Today we (mostly) finish 4.3.
**Read** Appendix C and section 4.4 for next class.
Work through recommended homework questions.

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

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

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

**Question:**
If $P$ is invertible, how do $\det A$ and $\det(P^{-1}AP)$ compare?

They are equal:
$$
\begin{aligned}
\det(P^{-1}AP) &= \det(P^{-1})\det(A)\det(P) \\
&= \frac{1}{\det (P)} \det(A) \det(P) = \det A.
\end{aligned}
$$

**Definition:**
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**.

By the fundamental theorem of invertible matrices, the solutions to the characteristic equation are exactly the eigenvalues.

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 the eigenspace
$E_\lambda = \null (A - \lambda I)$
by solving the system $(A - \lambda I) \vx = \vec 0$.

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

Therefore:

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

Also:

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

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

**Solution:** 1. Last time, we computed the characteristic polynomial:
$$
\det (A - \lambda I) = - \lambda^3 + 4 \lambda^2 - 5 \lambda + 2
$$
2. Then we found that
$$
- \lambda^3 + 4 \lambda^2 - 5 \lambda + 2 = - (\lambda - 1)^2 (\lambda - 2)
$$
with roots $\lambda = 1$ and $\lambda = 2$.

3. To find the $\lambda = 1$ eigenspace, we do row reduction: $$ \kern-4ex [A - I \mid 0\,] = \bmat{rrr|r} -1 & 1 & 0 & 0 \\ 0 & -1 & 1 & 0 \\ 2 & -5 & 3 & 0 \emat \lra{} \bmat{rrr|r} 1 & 0 & -1 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & 0 \emat $$ We find that $x_3 = t$ is free and $x_1 = x_2 = x_3$, so $$ E_1 = \left\{ \colll t t t \right\} = \span \left( \colll 1 1 1 \right) $$ So $\colll 1 1 1$ is a basis of the eigenspace corresponding to $\lambda = 1$. Check!

Finding a basis for $E_2$ is similar; see text.

A root $a$ of a polynomial $f$ implies that $f(x) = (x-a) g(x)$.
Sometimes, $a$ is also a root of $g(x)$, as we found above.
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.

In the previous example, $\lambda = 1$ has algebraic multiplicity 2 and $\lambda = 2$ has algebraic multiplicity 1.

We also define the **geometric multiplicity** of an eigenvalue $\lambda$
to be the dimension of the corresponding eigenspace.
In the previous example, $\lambda = 1$ has geometric multiplicity 1
(and so does $\lambda = 2$).

**Example 4.19:** Find the eigenvalues and eigenspaces of
$A = \bmat{rrr} -1 & 0 & 1 \\ 3 & 0 & -3 \\ 1 & 0 & -1 \emat$.
Do partially, on board.

In this case, we find that $\lambda = 0$ has algebraic multiplicity 2 and geometric multiplicity 2.

These multiplicities will be important in Section 4.4.

**Theorem 4.15:** The eigenvalues of a triangular matrix
are the entries on its main diagonal (repeated according to
their algebraic multiplicity).

**Example:** If $A = \bmat{rrr} 1 & 0 & 0 \\ 2 & 3 & 0 \\ 4 & 5 & 1 \emat$,
then
$$
\kern-6ex
\det(A - \lambda I) = \bdmat{ccc} 1-\lambda & 0 & 0 \\ 2 & 3-\lambda & 0 \\ 4 & 5 & 1-\lambda \edmat
= (1 - \lambda)^2 (3 - \lambda) ,
$$
so the eigenvalues are $\lambda = 1$ (with algebraic multiplicity 2)
and $\lambda = 3$ (with algebraic multiplicity 1).

**Question:**
What are the eigenvalues of a diagonal matrix?

The eigenvalues are the diagonal entries.

**Question:**
What are the eigenvalues of $\bmat{cc} 0 & 4 \\ 1 & 0 \emat$?

The characteristic polynomial is
$$
\bdmat{rr} -\lambda & 4 \\ 1 & -\lambda \edmat = \lambda^2 - 4 = (\lambda-2)(\lambda+2),
$$
so the eigenvalues are 2 and -2. Trick question.

**Question:**
How can we tell whether a matrix $A$ is invertible using eigenvalues?

$A$ is invertible if and only if 0 is not an eigenvalue,
because 0 being an eigenvalue is equivalent to $\null(A)$ being
non-trivial, which is equivalent to $A$ not being invertible,
by the fundamental theorem.

So we can extend the fundamental theorem with two new entries:

**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 $A$

Next: how to become a Billionaire using the material from this course.