda \edmat + a_{13} \bdmat{cc} a_{21} & \!a_{22} -\lambda \\ a_{31} & a_{32} \edmat $$ which is a degree 3 polynomial in $\lambda$.

Similarly, if $A$ is $n \times n$, $\det (A - \lambda I)$ will be a degree $n$
polynomial in $\lambda$.
It is called the **characteristic polynomial** of $A$, and
$\det (A - \lambda I) = 0$ is called the **characteristic equation**.

1. Compute the characteristic polynomial $\det(A - \lambda I)$.

2. Find the eigenvalues of $A$ by solving the characteristic equation
$\det(A - \lambda I) = 0$.

3. For each eigenvalue $\lambda$, find a basis for $E_\lambda = \null (A - \lambda I)$
by solving the system $(A - \lambda I) \vx = \vec 0$.

**Theorem:**
An $n \times n$ matrix $A$ has at most $n$ distinct eigenvalues.

**Example 4.18**: Find the eigenvalues and eigenspaces of
$A = \bmat{rrr} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 2 & -5 & 4 \emat$.

**Solution:** 1. On board, compute the characteristic polynomial:
$$
\det (A - \lambda I) = - \lambda^3 + 4 \lambda^2 - 5 \lambda + 2
$$
2. To find the roots, it is often worth trying a few small integers
to start. We see that $\lambda = 1$ works. So by the factor theorem,
we know $\lambda - 1$ is a factor:
$$
- \lambda^3 + 4 \lambda^2 - 5 \lambda + 2
= (\lambda - 1)(\query{-} \lambda^2 + \query{3} \lambda \toggle{+ \text{?}}{-2}\endtoggle)
$$
Now we need to find roots of $-\lambda^2 + 3 \lambda - 2$.
Again, $\lambda = 1$ works, and this factors as $-(\lambda - 1)(\lambda - 2)$.
So
$$
\kern-4ex
\det (A - \lambda I) = - \lambda^3 + 4 \lambda^2 - 5 \lambda + 2
= - (\lambda - 1)^2 (\lambda - 2)
$$
and the roots are $\lambda = 1$ and $\lambda = 2$.